JEE Type Practice Problems

Hint

Vertex of triangle is \((1,1)\) and mid-point of sides through this vertex are \((-1,2)\) and \((3,2)\)
\(\Rightarrow\) Vertex \(B\) and \(C[latex] come out to be [latex](-3,3)\) and \((5,3)\)
\(\therefore \quad\) Centroid is \(\frac{1-3+5}{3}, \frac{1+3+5}{3} \Rightarrow\left(1, \frac{7}{3}\right)\)

Hint

\(\because\) Area of \((\triangle O P R)=\) Area of \((\triangle P Q R)=\) Area of \((\triangle O Q R)\)
\(\therefore\) By geometry \(R\) should be the centroid of \(\triangle O P Q\)
\(\therefore \quad R \equiv\left(\frac{0+3+6}{3}, \frac{0+4+0}{3}\right) \equiv\left(3, \frac{4}{3}\right)\)

Hint

Given, the vertices of a right angled triangle are \(A(1, k), B(1,1)\) and \(C(2,1)\) and area of \(\triangle A B C=1\) sq unit
We know that, area of right angled triangle
\(
\begin{aligned}
& =\frac{1}{2} \times B C \times A B=\frac{1}{2}(1)|(k-1)| \\
& =\frac{1}{2}|k-1| \Rightarrow \pm(k-1)=2 \\
\Rightarrow \quad k & =-1,3 \\
\therefore \quad K & =\{-1,3\}
\end{aligned}
\)

Hint

Let \(P \equiv(1,0)\) and \(Q \equiv(-1,0)\)
Given that, \(\frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{3}\)
\(
\Rightarrow \quad 3 A P=A Q \text { or } 9(A P)^2=(A Q)^2
\)
Let \(\quad A \equiv(x, y)\), then
\(
\begin{aligned}
& a\left((x-1)^2+(y-0)^2\right)=(x+1)^2+(y-0)^2 \\
& \Rightarrow 8 x^2+8 y^2-20 x+8=0 \\
& \Rightarrow \quad x^2+y^2-\frac{5}{2} x+1=0 \dots(i)\\
&
\end{aligned}
\)
Circumcentre of \(\triangle A B C=\) Centre of circle Eq. (i)
\(
=\left(\frac{5}{4}, 0\right)
\)

Hint

From the figure, we have
\(
\begin{aligned}
a & =2, b=2 \sqrt{2}, c=2 \\
x_1 & =0, x_2=0, x_3=2 \quad(\because B C=a, C A=b, A B=c)
\end{aligned}
\)
Now, \(x\)-coordinate of incentre is given as
\(
\frac{a x_1+b x_2+c x_3}{a+b+c}
\)
\(\Rightarrow x\)-coordinate of incentre
\(
\begin{aligned}
& =\frac{2 \times 0+20 \sqrt{2} \times 0+2 \times 2}{2+2+2 \sqrt{2}} \\
& =\frac{2}{2+\sqrt{2}}=2-\sqrt{2}
\end{aligned}
\)

Hint

Total number of integral points inside the square
\(
O A B C=40 \times 40=1600 .
\)

Number of integral points on \(A C=\) Number of integral points on \(O B=40\) (namely \((1,1),(2,2),(3,3), \ldots,(40,40)\)
Hence, number of integral points inside the
\(
\triangle O A C=\frac{1600-40}{2}=780
\)
Alternate : \(x>0, y>0 x+y<41\) or \(0<x+y<41\)
\(\therefore\) Number of integral points inside the \(\triangle O A C={ }^{40} C_2=780\).

Hint

\(
\begin{aligned}
& \frac{1}{2}|| \begin{array}{ccc}
k & -3 k & 1 \\
5 & k & 1 \\
-k & 2 & 1
\end{array}||=28 \\
& \Rightarrow \quad 5 k^2+13 k+10= \pm 56 \\
& \Rightarrow \quad 5 k^2+13 k-46=0 \\
& \text { or } \quad 5 k^2+13 k+66=0 \\
& \therefore \quad 5 k^2+13 k-46=0 \\
& \text { and } \quad 5 k^2+13 k+66 \neq 0 & (\because D<0) \\
& \Rightarrow \quad k=2,-\frac{23}{5} \\
& \therefore \quad k=2, k \neq-\frac{23}{5} & (\because k \in I) \\
&
\end{aligned}
\)
\(\therefore\) Vertices an \(A \equiv(2,-6), B \equiv(5,2), C \equiv(-2,2)\).
Denote the points are \(\left(x_1, y_1\right),\left(x_2, y_2\right)\) and \(\left(x_3, y_3\right)\) from the
matrix \(P=\left[\begin{array}{ll}x_1-x_3 & y_1-y_3 \\ x_2-x_3 & y_2-y_3\end{array}\right]=\left[\begin{array}{cc}4 & -8 \\ 7 & 0\end{array}\right]\)
\(\therefore \quad \lambda=\frac{\overrightarrow{ R }_1 \cdot \overrightarrow{ R }_2}{|P|}=\frac{28}{56}=\frac{1}{2}\)
\(\therefore\) Circumcentre of triangle is \(\left(\frac{7+\frac{1}{2} \times-8}{2}, \frac{-4-\frac{1}{2} \times-3}{2}\right)\)
or \(\left(\frac{3}{2},-\frac{5}{4}\right)\) and centroid is \(\left(\frac{5}{3},-\frac{2}{3}\right)\)
then, orthocentre
\(
=\left(\frac{5}{3} \times 3-2 \times \frac{3}{2},-\frac{2}{3} \times 3+2 \times \frac{5}{4}\right) \text { or }\left(2, \frac{1}{2}\right) \text {. }
\)

Hint

The line passing through the intersection of lines \(a x+2 b y+3 b=0\) and \(b x-2 a y-3 a=0\) is
\(
\begin{aligned}
a x+2 b y+3 b+\lambda(b x-2 a y-3 a) & =0 \\
\Rightarrow(a+b \lambda) x+(2 b-2 a \lambda) y+3 b-3 \lambda a & =0
\end{aligned}
\)
As this line is parallel to \(X\)-axis.
\(
\begin{gathered}
\therefore \quad a+b \lambda=0 \Rightarrow \lambda=-\frac{a}{b} \\
\Rightarrow a x+2 b y+3 a-\frac{a}{b}(b x-2 a y-3 a)=0 \\
\Rightarrow a x+2 b y+3 b-a x+\frac{2 a^2}{b} y+\frac{3 a^2}{b}=0 \\
y\left(2 b+\frac{2 a^2}{b}\right)+3 b+\frac{3 a^2}{b}=0 \\
y\left(\frac{2 b^2+2 a^2}{b}\right)=-\left(\frac{3 b^2+3 a^2}{b}\right) \\
y=\frac{-3\left(a^2+b^2\right)}{2\left(b^2+a^2\right)}=\frac{-3}{2}
\end{gathered}
\)
So, it is \(\frac{3}{2}\) units below \(X\)-axis.

Hint

\(\because A\) is the mid-point of \(P Q\), therefore
\(
\begin{array}{ll}
& \frac{a+0}{2}=3, \frac{0+b}{2}=4 \\
\Rightarrow & a=6, b=8
\end{array}
\)
\(\therefore\) Equation of line is \(\frac{x}{6}+\frac{y}{8}=1\)
or \(4 x+3 y=24\)

Hint

Clearly for point \(P\),
\(
a^2-3 a<0 \text { and } a^2-\frac{a}{2}>0
\)
\(
\Rightarrow \quad \frac{1}{2}<a<3
\)

Hint

Point of intersection of \(L_1\) and \(L_2\) is \(A(0,0)\).
Also \(P(-2,-2), Q(1,-2)\)
\(\because A R\) is the bisector of \(\angle P A Q\), therefore \(R\) divides \(P Q\) in the same ratio as \(A P: A Q\).
Thus \(P R: R Q=A P: A Q=2 \sqrt{2}: \sqrt{5}\)
\(\therefore\) Statement I is true.
Statement II is clearly false.

Hint

Given : The coordinates of points \(P, Q, R\) are \((-1,0),(0,0)\), \((3,3 \sqrt{3})\) respectively.
Slope of equation \(Q R=\frac{y_2-y_1}{x_2-x_1}=\frac{3 \sqrt{3}}{3}\)
\(\Rightarrow \quad \tan \theta=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}\)
\(\Rightarrow \quad \angle R Q X=\frac{\pi}{3}\)
\(\therefore \quad \angle R Q P=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}\)
Let \(Q M\) bisects the \(\angle P Q R\),
\(\therefore \quad\) Slope of the line \(Q M=\tan \frac{2 \pi}{3}=-\sqrt{3}\)
\(\therefore\) Equation of line \(Q M\) is \((y-0)=-\sqrt{3}(x-0)\)
\(\Rightarrow \quad y=-\sqrt{3} x \Rightarrow \sqrt{3} x+y=0\)

Hint

(A) \(\because L_1, L_2, L_3\) are concurrent, then
\(
\left|\begin{array}{ccc}
1 & 3 & -5 \\
3 & -k & -1 \\
5 & 2 & -12
\end{array}\right|=0 \Rightarrow k=5
\)
(B) slope of \(\left(L_1\right)=\) slope of \(\left(L_2\right)\)
\(\Rightarrow \quad-\frac{1}{3}=\frac{3}{k} \therefore k=-9\)
and slope of \(\left(L_3\right)=\) slope of \(\left(L_2\right)\)
\(\Rightarrow \quad-\frac{5}{2}=\frac{3}{k} \quad \therefore k=-\frac{6}{5}\)
(C) Lines are not concurrent or not parallel, then
\(
\begin{aligned}
& k \neq 5, k \neq-9, k \neq-\frac{6}{5} \\
& \therefore \quad k=\frac{5}{6} \\
&
\end{aligned}
\)
(D) The given lines do not form a triangle if they are concurrent or any two of them are parallel.
\(
\therefore \quad k=5, k=-9, k=-\frac{6}{5}
\)

Hint

Slope of \(P Q=\frac{3-4}{k-1}=\frac{-1}{k-1}\)
Slope of perpendicular bisector of \(P Q=(k-1)\)
Also mid-point of \(P Q\left(\frac{k+1}{2}, \frac{7}{2}\right)\)
\(\therefore\) Equation of perpendicular bisector is
\(
\begin{array}{lc}
& y-\frac{7}{2}=(k-1)\left(x-\frac{k+1}{2}\right) \\
\Rightarrow & 2 y-7=2(k-1) x-\left(k^2-1\right) \\
\Rightarrow & 2(k-1) x-2 y+\left(8-k^2\right)=0 \\
\therefore & Y \text {-intercept }=-\frac{8-k^2}{-2}=-4 \\
\Rightarrow & 8-k^2=-8 \text { or } k^2=16 \Rightarrow k= \pm 4
\end{array}
\)

Hint

If the line \(p\left(p^2+1\right) x-y+q=0\)
and \(\quad\left(p^2+1\right)^2 x+\left(p^2+1\right) y+2 q=0\)
are perpendicular to a common line, then these lines must be parallel to each other,
\(
\begin{aligned}
& \therefore \quad m_1=m_2 \Rightarrow-\frac{p\left(p^2+1\right)}{-1}=-\frac{\left(p^2+1\right)^2}{p^2+1} \\
& \Rightarrow \quad\left(p^2+1\right)(p+1)=0 \\
& \Rightarrow \quad p=-1 \\
&
\end{aligned}
\)
\(\therefore p\) can have exactly one value.

Hint

Slope of line \(L=-\frac{b}{5}\)
Slope of line \(K=-\frac{3}{c}\)
Line \(L\) is parallel to line \(K\).
\(\Rightarrow \quad \frac{b}{5}=\frac{3}{c} \Rightarrow b c=15\)
\((13,32)\) is a point on \(L\).
\(
\begin{array}{ll}
\therefore & \frac{13}{5}+\frac{32}{b}=1 \Rightarrow \frac{32}{b}=-\frac{8}{5} \\
\Rightarrow & b=-20 \Rightarrow c=-\frac{3}{4}
\end{array}
\)
Equation of \(K: y-4 x=3\)
\(
\Rightarrow \quad 4 x-y+3=0
\)
Distance between \(L\) and \(K=\frac{|52-32+3|}{\sqrt{17}}\)
\(
=\frac{23}{\sqrt{17}}
\)

Hint

Let the slope of line \(L\) be \(m\). Then
\(
\left|\frac{m+\sqrt{3}}{1-\sqrt{3} m}\right|=\sqrt{3}
\)
\(
\begin{array}{ll}
\Rightarrow & m+\sqrt{3}= \pm(\sqrt{3}-3 m) \\
\Rightarrow & 4 m=0 \text { or } 2 m=2 \sqrt{3} \\
\Rightarrow & m=0 \text { or } m=\sqrt{3}
\end{array}
\)
\(\because L\) intersects \(X\)-axis,
\(\therefore \quad m=\sqrt{3}\)
\(\therefore\) Equation of \(L\) is \(y+2=\sqrt{3}(x-3)\)
or \(\sqrt{3} x-y-(2+3 \sqrt{3})=0\)

Hint

\(
L_1: y-x=0, L_2: 2 x+y=0, \quad L_3: y+2=0
\)
On solving the equation of lines \(L_1\) and \(L_2\), we get their point of intersection \((0,0)\) i.e. origin \(O\).
On solving the equation of lines \(L_1\) and \(L_3\),
we get
\(
P=(-2,-2)
\)
Similarly, we get
\(
Q=(-1,-2)
\)
We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [Angle Bisector Theorem of a Triangle]
\(
\therefore \quad \frac{P R}{R Q}=\frac{O P}{O Q}=\frac{\sqrt{(-2)^2+(-2)^2}}{\sqrt{(-1)^2+(-2)^2}}=\frac{2 \sqrt{2}}{\sqrt{5}}
\)

Hint

Let the joining points be \(A(1,1)\) and \(B(2,4)\).
Let point \(C\) divides line \(A B\) in the ratio \(3: 2\). So, by section formula we have
\(
C=\left(\frac{3 \times 2+2 \times 1}{3+2}, \frac{3 \times 4+2 \times 1}{3+2}\right)=\left(\frac{8}{5}, \frac{14}{5}\right)
\)
Since Line \(2 x+y=k\) passes through \(C\left(\frac{8}{5}, \frac{14}{5}\right)\)
\(\therefore C\) satisfies the equation \(2 x+y=k\).
\(
\Rightarrow \quad \frac{2+8}{5}+\frac{14}{5}=k \Rightarrow k=6
\)

Hint

Suppose \(B(0,1)\) be any point on given line and coordinate of \(A\) is \((\sqrt{3}, 0)\). So, equation of
Reflected ray is \(\frac{-1-0}{0-\sqrt{3}}=\frac{y-0}{x-\sqrt{3}}\)
\(
\Rightarrow \quad \sqrt{3} y=x-\sqrt{3}
\)

Hint

The intersection point of two lines is \(\left(\frac{-c}{a+b}, \frac{-c}{a+b}\right)\) Distance between \((1,1)\) and \(\left(\frac{-c}{a+b}, \frac{-c}{a+b}\right)<2 \sqrt{2}\)
\(
\begin{array}{lc}
\Rightarrow & 2\left(1+\frac{c}{a+b}\right)^2<8 \\
\Rightarrow & 1+\frac{c}{a+b}<2 \\
\Rightarrow & a+b-c>0
\end{array}
\)

Hint

Let \(P, Q, R\), be the vertices of \(\triangle P Q R\)
Since, \(P S\) is the median, \(S\) is mid-point of \(Q R\)
So,
\(
S=\left(\frac{7+6}{2}, \frac{3-1}{2}\right)=\left(\frac{13}{2}, 1\right)
\)
Now, slope of \(P S=\frac{2-1}{2-\frac{13}{2}}=-\frac{2}{9}\)
Since, required line is parallel to \(P S\) therefore slope of required line = slope of PS Now, eqn of line passing through \((1,-1)\) and having slope \(-\frac{2}{9}\) is
\(
\begin{aligned}
y-(-1) & =-\frac{2}{9}(x-1) \\
9 y+9 & =-2 x+2 \\
\Rightarrow \quad 2 x+9 y+7 & =0
\end{aligned}
\)

Hint

Given lines are
\(
\begin{aligned}
& 4 a x+2 a y+c=0 \\
& 5 b x+2 b y+d=0
\end{aligned}
\)

The point of intersection will be
\(
\begin{aligned}
& \frac{x}{2 a d-2 b c}=\frac{-y}{4 a d-5 b c}=\frac{1}{8 a b-10 a b} \\
\Rightarrow \quad & x=\frac{2(a d-b c)}{-2 a b}=\frac{b c-a d}{a b} \\
\Rightarrow \quad & y=\frac{5 b c-4 a d}{-2 a b}=\frac{4 a d-5 b c}{2 a b}
\end{aligned}
\)
\(\because\) Point of intersection is in fourth quadrant so \(x\) is positive and \(y\) is negative.
Also distance from axes is same
So \(\quad x=-y \quad(\because\) distance from \(X\)-axis is \(-y\) as \(y\) is negative)
\(
\frac{b c-a d}{a b}=\frac{5 b c-4 a d}{2 a b} \Rightarrow 3 b c-2 a d=0
\)

Hint

Let the point \(P\) be \((x, y)\)
Then
\(
d_1(P)=\left|\frac{x-y}{\sqrt{2}}\right| \text { and } d_2(P)=\left|\frac{x+y}{\sqrt{2}}\right|
\)
For \(P\) lying in first quadrant \(x>0, y>0\).
Also
\(
\begin{array}{ll}
\text { Also } & 2 \leq d_1(P)+d_2(P) \leq 4 \\
\Rightarrow & 2 \leq\left|\frac{x-y}{\sqrt{2}}\right|+\left|\frac{x+y}{\sqrt{2}}\right| \leq 4
\end{array}
\)
If \(x>y\), then
\(
2 \leq \frac{x-y+x+y}{\sqrt{2}} \leq 4 \text { or } \sqrt{2} \leq x \leq 2 \sqrt{2}
\)
If \(x<y\), then
\(
2 \leq \frac{y-x+x+y}{2} \leq 4 \text { or } \sqrt{2} \leq y \leq 2 \sqrt{2}
\)
The required region is the shaded region in the figure given below.

\(
\therefore \text { Required area }=(2 \sqrt{2})^2-(\sqrt{2})^2=8-2=6 \text { sq units }
\)

Hint

Total number of integral points inside the square \(O A B C\)
\(
=40 \times 40=1600
\)

Number of integral points on \(A C\)
\(
\begin{aligned}
& =\text { Number of integral points on } O B \\
& =40[\text { namely }(1,1),(2,2) \ldots(40,40)]
\end{aligned}
\)
\(\therefore\) Number of integral points inside the \(\triangle O A C\)
\(
=\frac{1600-40}{2}=780
\)

Hint

Let other two sides of rhombus are
\(
\begin{array}{r}
x-y+\lambda=0 \\
7 x-y+\mu=0
\end{array}
\)
then \(O\) is equidistant from \(A B\) and \(D C\) and from \(A D\) and \(B C\)
\(
\begin{array}{llrl}
\therefore & |-1+2+1| & =|-1+2+\lambda| \Rightarrow \lambda=-3 \\
\text { and } & |-7+2-5| & =|-7+2+\mu| \Rightarrow \mu=15
\end{array}
\)
\(\therefore\) Other two sides are
\(
\begin{array}{r}
x-y-3=0 \\
7 x-y+15=0
\end{array}
\)
On solving the equation of sides pairwise, we get the vertices as \(\left(\frac{1}{3}, \frac{-8}{3}\right),(1,2),\left(\frac{-7}{3}, \frac{-4}{3}\right),(-3,-6)\)

Hint

Let \(A\) be the area of small sector, then area of major sector is \(3 A\).
\(
\begin{aligned}
& \therefore \quad A+3 A+A+3 A=\pi r^2 \\
& \Rightarrow \quad A=\frac{\pi}{8} r^2 \\
& \Rightarrow \quad \frac{1}{2} r^2 \theta=\frac{\pi}{8} r^2 \quad\left[\because \text { area of sector }=\frac{1}{2} r^2 \theta\right] \\
& \therefore \quad \theta=\frac{\pi}{4} \\
&
\end{aligned}
\)
\(\therefore\) Angle between lines represented by
\(
\begin{aligned}
& a x^2+2(a+b) x y+b y^2=0 \text { is } \frac{\pi}{4} \text {. } \\
& \Rightarrow \quad \tan \frac{\pi}{4}=\frac{2 \sqrt{(a+b)^2-a b}}{|a+b|} \\
& \Rightarrow \quad 1=\frac{2 \sqrt{(a+b)^2-a b}}{|a+b|} \\
& \Rightarrow \quad(a+b)^2=4(a+b)^2-4 a b \\
& \therefore \quad 3 a^2+2 a b+3 b^2=0 \\
&
\end{aligned}
\)

Hint

The equation of the bisectors of the lines \(x y=0\) are \(y= \pm x\).
Putting \(y= \pm x\) in
\(
\begin{aligned}
m y^2+\left(1-m^2\right) x y-m x^2 & =0 \text {, we get } \\
\pm\left(1-m^2\right) x^2 & =0 \\
\Rightarrow \quad m^2 & =1
\end{aligned}
\)

Hint

(b) Let \(A \equiv(a \cos t, a \sin t), B \equiv(b \sin t,-b \cos t)\) and \(C \equiv(1,0)\)
\(\therefore\) Centroid \(\equiv\left(\frac{a \cos t+b \sin t+1}{3}, \frac{a \sin t-b \cos t}{3}\right)\)
Let \(\quad x=\frac{a \cos t+b \sin t+1}{3} \Rightarrow 3 x-1=a \cos t+b \sin t \dots(i)\)
and \(y=\frac{a \sin t-b \cos t}{3} \Rightarrow 3 y=a \sin t-b \cos t \dots(ii)\)
On squaring and adding Eqs. (i) and (ii), we get
\(
(3 x-1)^2+(3 y)^2=a^2+b^2
\)
which is locus of centroid.

Hint

\(
\begin{aligned}
& \text { Let } \quad O \equiv(0,0), A \equiv(2,0) \\
& \text { and } B \equiv(1, \sqrt{3}) \\
& \because \quad O A=2=O B=A B \\
& \Rightarrow \triangle O A B \text { is an equilateral. } \\
& \therefore \text { Incentre }=\text { Centroid } \\
& \equiv\left(\frac{0+2+1}{3}, \frac{0+0+\sqrt{3}}{3}\right) \\
& \therefore \quad \text { Incentre } \equiv\left(1, \frac{1}{\sqrt{3}}\right) \\
&
\end{aligned}
\)

Hint

Let orthocentre
\(
H \equiv(\alpha, \beta)
\)
\(\therefore\) Slope of \(B H \times\) slope of \(O A=-1\)
\(
\Rightarrow \quad\left(\frac{\beta-4}{\alpha-3}\right) \times\left(\frac{0}{4}\right)=-1
\)
\(
\begin{aligned}
\therefore & \alpha-3 & =0 \\
\Rightarrow & \alpha & =3 \dots(i)
\end{aligned}
\)
and slope of \(A H \times\) slope of \(O B=-1\)
\(
\Rightarrow \quad\left(\frac{\beta-0}{\alpha-4}\right) \times\left(\frac{4}{3}\right)=-1
\)
From Eq. (i) ,
\(
\beta=\frac{3}{4}
\)
Hence, orthocentre is \(\left(3, \frac{3}{4}\right)\).

Hint

(a) Let common ratio of GP is \(r\), then \(x_2=x_1 r, x_3=x_1 r^2, y_2=y_1 r\) and \(y_3=y_1 r^2\).
Let \(A \equiv\left(x_1, y_1\right), B \equiv\left(x_2, y_2\right)\) and \(C \equiv\left(x_3, y_3\right)\)
\(
\begin{aligned}
\therefore \text { Area of } \triangle A B C & =\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right| \\
& =\frac{1}{2}\left|\begin{array}{ccc}
x_1 & y_1 & 1 \\
x_1 r & y_1 r & 1 \\
x_1 r^2 & y_1 r^2 & 1
\end{array}\right|
\end{aligned}
\)
\(
=\frac{1}{2}\left|x_1 y_1\right||| \begin{array}{ccc}
1 & 1 & 1 \\
r & r & 1 \\
r^2 & r^2 & 1
\end{array}||=0 \text { ( } \because C_1 \text { and } C_2 \text { are identical) }
\)
\(\Rightarrow\) Points \(A, B, C\) are collinear.

Hint

(a) Since, the image of \((h, k)\) w.r.t. \(Y\)-axis is \((-h, k)\).
\(\therefore\) Coordinate of \(A\) are \((-2,-1)\).
If \((X, Y)\) are the coordinates of \(A\) w.r.t. the new coordinate axes obtained by turning the axes through an angle \(45^{\circ}\) in the clockwise direction, then
\(
\begin{aligned}
X & =-2 \cos \left(-45^{\circ}\right)-\sin \left(-45^{\circ}\right) \\
& =-\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}}=-\frac{1}{\sqrt{2}}
\end{aligned}
\)
and
\(
\begin{aligned}
Y & =2 \sin \left(-45^{\circ}\right)-\cos \left(-45^{\circ}\right) \\
& =-\frac{2}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\frac{3}{\sqrt{2}}
\end{aligned}
\)
\(\therefore\) Required coordinates are \(\left(-\frac{1}{\sqrt{2}},-\frac{3}{\sqrt{2}}\right)\).

Hint

If \(a\) be the side of the square, then diagonal \(d=a \sqrt{2}\) by hypothesis
\(
a_n=\sqrt{2} a_{n+1}
\)
\(
\begin{aligned}
a_{n+1} & =\frac{a_n}{\sqrt{2}}=\frac{a_{n-1}}{(\sqrt{2})^2}=\frac{a_{n-2}}{(\sqrt{2})^3}=\ldots=\frac{a_1}{(\sqrt{2})^n} \\
a_n & =\frac{a_1}{(\sqrt{2})^{n-1}}=\frac{10}{(2)^{(n-1) / 2}}
\end{aligned}
\)
\(
\begin{aligned}
& \because \quad \text { Area of } S_n<1 \Rightarrow a_n^2<1 \\
& \Rightarrow \quad \frac{100}{2^{n-1}}<1 \\
& \Rightarrow \quad 2^{n-1}>100>2^6 \\
& \Rightarrow \quad n-1>6 \\
& \Rightarrow \quad n>7 \\
& \therefore \quad n=8,9,10, \ldots \\
&
\end{aligned}
\)

Hint

(a,c,d) Let \(A \equiv\left(x_1, y_1\right), B \equiv\left(x_2, y_2\right)\) and \(C \equiv\left(x_3, y_3\right)\) be the vertices of triangle \(A B C\). Given \(x_1, y_1, x_2, y_2, x_3, y_3\) be all integers.
Now, area of \(\triangle A B C=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\) \(=\) Rational ….(i)

If \(\triangle A B C\) is equilateral then,
\(
\text { Area of } \begin{aligned}
\triangle A B C & =\frac{\sqrt{3}}{4}(\text { side })^2 \\
& =\frac{\sqrt{3}}{4}\left\{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2\right\} \\
& =\text { Irrational } \dots(ii)
\end{aligned}
\)
It is clear from Eqs. (i) and (ii), \(\triangle A B C\) can not be equilateral.

Hint

(b,c,d) Let vertex of the \(\triangle A B C\) be \(A(x, y)\)
\(
\begin{aligned}
& \therefore \quad A B=A C \\
& \Rightarrow \quad(A B)^2=(A C)^2 \\
& \Rightarrow \quad(x-1)^2+(y-3)^2=(x+2)^2+(y-7)^2 \\
& \Rightarrow \quad 6 x-8 y+43=0 \dots(i) \\
&
\end{aligned}
\)
Here, use observe that the coordinates \(\left(-\frac{1}{2}, 5\right),\left(\frac{5}{6}, 6\right)\) and \(\left(-7, \frac{1}{8}\right)\) satisfy the Eq. (i).

Hint

\(\alpha, \beta\) are the roots of \(x^2-6 a x+2=0\)
\(
\begin{aligned}
& \therefore \quad \alpha+\beta=6 a \dots(i) \\
& \text { and } \quad \alpha \beta=2 \dots(ii) \\
&
\end{aligned}
\)
Again, \(\beta, \gamma\) are the roots of \(x^2-6 b x+3=0\)
\(
\begin{aligned}
& \therefore & \beta+\gamma & =6 b \dots(iii) \\
& \text { and } & \beta \gamma & =3 \dots(iv)
\end{aligned}
\)
Again, \(\gamma, \alpha\) are the roots of \(x^2-6 c x+6=0\)
\(
\begin{array}{ll}
\therefore & \gamma+\alpha=6 c \dots(v) \\
\text { and } & \gamma \alpha=6 \dots(vi)
\end{array}
\)
from Eqs. (ii), (iv) and (vi), we get
\(
\begin{array}{lc}
& \alpha \beta \cdot \beta \gamma \cdot \gamma \alpha=2 \cdot 3 \cdot 6 \\
\Rightarrow & \alpha \beta \gamma=6 \\
\therefore & \alpha=2, \beta=1, \gamma=3
\end{array}
\)
Adding Eqs. (i), (iii) and (v), we get
\(
\begin{aligned}
& 2(\alpha+\beta+\gamma)=6(a+b+c) \\
& a+b+c=\frac{1}{3}(2+1+3)=2
\end{aligned}
\)

Hint

Since, the given points are collinear, then
\(
\left|\begin{array}{ccc}
-2 & 0 & 1 \\
-1 & \frac{1}{\sqrt{3}} & 1 \\
\cos \theta & \sin \theta & 1
\end{array}\right|=0
\)
\(
\begin{aligned}
& \Rightarrow \quad-2\left(\frac{1}{\sqrt{3}}-\sin \theta\right)-0+1\left(-\sin \theta-\frac{\cos \theta}{\sqrt{3}}\right)=0 \\
& \Rightarrow \quad \sqrt{3} \sin \theta-\cos \theta=2 \\
& \Rightarrow \quad \frac{\sqrt{3}}{2} \sin \theta-\frac{1}{2} \cos \theta=1 \\
& \text { or } \\
& \sin \left(\theta-\frac{\pi}{6}\right)=1 \\
& \text { or } \\
& \theta-\frac{\pi}{6}=2 n \pi+\frac{\pi}{2} \\
& \text { for } n=0 \text {, } \\
& \theta=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2 \pi}{3} \in[0,2 \pi] \\
&
\end{aligned}
\)
Number of values \(\theta\) is 1 .

Hint

\(\because(\alpha, \beta)\) lies on \(y=2 x+3\)
then \(\quad \beta=2 \alpha+3\)
Thus, the coordinates of \(A\) are \((\alpha, 2 \alpha+3) \dots(i)\)
\(
\begin{aligned}
\Delta & =\frac{1}{2}\left|\left\{\left|\begin{array}{cc}
\alpha & 2 \alpha+3 \\
1 & 2
\end{array}\right|+\left|\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right|+\left|\begin{array}{cc}
2 & 3 \\
\alpha & 2 \alpha+3
\end{array}\right|\right\}\right| \\
& =\frac{1}{2}|2 \alpha-(2 \alpha+3)+3-4+4 \alpha+6-3 \alpha| \\
& =\frac{1}{2}|\alpha+2|
\end{aligned}
\)
\(
\begin{array}{ll}
\text { but } & {[\Delta]=2} \\
\Rightarrow & {\left[\frac{1}{2}|\alpha+2|\right]=2} \\
\Rightarrow & 2 \leq \frac{|\alpha+2|}{2}<3 \\
\Rightarrow & 4 \leq|\alpha+2|<6 \\
\Rightarrow & 4 \leq \alpha+2<6 \\
\text { and } & -6<\alpha+2 \leq-4 \\
\Rightarrow & 2 \leq \alpha<4 \text { and }-8<\alpha \leq-6 \\
\because & \alpha \in I \\
\therefore & \alpha=2,3,-7,-6
\end{array}
\)
Hence, possible coordinates of \(A\) are \((2,7),(3,9),(-7,-11)\) and \((-6,-9)\).

Hint

Let straight line makes an angle \(\theta\) with positive direction of \(X\)-axis,
then
\(
\tan \theta=\frac{3}{4}
\)
\(
\therefore \quad \sin \theta=\frac{3}{5} \text { and } \cos \theta=\frac{4}{5}
\)

\(\therefore\) Equation of the straight line through \(A(3,2)\) in parametric form is
\(
\begin{gathered}
\frac{x-3}{\cos \theta}=\frac{y-2}{\sin \theta}= \pm 5 \\
\therefore \quad x=3 \pm 5 \cos \theta=3 \pm 5 \times \frac{4}{5}=3 \pm 4=7 \text { or }-1 \\
\text { and } \quad y=2 \pm 5 \sin \theta=2 \pm 5 \times \frac{3}{5}=2 \pm 3=5 \text { or }-1
\end{gathered}
\)
Hence, the coordinates of the points are \((7,5)\) and \((-1,-1)\).

Hint

\(\because\) Line makes an angle \(\frac{3 \pi}{4}\) with the negative direction of \(X\)-axis.
\(\therefore\) Line makes an angle \(\frac{\pi}{4}\) with the positive direction of \(X\)-axis.

\(\therefore\) The equation of the line through \((1,2)\) in parametric form is
i.e.
\(
\frac{x-2}{\cos \left(\frac{\pi}{4}\right)}=\frac{y-3}{\sin \left(\frac{\pi}{4}\right)}=r
\)
i.e. \(\quad \frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y-3}{\frac{1}{\sqrt{2}}}=r \dots(i)\)
\(\therefore \quad x=2+\frac{r}{\sqrt{2}}[latex] and [latex]y=3+\frac{r}{\sqrt{2}}\)
Let the line (i) meet the line \(x+y-7=0\) in \(P\)
\(\therefore\) Coordinates of \(P\left(2+\frac{r}{\sqrt{2}}, 3+\frac{r}{\sqrt{2}}\right)\) lies on \(x+y-7=0\)
then \(\quad 2+\frac{r}{\sqrt{2}}+3+\frac{r}{\sqrt{2}}-7=0\)
or \(\quad \frac{2 r}{\sqrt{2}}=2\) or \(r=\sqrt{2}\)
\(\therefore \quad A P=\sqrt{2}\)

Hint

Since, slope of the line \(2 x-2 y+5=0\) is 1 , its makes an angle \(\frac{\pi}{4}\) with positive direction of \(X\)-axis.
The equation of the line through \((2,3)\) and making an angle \(\frac{\pi}{4}\) in parametric form
\(
\frac{x-2}{\cos \left(\frac{\pi}{4}\right)}=\frac{y-3}{\sin \left(\frac{\pi}{4}\right)}=r \text { or } \frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y-3}{\frac{1}{\sqrt{2}}}=r
\)
This point lies on the line \(2 x-3 y+9=0\)
\(
\begin{aligned}
\Rightarrow & 2\left(2+\frac{r}{\sqrt{2}}\right)-3\left(3+\frac{r}{\sqrt{2}}\right)+9 & =0 \\
\Rightarrow & -\frac{r}{\sqrt{2}}+4 & =0
\end{aligned}
\)
\(
r=4 \sqrt{2}
\)

Hint

Slope of line \(y-\sqrt{3} x+3=0\) is \(\sqrt{3}\)
If line makes an angle \(\theta\) with \(X\)-axis, then \(\tan \theta=\sqrt{3}\)
\(
\begin{aligned}
\therefore & \theta=60^{\circ} \\
\frac{x-\sqrt{3}}{\cos 60^{\circ}} & =\frac{y-0}{\sin 60^{\circ}}=r
\end{aligned}
\)

\(\Rightarrow\left(\sqrt{3}+\frac{r}{2}, \frac{r \sqrt{3}}{2}\right)\) be a point on the parabola \(y^2=x+2\)
then, \(\frac{3}{4} r^2=\sqrt{3}+\frac{r}{2}+2 \Rightarrow 3 r^2-2 r-4(2+\sqrt{3})=0\)
\(\therefore \quad P A \cdot P B=r_1 r_2=\left|\frac{-4(2+\sqrt{3})}{3}\right|=\frac{4(2+\sqrt{3})}{3}\)

Hint

\(
\begin{aligned}
& \text { (By special corollary (Case-II)) } \\
& \because \quad A \equiv(1,1) \\
& \therefore \quad z_A=1+i \text {, where } i=\sqrt{-1} \\
& \text { and } \\
& C \equiv(-2,-1) \\
& \therefore \quad z_C=-2-i \\
&
\end{aligned}
\)

then centre of square \(E \equiv\left(-\frac{1}{2}, 0\right)\)
\(\therefore \quad z_E=-\frac{1}{2}\)
Now, in \(\triangle A E B,(E A=E B)\)
\(
\begin{aligned}
& \frac{z_B-z_E}{z_A-z_E}=e^{i \frac{\pi}{2}}=i \\
& \Rightarrow \quad z_B+\frac{1}{2}=i\left(1+i+\frac{1}{2}\right) \\
& \therefore \quad z_B=-\frac{3}{2}+\frac{3}{2} i \\
& B \equiv\left(-\frac{3}{2}, \frac{3}{2}\right) \\
& D \equiv\left(-1+\frac{3}{2},-\frac{3}{2}\right) \\
& D \equiv\left(\frac{1}{2},-\frac{3}{2}\right) \\
&
\end{aligned}
\)
Hence, equation of other diagonal \(B D\) is
\(
\begin{gathered}
y-0=\frac{\frac{3}{2}-0}{-\frac{3}{2}+\frac{1}{2}}\left(x+\frac{1}{2}\right) \\
\Rightarrow \quad 6 x+4 y+3=0
\end{gathered}
\)

Hint

(c) Since \(A B\) is parallel to \(y=x\).
\(\therefore\) Equation of \(A B\) is \(y=x+a\)
\(\because \quad A\) lies on \(y=1\)
\(\therefore \quad A \equiv(1-a, 1)\)

Again, \(B\) lies on \(x=2\)
\(
\therefore \quad B=(2,2+a)
\)
\(\Rightarrow\) Equation of \(A D\) is
\(
y-1=-[x-(1-a)] \text { or } y=2-x-a
\)
\(\because D\) lies on \(x=-2\)
\(\therefore \quad D \equiv(-2,4-a)\)
Let \(C \equiv(h, k)\)
\(\because\) Diagonals of rectangle bisects to each other
\(
\begin{aligned}
& \therefore \quad h+1-a=2-2 \Rightarrow a=1+h \\
& \text { and } k+1=2+a+4-a \\
& \Rightarrow \quad k=5 \\
&
\end{aligned}
\)
\(\therefore\) Locus of \(C\) is \(y=5\)

Hint

(a)
\(
\because \quad(\lambda+1)^2 x+\lambda y-2 \lambda^2-2=0
\)
\(
\begin{array}{ll}
\text { or } & \left(\lambda^2+2 \lambda+1\right) x+\lambda y-2 \lambda^2-2=0 \\
\text { or } & \left(\lambda^2+1\right)(x-2)+\lambda(2 x+y)=0
\end{array}
\)
\(
\left(\lambda^2+1\right)(x-2)+\lambda(2 x+y)=0
\)
\(\therefore\) For fixed point
\(
x-2=0 \text { and } 2 x+y=0
\)
\(\therefore\) Fixed point is \((2,-4)\)
\(\therefore \quad\) Equation of required line is \(y+4=2(x-2)\)
or \(y=2 x-8\)

Hint

(a) Since the point \(P\left(a, a^2\right)\) lies on \(y=x^2\)
\(
\begin{array}{ll}
\text { Solving, } & y=x^2 \\
\text { and } & x+y=2 \text {, we get } \\
& x^2+x-2=0 \\
\Rightarrow & (x+2)(x-1)=0 \\
\Rightarrow & x=-2,1
\end{array}
\)
It is clear from figure,
\(
A \equiv(1,1)
\)
also \(a>0\) for I quadrant.
\(
\therefore \quad a \in(0,1)
\)

Hint

(b) Equation of line \(\frac{a x}{c-1}+\frac{b y}{c-1}+1=0\) has two independent parameters. It can pass through a fixed point if it contains only one independent parameter. Now there must be one relation between \(\frac{a}{c-1}\) and \(\frac{b}{c-1}\) independent of \(a, b\) and \(c\) so that \(\frac{a}{c-1}\) can be expressed in terms of \(\frac{b}{c-1}\) and straight line contains only one independent parameter. Now that given relation can be expressed as \(\frac{5 a}{c-1}+\frac{4 b}{c-1}=\frac{t-20 c}{c-1}\). RHS is independent of \(c\) if \(t=20\).

Hint

(d) The three lines are concurrent if
\(
\left|\begin{array}{ccc}
a & a m & 1 \\
b & m+1 & 1 \\
c & (m+2) c & 1
\end{array}\right|=0
\)

Applying \(C_2 \rightarrow C_2-m C_1\), then
\(
\left|\begin{array}{ccc}
a & 0 & 1 \\
b & b & 1 \\
c & 2 c & 1
\end{array}\right|=0
\)
or \(a(b-2 c)-0+1(2 b c-b c)=0\)
or \(b=\frac{2 a c}{a+c}\), which is independent of \(m\).
\(\therefore a, b, c\) are in HP for all \(m\).

Hint

(a) Reflected ray is \(X\)-axis.
the angle between \(x=1\) and \(x+y=1\) is 45 degrees.
as angle \(i=\) angle \(r\)
so, angle \(r=45\) degrees
putting the value in the formula \(\tan \theta=\left(m_1-m_2 / 1+m_1 m_2\right)\)
we get \(m=0\)
\(
\therefore \text { Equation } y=0
\)

Hint

(b) Given line is
\(
\begin{array}{ll}
& \frac{x}{a}+\frac{y}{b}=1 \dots(i)\\
\therefore & A \equiv(a, 0), B \equiv(0, b)
\end{array}
\)
\(
\begin{aligned}
& \because & \text { Area of }(\triangle A O B) & =\Delta \\
& \therefore & \frac{1}{2}|a b| =\Delta \Rightarrow a b=2 \Delta \quad(\because a b>0)
\end{aligned}
\)
Since, the line (i) passes through the point \(P(\alpha, \beta)\).
\(
\therefore \quad \frac{\alpha}{a}+\frac{\beta}{b}=1 \Rightarrow \frac{\alpha}{a}+\frac{a \beta}{2 \Delta}=1 \quad\left(\because b=\frac{2 \Delta}{a}\right)
\)
\(
\begin{aligned}
& \Rightarrow \quad \alpha^2 \beta-2 a \Delta+2 \Delta \alpha=0 \\
& \because a \text { is real } \\
& \therefore \quad D \geq 0 \\
& \Rightarrow \quad 4 \Delta^2-4 \beta(2 \Delta \alpha) \geq 0 \text { or } \Delta \geq 2 \alpha \beta \\
&
\end{aligned}
\)
\(\therefore\) Least value of \(\Delta\) is \(2 \alpha \beta\).

Hint

(b) \(|P A-P B| \leq|A B|\)
Maximum value of \(|P A-P B|\) is \(|A B|\), which is possible only when \(P, A, B\) are collinear
if \(P(x, y)\), then equation \(A B\) is
\(
\frac{x}{2}+\frac{y}{2}=1
\)
\(
x+y=2 \dots(i)
\)
Now solving Eq. (i)
and \(2 x+3 y+1=0 \dots(ii)\)
Then, we get,
\(
\begin{array}{rlrl}
& x=7, y & =-5 \\
\therefore & P & \equiv(7,-5)
\end{array}
\)

Hint

(b) The system of straight lines \(a(2 x+y-3)+b(3 x+2 y-5)=0)\) passes through the point of intersection of the lines \(2 x+y-3=0\) and \(3 x+2 y-5=0\) i.e. \((1,1)\)
The line of this family which is farthest from \((4,-3)\) is the line through \((1,1)\) and perpendicular to the line joining \((1,1)\) and \((4,-3)\)
\(\therefore\) The required line is
\(
y-1=\frac{3}{4}(x-1)
\)
or \(3 x-4 y+1=0\)

Hint

\(
( a , b , c , d )
\)
Let \(P Q R S\) be a square inscribed in \(\triangle A B C\) and
\(
P Q=Q R=R S=S P=\lambda \text { (say) }
\)
Now equation of \(A B\) is
\(
x-2 y=0 \dots(i)
\)
and equation of \(B C\) is
\(
x+y-3=0 \dots(ii)
\)
\(\because S\) lies on \(A B\), then
\(
a-2 \lambda=0 \dots(iii)
\)
and \(R\) lies on \(B C\), then
\(
a+\lambda+\lambda-3=0 \text { or } a+2 \lambda-3=0 \dots(iv)
\)
From Eqs. (iii) and (iv), we get \(a=\frac{3}{2}, \lambda=\frac{3}{4}\)
Hence,
\(
\begin{aligned}
& P \equiv\left(\frac{3}{2}, 0\right), \quad Q \equiv\left(\frac{9}{4}, 0\right), \\
& R \equiv\left(\frac{9}{4}, \frac{3}{2}\right), \quad S=\left(\frac{3}{2}, \frac{3}{4}\right)
\end{aligned}
\)

Hint

\(
\because A, B, A^{\prime}, B^{\prime} \text { are concyclic then, }
\)

\(
\begin{aligned}
O A \cdot O A^{\prime} & =O B \cdot O B^{\prime} \\
(a) \cdot\left(-a^{\prime}\right) & =(b) \cdot\left(-b^{\prime}\right) \\
a a^{\prime} & =b b^{\prime} \dots(i)
\end{aligned}
\)
The equation of altitude through \(A^{\prime}\) is
\(
y-0=\frac{a}{b}\left(x+a^{\prime}\right)
\)
It intersects the altitude
\(
x=0 \text { at } y=\frac{a a^{\prime}}{b}
\)
\(
\therefore \text { Orthocentre is }\left(0, \frac{a a^{\prime}}{b}\right) \text { or }\left(0, b^{\prime}\right) \text { [from Eq. (i)] }
\)

Hint

Let the slope of \(u=0\) be \(m\), then the slope of \(v=0\) is \(\frac{9 m}{2}\).
Therefore,
\(
\left|\frac{m-\frac{9 m}{2}}{1+m \times \frac{9 m}{2}}\right|=\frac{7}{9}
\)
\(
\left|\frac{-7 m}{2+9 m^2}\right|=\frac{7}{9}
\)
\(
\begin{array}{ll}
\Rightarrow & 9 m^2+9 m+2=0 \text { or } 9 m^2-9 m+2=0 \\
\Rightarrow & m=-\frac{2}{3},-\frac{1}{3} \text { or } m=\frac{2}{3}, \frac{1}{3}
\end{array}
\)
Therefore, the equation of lines are
(i) \(2 x+3 y=0\) and \(3 x+y=0\)
(ii) \(x+3 y=0\) and \(3 x+2 y=0\)
(iii) \(2 x=3 y\) and \(3 x=y\)
(iv) \(x=3 y\) and \(3 x=2 y\)

Hint

\((a, b)\)
Here, \(\angle C O A=30^{\circ}\)
Let \(O A=A B=B C=C O=x\)
\(\because\) Area of rhombus \(O A B C\)
\(
\begin{aligned}
& =2 \times \frac{1}{2} \times x \times x \sin 30^{\circ} \\
& =\frac{x^2}{2}=2 \text { [given] }
\end{aligned}
[latex][latex]
\therefore \quad x=2
\)

Coordinates of \(A\) and \(C\) are \((\sqrt{3}, 1)\) and \((1, \sqrt{3})\) in I quadrant and in III quadrant are \((-\sqrt{3},-1)\) and \((-1,-\sqrt{3})\)
Hence, coordinates of \(B\) are \((\sqrt{3}+1, \sqrt{3}+1)\) and \((-\sqrt{3}-1,-\sqrt{3}-1)\)

Hint

\(
\because \quad A B=A P=B P=2 \sqrt{2}
\)
\(\therefore\) Coordinates of \(P\) are \(\left(3+2 \sqrt{2} \cos 105^{\circ}, 2+2 \sqrt{2} \sin 105^{\circ}\right)\)

or \(\quad(3-(\sqrt{3}-1), 2+\sqrt{3}+1)\)
or \((4-\sqrt{3}, 3+\sqrt{3})\)
If \(P\) below \(A B\), then coordinates of \(P\) are
\(
\begin{array}{ll}
& \left(3+2 \sqrt{2} \cos 15^{\circ}, 2-2 \sqrt{2} \sin 15^{\circ}\right) \\
\text { or } & {[(3+\sqrt{3}+1,2-(\sqrt{3}-1)]} \\
\text { or } & (4+\sqrt{3}, 3-\sqrt{3})
\end{array}
\)

Hint

Let the equation of line ‘ \(L\) ‘ through origin is
\(
\begin{array}{rlrl}
& & \frac{x-0}{\cos \theta} & =\frac{y-0}{\sin \theta}=r \\
& \therefore & P & \equiv(r \cos \theta, r \sin \theta) \\
\text { Let } & O A & =r_1 \text { and } O B=r_2 \\
& \therefore & A & \equiv\left(r_1 \cos \theta, r_1 \sin \theta\right) \\
& \text { and } & B & \equiv\left(r_2 \cos \theta, r_2 \sin \theta\right)
\end{array}
\)
\(
\begin{aligned}
& \text { A lies on } \quad L_1: y-x-10=0 \\
& \therefore \quad r_1 \sin \theta-r_1 \cos \theta-10=0 \\
& \Rightarrow \quad r_1=\frac{10}{\sin \theta-\cos \theta} \dots(i)
\end{aligned}
\)
\(
B \text { lies on } L_1: y-x-20=0
\)
\(
\begin{array}{ll}
\therefore & r_2 \sin \theta-r_2 \cos \theta-20=0 \\
\Rightarrow & r_2=\frac{20}{\sin \theta-\cos \theta} \dots(ii)
\end{array}
\)
\(
\begin{aligned}
& \frac{1}{(O P)^2}=\frac{1}{(O A)^2}+\frac{1}{(O B)^2} \\
& \frac{1}{r^2}=\frac{1}{r_1^2}+\frac{1}{r_2^2} \\
& \frac{1}{r^2}=\frac{(\sin \theta-\cos \theta)^2}{100}+\frac{(\sin \theta-\cos \theta)^2}{400} \\
& 400=4(r \sin \theta-r \cos \theta)^2+(r \sin \theta-r \cos \theta)^2
\end{aligned}
\)
Locus of \(P\) is \((y-x)^2=80\)

Hint

First, we construct the graph of the given quadrilateral.

It is clear from the graph that there are six points lying inside the quadrilateral.

Hint

(2) Given family of lines is
\(
\begin{aligned}
x(1+\lambda)+\lambda y+2+\lambda & =0 \\
\Rightarrow \quad(x+2)+\lambda(x+y+1) & =0
\end{aligned}
\)
for common point or fixed point
\(
\quad \begin{array}{r}
x+2=0 \\
\text { and } x+y+1=0
\end{array}
\)
\(
x=-2, y=1
\)
Common point is \((-2,1)\)
\(
\begin{aligned}
d & =\max \{|-2|,|1|\} \\
& =\max \{2,1\}=2
\end{aligned}
\)

Hint

(c) \(\because L_1, L_3\) are parallel.
\(\therefore\) Distance between \(L_1\) and \(L_3=\frac{1}{\sqrt{\left(\frac{1}{9}+\frac{1}{16}\right)}}=\frac{12}{5}\) and \(L_2, L_4\) are parallel.
\(\therefore\) Distance between \(L_2\) and \(L_4=\frac{1}{\sqrt{\left(\frac{1}{16}+\frac{1}{9}\right)}}=\frac{12}{5}\)
\(\therefore\) Distance between \(L_1\) and \(L_3=\) Distance between \(L_2\) and \(L_4\).
\(\therefore\) Quadrilateral formed by \(L_1, L_2, L_3, L_4\) is a rhombus. Hence, statement I is true and statement II is false.

Hint

\(
\begin{aligned}
& L_1 \equiv 3 x+4 y=0 \\
& L_2 \equiv 5 x-12 y=0 \text { and } L_3 \equiv y-15=0
\end{aligned}
\)
Length of \(\perp\) from \(P\) to \(L_1=\frac{3+32}{5}=7\)
Length of \(\perp\) from \(P\) to \(L_2=\frac{|5-96|}{13}=7\)
and Length of \(\perp\) from \(P\) to \(L_3=\frac{|8-15|}{1}=7\)
\(\therefore \quad\) Statement II is true

Also,
\(
\text { Area of } \begin{aligned}
\triangle O P A & =\frac{1}{2} \times O A \times 7 \\
& =\frac{1}{2} \times 39 \times 7=\Delta_1
\end{aligned}
\)
Area of \(\triangle O P B=\frac{1}{2} \times O B \times 7\)
\(
=\frac{1}{2} \times 25 \times 7=\Delta_2
\)
and
\(
\text { Area of } \begin{aligned}
\triangle A P B & =\frac{1}{2} \times A B \times 7 \\
& =\frac{1}{2} \times 56 \times 7=\Delta_3
\end{aligned}
\)
\(
\begin{aligned}
\therefore \quad \Delta_1+\Delta_2+\Delta_3 & =\frac{7}{2}(39+25+56) \\
& =\frac{7 \times 120}{2}=\frac{1}{2} \times 56 \times 15=\text { Area of } \triangle A O B
\end{aligned}
\)
\(\Rightarrow P\) inside the triangle.
Hence, both statements are true and statement II is not correct explanation of statement I.

Hint

Polar co-ordinates of any point are \((r, \theta)\), where \(r=\sqrt{x^2+y^2}\) and \(\theta=\tan ^{-1}\left(\frac{y}{x}\right)\). \(x=\sqrt{3} ; y=1\)
Let their polar co-ordinates be \((r, \theta) \Rightarrow x=r \cos \theta ; y=r \sin \theta\)
\(
\begin{array}{ll}
\text { So } r \Rightarrow \sqrt{x^2+y^2} & r=\sqrt{3+1} \\
\theta \Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=2 & \theta \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} \\
\therefore(r, \theta)=\left(2, \frac{\pi}{6}\right) . &
\end{array}
\)

Hint

The distance between two points \(=\sqrt{r_1^2+r_2^2-2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)}\). Therefore,
\(
P Q=\sqrt{r_1^2+r_2^2-2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)}=\sqrt{4+9-2.2 \cdot 3 \cos \left(-\frac{\pi}{6}-\frac{\pi}{6}\right)}==\sqrt{4+9-12 \cos \left(\frac{\pi}{3}\right)}=\sqrt{13-12 \cdot \frac{1}{2}}=\sqrt{7}
\)

Hint

Given, abscissa = ordinate. Therefore distance can be found by considering the co-ordinates of required point be \(P ( k , k )\).
Now given \(PA = PB \Rightarrow \sqrt{(k-1)^2+k^2}=\sqrt{k^2+(k-3)^2}\)
\(
2 k^2-2 k+1=2 k^2-6 k+9 \Rightarrow 4 k=8 \Rightarrow k=2
\)

Hint

Distance formula of two points can be used to prove \(A B^2+ BC ^2+ CA ^2=3\left(G A^2+G B^2+G C^2\right)\). In triangle \(A B C\), let \(B\) be the origin and \(B C\) the \(x\)-axis. Let \(A\) be \((h, k)\) and
\(
C \text { be }(a, 0) \text {. Then centroid } G \text { is }\left(\frac{a+h}{3}, \frac{k}{3}\right) \text {. }
\)
LHS
\(
\begin{aligned}
& =A B^2+B C^2+C A^2=(h-0)^2+(k-0)^2+a^2+(h-a)^2+(k-0)^2 \\
& =2 h^2+2 k^2+2 a^2-2 a h
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \quad \text { RHS } \\
& =3\left[\left(\frac{a+h}{3}-h\right)^2+\left(\frac{k}{3}-k\right)^2+\left(\frac{a+h}{3}-0\right)^2+\left(\frac{k}{3}-0\right)^2+\left(\frac{a+h}{3}-a\right)^2+\left(\frac{k}{3}-0\right)\right] \\
& =1 / 3\left[(a-2 h)^2+4 k^2+(a+h)^2+k^2+(h-2 a)^2+k^2\right]=2 h^2+2 k^2+2 a^2-2 a h
\end{aligned}
\)
Hence, it is equal on both sides.

Hint

Area of the triangle formed by the points \(\left(x_1, y_1\right),\left(x_2, y_2\right)\) and \(\left(x_3, y_3\right)\) is \(\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|\).
Substituting the given co-ordinates, we can obtain area of given triangle.
\(
D=\frac{1}{2}\left|\begin{array}{ccc}
p^2 & -p & 1 \\
q^2 & q & 1 \\
r^2 & -r & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
p^2-q^2 & -(p+q) & 0 \\
q^2-r^2 & q+r & 0 \\
r^2 & -r & 1
\end{array}\right|=\frac{1}{2}(p+q)(q+r)\left|\begin{array}{ccc}
p-q & -1 & 0 \\
q-r & 1 & 0 \\
r^2 & -r & 1
\end{array}\right|
\)
\(
=\frac{1}{2}(p+q)(q+r)[(p-q)+(q-r)]=\frac{1}{2}(p+q)(q+r)(p-r)
\)

Hint

Here we can obtain the equation of locus of given point by using given condition and distance formula of two points.
Let the co-ordinates of such a point be \(( x , y )\). Draw \(PM \perp\) to \(x\)-axis.
Hence, \(P M=y\)
\(PN =2 PM\) (given)

i.e. \((x-0)^2+(y-1)^2=4 y^2\)
i.e. \(x^2-3 y^2-2 y+1=0\).

Hint

The centroid \((h, k)\) of a triangle formed by points \(\left(x_1, y_1\right),\left(x_2, y_2\right)\) and \(\left(x_3, y_3\right)\) will be \(h =\frac{ x _1+ x _2+ x _3}{3}\) and \(k =\frac{ y _1+ y _2+ y _3}{3}\)
(A) If \((h, k)\) is the centroid, then
\(
h=\frac{a \cos t+b \sin t+1}{3}, k=\frac{a \sin t-b \cos t+0}{3} \Rightarrow(3 h-1)^2+(3 k)^2=(a \cos t+b \sin t)^2+(a \sin t t-b \cos t)^2=a^2+b^2
\)
\(\therefore\) Locus of \((h, k)\) is \((3 x-1)^2+(3 y)^2=a^2+b^2\)

Hint

We know that, \(\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|\), where \(m_1\) and \(m_2\) are the slope of lines and \(\theta\) is the angle between them. Let \(m_1=\frac{1}{2}, m_2=m\) and \(\theta=\frac{\pi}{4}\) So, \(\tan \frac{\pi}{4}=\left|\frac{m-(1 / 2)}{1+(1 / 2) m}\right| \Rightarrow 1= \pm \frac{m-(1 / 2)}{1+(1 / 2) m} \Rightarrow m=3\) or \(-(1 / 3)\)

Hint

Given two lines are perpendicular to each other. Therefore, product of their slope will be -1 .
Slope of the line through the points \((-2,6)\) and \((4,8)\) is \(m_1=\frac{8-6}{4-(-2)}=\frac{2}{6}=\frac{1}{3}\)
Slope of the line through the points \((8,12)\) and \((x, 24)\) is \(m_2=\frac{24-12}{x-8}=\frac{12}{x-8}\)
Since two lines are perpendicular \(m_1 m_2=-1\)
\(
\Rightarrow \frac{1}{3} \times \frac{12}{x-8}=-1 \quad \Rightarrow x=4
\)

Hint

By using formula \(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r\), we can obtain co-ordinates of point. The equation of line
\(
\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}= \pm r \Rightarrow \frac{x-2}{\cos 30^{\circ}}=\frac{y-3}{\sin 30^{\circ}}= \pm 4 \Rightarrow x=2 \pm 2 \sqrt{3}, y=3 \pm 2
\)

So, co-ordinate of two points are \((2 \pm 2 \sqrt{3}, 3 \pm 2)\)

Hint

(d) The line passing through the third vertex and orthocentre must be perpendicular to line through \((-2,3)\) and \((5,-1)\). Therefore, product of their slope will be -1 .
Given the two vertices \(B(-2,3)\) and \(C(5,-1)\); let \(H(0,0)\) be the orthocentre; \(A(h, k)\) the third vertex.
Then, the slope of the line through \(A\) and \(H\) is \(k / h\), while the line through \(B\) and \(C\) has the slope \((-1-3) /(5+2)=-4 / 7\). By the property of the orthocentre, these two lines must be perpendicular,
So we have \(\left(\frac{k}{h}\right)\left(-\frac{4}{7}\right)=-1 \Rightarrow \frac{k}{h}=\frac{7}{4}\)
Also \(\frac{5-2+ h }{3}+\frac{-1+3+ k }{3}=7 \Rightarrow h + k =16\)
Which is not satisfied by the points given in (a), (b) or (c).

Hint

By using \(x \cos \alpha+y \sin \alpha=P\), we can solve this problem. Here \(\alpha=30^{\circ}\) and \(P=3\). So equation is \(x \cos 30^{\circ}+y \sin 30^{\circ}=3 x \frac{\sqrt{3}}{2}+\frac{y}{2}=3 \quad \Rightarrow \sqrt{3} x+y=6\)

Hint

By using formula of distance between two parallel line, i.e. \(\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|\), we can find the diameter of given circle.
These are two parallel lines
\(
\begin{aligned}
& 3 x-4 y+10=0 \dots(i)\\
& 6 x-8 y+30=0 \dots(ii)
\end{aligned}
\)
Dividing second equation by 2 gives \(\quad 3 x-4 y+15=0 ; \therefore \quad d=\left|\frac{15-10}{\sqrt{3^2+4^2}}\right|=1\)

Hint

Given equation is \(a x+b y+c=0 \dots(i)\)
Since \(a, b\) and \(c\) are in A.P.
\(
\begin{aligned}
& \therefore b =\frac{a+c}{2} \\
& \Rightarrow a + c =2 b \\
& \Rightarrow a -2 b + c =0 \dots(ii)
\end{aligned}
\)
Comparing equation (i) with eq. (ii) we get,
\(
x=1, y=-2
\)
So, the line will pass through \((1,-2)\).

Hint

(C) Here the three lines are \(x=0, y=0\) and \(x+y=1\).
Since the triangle formed by the line \(x=0, y=0\) and \(x+y=1\) is right angled, the orthocentre lies at the vertex \((0,0)\), the point of intersection of the perpendicular lines \(x=0\) and \(y=0\).

Hint

(b) By using intercept form of equation of line, we will get equation of line before and after rotation. As their perpendicular length from the origin does not change, by using distance formula the result can be obtained.
Equation of the line \(L\) in the two co-ordinate system is \(\frac{x}{a}+\frac{y}{b}=1, \frac{X}{p}+\frac{Y}{q}=1\) Where \((x, y)\) are the new co-ordinates of a point \((x, y)\) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed:
\(
\frac{1}{\sqrt{\left(1 / a^2\right)+\left(1 / b^2\right)}}=\frac{1}{\sqrt{\left(1 / p^2\right)+\left(1 / q^2\right)}} \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2} \Rightarrow \frac{1}{a^2}-\frac{1}{p^2}+\frac{1}{b^2}-\frac{1}{q^2}=0
\)

Hint

By considering the required ratio be \(\lambda: 1\), and using section formula, we can solve above problem.
The point of division \(P\) is internal as \(A\) and \(B\) lie on opposite sides of given line.
Let required ratio be \(\lambda: 1\).
Since, \(P\left(\frac{8 \lambda+3}{\lambda+1}, \frac{9 \lambda-1}{\lambda+1}\right)\) lies on \(y-x+2=0\),
\(
\therefore \frac{9 \lambda-1}{\lambda+1}-\frac{8 \lambda+3}{\lambda+1}+2=0 \text { or } \lambda=\frac{2}{3}
\)

Hint

Using \(x=\frac{a x_1+b x_2+c x_3}{a+b+c}, \quad y=\frac{a y_1+b y_2+c y_3}{a+b+c}\) we can obtain the incentre.
\(
\begin{aligned}
& a=B C=\sqrt{(5+2)^2+(5-4)^2}=5 \sqrt{2} \\
& b=C A=\sqrt{(5-4)^2+(5+2)^2}=5 \sqrt{2} \\
& C=A B=\sqrt{(-2-4)^2+(4+2)^2}=6 \sqrt{2}
\end{aligned}
\)
If incentre I is \((\bar{x}, \bar{y})\), then,
\(
\bar{x}=\frac{a x_1+b x_2+c x_3}{a+b+c}
\)
\(
\begin{aligned}
& =\frac{20 \sqrt{2}-10 \sqrt{2}+30 \sqrt{2}}{5 \sqrt{2}+5 \sqrt{2}+6 \sqrt{2}}=\frac{5}{2} \\
& \bar{y}=\frac{a y_1+b y_2+c y_3}{a+b+c}
\end{aligned}
\)
\(
=\frac{-10 \sqrt{2}+20 \sqrt{2}+30 \sqrt{2}}{5 \sqrt{2}+5 \sqrt{2}+6 \sqrt{2}}=\frac{5}{2}
\)

Hint

Let coordinates of \(P\) be \(( t , a )\) and \(R\) be \(\left( x _1, y _1\right)\)
Slope of \(PQ = m \quad\) (given)
Slope of \(P S=-1 /(\) slope of \(P Q)=-1 / m\)
Equation of \(P_\theta=y-a=m(x-t) \dots(i)\)
As \(Q\) lies on \(x=b\) line, put \(x=b\) in (i) to get \(Q\).
\(
\Rightarrow \quad Q=[b, a+m(b-t)]
\)
Equation of \(P S \equiv y-a=-1 / m(x-t) \dots(ii)\)
As \(S\) lies on \(x =- b\) line, put \(x =- b\) in (ii) to get \(S\).
\(
\Rightarrow \quad S=[-b, a+1 / m(b+t)]
\)
Slope of \(R S=\frac{y_1-a \frac{-1}{m}(b+t)}{x_1+b}=m \dots(iii)\)
\(
\begin{aligned}
& \Rightarrow \quad b+t=m\left(y_1-a\right)-m^2(x+b) \\
& \text { Slope of } R Q=\frac{y_1-a-m(b-t)}{x_1-b}=-\frac{1}{m} \\
& \Rightarrow \quad \frac{m\left(y_1-a\right)+\left(x_1-b\right)}{m^2}=b-t
\end{aligned}
\)
Add (iii) and (iv) to eliminate t
\(
2 b=m\left(y_1-a\right)-m^2(x+b)+\frac{m\left(y_1-a\right)+\left(x_1-b\right)}{m^2}
\)
Locus is: \(m y+\left(1-m^2\right) x-a m-b\left(1+m^2\right)=0\)

Hint

For isosceles triangle \(A B C, A D\) is perpendicular bisector of side \(B C\) and it also bisects angle BAC. Hence by using equation of bisector formula, i.e.
\(
\frac{a x+b y+c}{\sqrt{a^2+b^2}}= \pm \frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}
\)
we can obtain slope of AD.
Equations of \(A B\) and \(A C\) are \(7 x-y+3=0\) and \(-x-y+3=0\), respectively.
\(
a_1=7 ; b_1=-1, c_1=3 ;
\)
\(
a _2= b _2=-1, c _2=3
\)
As \(c _1>0, c _2>0\) and \(a _1 a _2+ b _1 b _2=-6<0\)
Equation of the bisector of the acute angle BAD is
\(
\frac{7 x-y+3}{\sqrt{49+1}}=\frac{-x-y+3}{\sqrt{2}}, \text { i.e. } 3 x+y=3
\)
As slope of AD is -3 , slope of \(B C\) is \(\frac{1}{3}\) Equation of \(BC\) through \(P (1,-10)\) is \(y+10=\frac{1}{3}(x-1)\) or \(x-3 y=31\)

Hint

Equation of any line passing through the intersection of two other lines will be \(L_1+\lambda L_2=0\). Therefore, by using perpendicular distance formula of point to line, i.e
\(
p =\frac{\left| ax _1+ by _1+ c \right|}{\sqrt{ a ^2+ b ^2}}
\)
we can obtain required equation of line.
Any line through the point of intersection of given lines is
\(
\begin{aligned}
& x-3 y+1+\lambda(2 x+5 y-9)=0 \\
& \operatorname{or}(1+2 \lambda) x+(-3+5 \lambda) y+1-9 \lambda=0 \\
& \sqrt{5}=\frac{|0+0+1-9 \lambda|}{\sqrt{(1+2 \lambda)^2+(-3+5 \lambda)^2}}
\end{aligned}
\)
Squaring and simplifying, we get \(\lambda=\frac{7}{8}\).
Hence, required line has the equation \(2 x+y-5=0\).

Hint

We know \(\tan \phi=\left|\frac{2 \sqrt{h^2-a b}}{(a+b)}\right|\). Solving it, angle \(\phi\) can be obtained.
\(
\begin{aligned}
& x^2\left(\cos ^2 \theta-\sin ^2 \alpha\right)-2 x y \cos \theta \sin \theta+y^2\left(\sin ^2 \theta-\sin ^2 \alpha\right)=0 \\
& a=\cos ^2 \theta-\sin ^2 \alpha, 2 h=-2 \cos \theta \sin \theta, \\
& b=\sin ^2 \theta-\sin ^2 \alpha
\end{aligned}
\)
\(
\begin{aligned}
& \tan \phi=\left|\frac{2 \sqrt{h^2-a b}}{(a+b)}\right| \\
& =\frac{2 \sqrt{\cos ^2 \theta \sin ^2 \theta-\left(\cos ^2 \theta-\sin ^2 \alpha\right)\left(\sin ^2 \theta-\sin ^2 \alpha\right)}}{\left|\left(\cos ^2 \theta-\sin ^2 \alpha\right)+\sin ^2 \theta-\sin ^2 \alpha\right|} \\
& =\left|\frac{2 \sin \alpha \cos \alpha}{\cos 2 \alpha}\right|=\left|\frac{\sin 2 \alpha}{\cos 2 \alpha}\right|=\tan 2 \alpha \therefore \phi=2 a
\end{aligned}
\)

Hint

Step 1: Finding the point \(P\)
The straight line is passing through the origin \(O\), So the equation is \(y=m x \quad \ldots(1)\)
The equations are
\(
\begin{array}{ll}
4 x+2 y=9 & \ldots(2) \\
2 x+y=-6 & \ldots(3)
\end{array}
\)
Now substitute equation (1) in equation (2),
\(
\begin{aligned}
\Rightarrow \quad 2 x+y=\frac{9}{2}\Rightarrow \quad 2 x+m x=\frac{9}{2} \Rightarrow \quad x (2+ m )=\frac{9}{2} \\
x =\frac{9}{2(2+ m )}
\end{aligned}
\)
substitute the value of \(x\) in \(y\) of equation (1),
\(
\Rightarrow y=\frac{9 m}{2(2+m)}
\)
So the point of intersection of the line is \(P\left(\frac{9}{2(2+m)}, \frac{9 m}{2(2+m)}\right)\)
Step 2: Finding the point \(Q\)
Now substitute the equation (1) in equation (3),
\(
\begin{aligned}
\Rightarrow & 2 x+m x & =-6 \Rightarrow & x(2+m) & =-6 \Rightarrow & & x=\frac{-6}{2+m}
\end{aligned}
\)
substitute the value of \(x\) in \(y\) of equation (1),
\(
\Rightarrow y=\frac{-6 m}{2+m}
\)
So the point of intersection of the line is \(Q\left(\frac{-6}{2+m}, \frac{-6 m}{2+m}\right)\)
Step 3: Finding the ratio
The ratio can be determined by using the section formula, \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y 1}{m+n}\right)\) where \(m\) and \(n\) be the ratios.
Let us assume \(\lambda: 1\) be the ratio then, \(\left(\frac{m x_2+n x_1}{m+n}\right)=0\)
\(
\begin{array}{lr}
\Rightarrow & \frac{-6 \lambda}{2+m}+\frac{9(1)}{2 m+4}=0 \\
\Rightarrow+1 & \frac{-6 \lambda}{2+m}+\frac{9}{2(m+2)}=0 \\
\Rightarrow & -12 \lambda+9=0 \\
\Rightarrow & -12 \lambda=-9 \\
\Rightarrow & \lambda=\frac{3}{4}
\end{array}
\)

Hint

The vertices, \(O(0,0), A\left(\frac{1}{m-n}, \frac{m}{m-n}\right), B(0,1)\)
Area (parallelogram \(O A B C)=2\) area \((\triangle O A B)\)
\(
=2 \times \frac{1}{2}\left|\left[0\left(\frac{m}{m-n}-1\right)+\frac{1}{m-n}(1-0)+0\left(0-\frac{m}{m-n}\right)\right]\right|
\)
\(
=\frac{1}{|m-n|}
\)

Hint

(a) \(3 x+4 y=9\) and \(y=m x+1\) are two lines.
On equating the value of \(y\) from both equations to get the \(x\) co-ordinate of the point of intersection,
\(
\begin{aligned}
& 3 x+4(m x+1)=9 \Rightarrow(3+4 m) x=5 \\
& \Rightarrow \quad x=\frac{5}{3+4 m}
\end{aligned}
\)
For \(x\) to be an integer \(3+4 m\) should be a divisor of 5 i.e., 1,-1,5 or -5.
\(
\begin{aligned}
& 3+4 m=1 \Rightarrow m=-1 / 2 \text { (not integer) } \\
& 3+4 m=-1 \quad \Rightarrow m=-1 \text { (integer) } \\
& 3+4 m=5 \Rightarrow m=1 / 2 \text { (not an integer) } \quad 3+4 m=-5\\
& \Rightarrow m=-2 \text { (integer) } \\
& \therefore \quad \text { There are } 2 \text { integral values of } m \text {. } \\
&
\end{aligned}
\)

Hint

Let \((1, \sqrt{3}),(0,0)\) and \((2,0)\) are the coordinates of vertices A, \(O , B\) of \(\triangle A B C \).
\(\therefore \quad AO = OB = AB\). So, it is an equilaterial triangle and the incentre coincides with centroid.
\(
\therefore \quad \text { Incentre }=\left(\frac{0+1+2}{3}, \frac{0+0+\sqrt{3}}{3}\right)=\left(1, \frac{1}{\sqrt{3}}\right)
\)

Hint

\(
\text { (a) Since } x_1, x_2, x_3 \text { are in G.P. and } y_1, y_2, y_3 \text { are also in G.P. }
\)
Common ratio of both G.P.’s are same.
Let it be \(r\). Then
\(x_2=x_1 r, x_3=x_1 r^2\) and so is \(y_2=y_1 r, y_3=y_1 r^2\)
\(
\Delta=\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=r . r^2\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_1 & y_1 & 1 \\
x_1 & y_1 & 1
\end{array}\right|=0
\)
\(\therefore\) The points lie on a line, i.e., they are collinear.

Hint

Given lines are \(x+y=1\) and \(x y=0\)
when \(x =0\), then \(y =1\)
when \(x=1\), then \(y=0\)
\(\therefore(0,1)\) and \((1,0)\) are the vertices of triangle. Clearly, triangle is right – angled isosceles. Orthocentre of rightangled triangle is same as the vertex of right angle.
Therefore Orthocentre is \((0,0)\).

Hint

(a) Let variable point is \(P\) and fixed point \(S(-2,0)\), then \(P S=\frac{2}{3} P M\) where \(P M\) is the perpendicular distance of point \(P\) from given line \(x=-9 / 2\)
\(\therefore\) By definition of ellipse, \(P\) describes an ellipse with eccentricity \(e=\frac{2}{3}<1\)

Hint

(a) Let the two perpendicular lines be the co-ordinate axes. Let \((x, y)\) be the point sum of whose distances from two axes is 1 then we must have
\(
|x|+|y|=1 \quad \text { or } \quad \pm x \pm y=1
\)
These are the four lines
\(
x+y=1, x-y=1,-x+y=1,-x-y=1
\)
Any two adjacent sides are perpendicular to each other. Also each line is equidistant from origin. Therefore figure formed i.e., locus of the point is a square.

Hint

(d) Given :
\(
\begin{aligned}
& \quad P=(1,0), Q=(-1,0), R=(2,0) \\
& \text { Let } \quad S=(x, y) \\
& \text { Now, } S Q^2+S R^2=2 S P^2 \\
& \Rightarrow \quad(x+1)^2+y^2+(x-2)^2+y^2=2\left[( x -1)^2+y^2\right]
\end{aligned}
\)
\(
\Rightarrow 2 x^2+2 y^2-2 x+5=2 x^2+2 y^2-4 x +2
\)
\(\Rightarrow 2 x+3=0 \Rightarrow x=-3 / 2\), which is the locus of point \(S\).
This locus is a straight line parallel to \(y\)-axis.

Hint

(a) Solving the given equations of lines pairwise, we get the vertices of the triangle as
\(
A(-2,2), B(2,-2) \text { and } C(1,1)
\)

Then
\(
\begin{aligned}
& A B=\sqrt{16+16}=4 \sqrt{2} \\
& B C=\sqrt{1+9}=\sqrt{10} \text { and } C A=\sqrt{9+1}=\sqrt{10}
\end{aligned}
\)
\(\therefore \quad\) The triangle is isosceles.

Hint

(c) Reflection about the line \(y=x\), changes the point \((4,1)\) to \((1,4)\). On translation of \((1,4)\) through a distance of 2 units along positive direction of \(x\)-axis, the point becomes \((1+2,4)\), i.e., \((3,4)\).

On rotation about origin through an angle \(\pi / 4\) the point \(P\) takes the position \(P^{\prime}\) such that \(O P=O P^{\prime}\)
Also \(O P=5=O P^{\prime}\) and \(\cos \theta=\frac{3}{5} \cdot \sin \theta=\frac{4}{5}\)
Now, \(x=O P^{\prime} \cos \left(\frac{\pi}{4}+\theta\right)\)
\(
\begin{gathered}
=5\left(\cos \frac{\pi}{4} \cos \theta-\sin \frac{\pi}{4} \sin \theta\right)=5\left(\frac{3}{5 \sqrt{2}}-\frac{4}{5 \sqrt{2}}\right)=-\frac{1}{\sqrt{2}} \\
y=O P^{\prime} \sin \left(\frac{\pi}{4}+\theta\right)=5\left(\sin \frac{\pi}{4} \cos \theta+\cos \frac{\pi}{4} \sin \theta\right) \\
=5\left(\frac{3}{5 \sqrt{2}}+\frac{4}{5 \sqrt{2}}\right)=\frac{7}{\sqrt{2}}
\end{gathered}
\)
\(
\therefore \quad P^{\prime}=\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)
\)

Hint

Let \(B D\) be the bisector of \(\angle A B C\).
Then \(A D: D C=A B: B C\)
And
\(
\begin{aligned}
& A B=\sqrt{(5+1)^2+(1+7)^2}=10 \\
& B C=\sqrt{(5-1)^2+(1-4)^2}=5 \\
\therefore \quad & A D: D C=2: 1
\end{aligned}
\)
\(\therefore \quad\) By section formula coordinate of \(D\) is \(\left(\frac{1}{3}, \frac{1}{3}\right)\) Therefore equation of \(B D\) is
\(
\begin{aligned}
& y-1=\frac{1 / 3-1}{1 / 3-5}(x-5) \Rightarrow y-1=\frac{-2 / 3}{-14 / 3}(x-5) \\
& \Rightarrow \quad x-7 y+2=0
\end{aligned}
\)

Hint

The equations of sides of triangle \(A B C\) are
\(
\begin{aligned}
& A B: x+y=1 \\
& B C: 2 x+3 y=6 \\
& C A: 4 x-y=-4
\end{aligned}
\)
Solving these equations pairwise, we get the vertices of the triangle as \(A(-3 / 5,8 / 5), B(-3,4)\) and \(C(-3 / 7,16 / 7)\).
Let \(A D \perp B C\) as shown in the figure. Any line perpendicular to \(B C\) is \(3 x-2 y+\lambda=0\)
As it passes through the point \(A(-3 / 5,8 / 5)\)
\(
\therefore \quad \frac{-9}{5}-\frac{16}{5}+\lambda=0 \Rightarrow \lambda=5
\)
\(\therefore\) Equation of altitude \(A D\) is
\(
3 x-2 y+5=0 \dots(i)
\)
Any line perpendicular to side \(A C\) is \(x+4 y+\mu=0\)
As it passes through the point \(B(-3,4)\)
\(
\therefore \quad-3+16+\mu=0 \Rightarrow \mu=-13
\)
\(\therefore \quad\) Equation of altitude \(B E\) is \(x+4 y-13=0 \dots(ii)\)
Now orthocentre of a triangle is the point of intersection of the altitudes of the triangle.
On solving the equation (i) of \(AD\) and (ii) of \(BE\), we get \(x=3 / 7, y=22 / 7\)
\(\therefore \quad\) orthocentre \(=(3 / 7,22 / 7)\)
As both the co-ordinates are positive, orthocentre lies in the first quadrant.

Hint

Given that,
\(
\begin{aligned}
& | AP – BP |=6 \\
& \Rightarrow AP – BP = \pm 6 \\
& \Rightarrow AP = BP \pm 6
\end{aligned}
\)
The point
\(
\begin{aligned}
& A \left( x _1, y _1\right)=( o , 4) \\
& B \left( x _2, y _2\right)=(4,0)
\end{aligned}
\)
Using distance formula,
\(
{\sqrt{x^2+(y-4)^2}}=\sqrt{x^2+(y+4)^2} \pm 6
\)
On squaring both side and we get,
\(
\begin{aligned}
& \Rightarrow 16 y^2+81+72 y=9\left(x^2+(y+4)^2\right) \\
& \Rightarrow 16 y^2+81+72 y=9\left(x^2+y^2+16+8 y\right) \\
& \Rightarrow 16 y^2+81+72 y=9 x^2+9 y^2+144+72 y \\
& \Rightarrow 16 y^2+81=9 x^2+9 y^2+144 \\
& \Rightarrow 16 y^2+81-9 y^2=9 x^2+144 \\
& \Rightarrow 7 y^2+81=9 x^2+144 \\
& \Rightarrow 7 y^2-9 x^2=144-81 \\
& \Rightarrow 7 y^2-9 x^2=63 \\
& \Rightarrow \frac{7 y^2}{63}-\frac{9 x^2}{63}=1 \\
& \Rightarrow \frac{y^2}{9}-\frac{x^2}{7}=1
\end{aligned}
\)

Hint

Given curve : \(|x|+|y|=1\)
This curve represents four lines :
\(
\begin{aligned}
& \qquad x+y=1, x-y=1,-x+y=1 \\
& \text { and }-x-y=1 \\
& \text { These enclose a square of } \\
& \text { side = Distance between }
\end{aligned}
\)
opposite sides \(x+y=1\) and
\(
x+y=-1
\)
\(
\therefore \quad \text { Side }=\frac{1+1}{\sqrt{1+1}}=\sqrt{2}
\)

\(
\therefore \text { Required area }=(\text { side })^2=2 \text { sq. units }
\)

Hint

(True) Intersection point of \(x+2 y-10=0\) and \(2 x+y+5=0\) is \(\left(\frac{-20}{3}, \frac{25}{3}\right)\) which clearly satisfies the line \(5 x+4 y=0\). Hence the given statement is true.

Hint

(d) Slope of \(x+3 y=4\) is \(-1 / 3\) and slope of \(6 x-2 y=7\) is 3 .

Since, product of the two slopes is -1 , which shows that both diagonals are perpendicular. Hence \(P Q R S\) must be a rhombus.

Hint

(c) \(P Q R S\) will represent a parallelogram if and only if the mid-point of \(P R\) is same as that of the mid-point of QS. i.e., if and only if \(\frac{1+5}{2}=\frac{4+a}{2}\) and \(\frac{2+7}{2}=\frac{6+b}{2} \Rightarrow a=2\) and \(b=3\).

Hint

(a, c) Substituting the co-ordinates of the points \((1,3),(5,0)\) and \((-1\), 2) in \(3 x+2 y\), we obtain the value 8,15 and 1 which are all +ve. Therefore, all the points lying inside the triangle formed by given points satisfy \(3 x+2 y \geq 0\).
\(\therefore\) (a) is correct.
Substituting the co-ordinates of the given points in \(2 x+y-13\), we find the values \(-8,-3\) and -13 which are all \(-v e\).
\(\therefore\) (b) is not correct.

Again substituting the given points in \(2 x-3 y-12\) we get \(-19,-2\), 20 which are all -ve.
It follows that all points lying inside the triangle formed by given points satisfy \(2 x-3 y-12 \leq 0\).
\(\therefore\) (c) is correct.
Finally substituting the co-ordinates of the given points in \(-2 x+\) \(y\), we get \(1,-10\) and 4 which are not all +ve.
\(\therefore \quad\) (d) is not correct.
Therefore, (a) and (c) are the correct answers.

Hint

(e) Let \(A(0,8 / 3), B(1,3)\) and \(C(82,30)\).

Now, slope of line \(A B=\frac{3-8 / 3}{1-0}=\frac{1}{3}\)
Slope of line \(B C=\frac{30-3}{82-1}=\frac{27}{81}=\frac{1}{3}\)
\(\Rightarrow A B \| B C\) and \(B\) is common point.
\(\therefore \quad A, B, C\) are collinear.

Hint

Equation of the line passing through \(P(h, k)\) and parallel to \(x\)-axis is \(y=k \dots(i)\).
Other two given lines are
\(y=x \dots(ii)\)
and \(x+y=2 \dots(iii)\)
Let \(A B C\) be the formed by the lines (i), (ii) and (iii), as shown in the figure.

On solving the three equations pairwise we get the co-ordinates of vertices \(A, B\) and \(C\) as \(A(k, k), B(1,1)\) and \(C(2-k, k)\)
\(\therefore \quad\) Area of \(\triangle B C=\frac{1}{2}\left|\begin{array}{ccc}k & k & 1 \\ 1 & 1 & 1 \\ 2-k & k & 1\end{array}\right|=4 h^2\) Applying \(C_1 \rightarrow C_1-C_2\)
\(
\begin{aligned}
& \frac{1}{2}\left|\begin{array}{ccc}
0 & k & 1 \\
0 & 1 & 1 \\
2-2 k & k & 1
\end{array}\right|=4 h^2 \\
& \Rightarrow \frac{1}{2}|(2-2 k)(k-1)|=4 h^2 \Rightarrow(k-1)^2=4 h^2 \\
& \Rightarrow k-1=2 h \quad \text { or } \quad k-1=-2 h \\
& \Rightarrow k=2 h+1 \quad \text { or } k=-2 h+1 \\
& \therefore \quad \text { Locus of }(h, k) \text { is, } y=2 x+1 \text { or } y=-2 x+1 .
\end{aligned}
\)

Hint

Let slope of the given line be \(m\).
Then equation of the line is
\(y-2=m(x-8)\), where \(m<0\)
\(
\Rightarrow \quad P \equiv\left(8-\frac{2}{m}, 0\right) \text { and } Q \equiv(0,2-8 m)
\)
\(
\text { Now, } O P+O Q=\left|8-\frac{2}{m}\right|+|2-8 m|
\)
\(
=10+\frac{2}{-m}+8(-m) \geq 10+2 \sqrt{\frac{2}{-m} \times 8(-m)} \geq 18
\)

Hint

Equation of the line \(A B\) is
\(
\frac{x}{7}-\frac{y}{5}=1 \Rightarrow 5 x-7 y-35=0
\)
Equation of line \(P Q \perp A B\) is \(7 x+5 y+\lambda=0\) which meets \(x\) and \(y\) axis at points \(P(-\lambda / 7,0)\) and \(Q(0,-\lambda / 5)\) respectively.

Equation of \(A Q\) is
\(
\frac{x}{7}+\frac{y}{-\lambda / 5}=1 \Rightarrow \lambda x-35 y-7 \lambda=0 \dots(i)
\)
Equation of \(B P\) is
\(
\frac{-7 x}{\lambda}-\frac{y}{5}=1 \Rightarrow 35 x+\lambda y+5 \lambda=0 \dots(ii)
\)
Locus of \(R\) the point of intersection of (i) and (ii) can be obtained by eliminating \(\lambda\) from these equations as follows
\(
\begin{aligned}
& 35 x+(5+y)\left(\frac{35 y}{x-7}\right)=0 \\
\Rightarrow & 35 x(x-7)+35 y(5+y)=0 \Rightarrow x^2+y^2-7 x+5 y=0
\end{aligned}
\)

Hint

A being on \(y\)-axis, consider its co-ordinates as \((0, a )\). The diagonals intersect at \(P(1,2)\).

Again we know that diagonals will be parallel to the angle bisectors of the two lines \(y=x+2\) and \(y=7 x+3\)
\(
\begin{aligned}
& \text { i.e., } \frac{x-y+2}{\sqrt{2}}= \pm \frac{7 x-y+3}{5 \sqrt{2}} \\
& \Rightarrow 5 x-5 y+10= \pm(7 x-y+3) \\
& \Rightarrow 2 x+4 y-7=0 \text { and } 12 x-6 y+13=0
\end{aligned}
\)
Slope of \(2 x+4 y-7=0\) is \(m_1=-1 / 2\)
and slope of \(12 x-6 y+13=0\) is \(m_2=2\)
Let diagonal \(d_1\) be parallel to \(2 x+4 y-7=0\) and diagonal \(d_2\) be parallel to \(12 x-6 y+13=0\). The vertex \(A\) could be on any of the two diagonals. Hence slope of \(A P\) is either \(-1 / 2\) or 2 .
\(
\begin{aligned}
& \Rightarrow \frac{2-a}{1-0}=2 \text { or } \frac{-1}{2} \\
& \therefore a=0 \quad \text { or } \frac{5}{2}
\end{aligned}
\)
\(\therefore \quad\) Possible co-ordinates of \(A\) is \((0,0)\) or \((0,5 / 2)\)

Hint

Let \(O\) be the centre of the circle. \(M\) is the mid point of \(A B\). Then
\(
O M \perp A B
\)
Let \(O M\) when produced meets the circle at \(P\) and \(Q\).
\(\therefore \quad P Q\) is a diameter perpendicular to \(A B\) and passing through \(M\).
\(
M=\left(\frac{-3+5}{2}, \frac{4+4}{2}\right)=(1,4)
\)
Slope of \(A B=\frac{4-4}{5+3}=0\)
\(\therefore \quad P Q\), being perpendicular to \(A B\), is a line parallel to \(y\)-axis passing through \((1,4)\).
\(\therefore \quad\) Its equation is
\(
x=1 \dots(i)
\)
Also eq. of one of the diameter given is
\(
4 y=x+7 \dots(ii)
\)
On solving (i) and (ii), we get co-ordinates of centre \(O\) as \((1,2)\)
Let co-ordinates of \(D\) be \((\alpha, \beta)\)
Since \(O\) is mid point of \(B D\),
\(
\therefore\left(\frac{\alpha+5}{2}, \frac{\beta+4}{2}\right)=(1,2) \Rightarrow \alpha=-3, \beta=0
\)
\(\therefore\) Co-ordinate of \(D=(-3,0)\)
Now \(A D=\sqrt{(-3+3)^2+(4-0)^2}=4\)
and \(A B=\sqrt{(5+3)^2+(4-4)^2}=8\)
\(\therefore\) Area of rectangle \(A B C D=A B \times A D=8 \times 4\)
\(=32\) square units.

Hint

Area of \(\triangle A B C=\frac{1}{2}\left|\begin{array}{ccc}6 & 3 & 1 \\ -3 & 5 & 1 \\ 4 & -2 & 1\end{array}\right|\)
\(
=\frac{1}{2}[6(7)+3(5)+4(-2)]=\frac{49}{2}
\)
\(
\text { Area of } \triangle A B C=\frac{1}{2}\left|\begin{array}{ccc}
x & y & 1 \\
-3 & 5 & 1 \\
4 & -2 & 1
\end{array}\right|
\)
\(
=\frac{1}{2}(7 x+7 y-14)-\frac{7}{2}|x+y-2|
\)
\(
\text { Now, } \frac{\operatorname{ar}(\triangle P B C)}{\operatorname{ar}(\triangle A B C)}=\frac{\frac{7}{2}|x+y-2|}{\frac{49}{2}}=\left|\frac{x+y-2}{7}\right|
\)

Hint

Slope
\(
\text { of } \begin{aligned}
B C & =\frac{a\left(t_1+t_3\right)-a\left(t_2+t_3\right)}{a t_1 t_3-a t_2 t_3} \\
& =\frac{a\left(t_1+t_3-t_2-t_3\right)}{a t_3\left(t_1-t_2\right)}=\frac{1}{t_3}
\end{aligned}
\)
\(\therefore \quad\) Slope of \(A D=-t_3\)
\(\therefore\) Equation of \(A D\),
\(
\begin{aligned}
& y-a\left(t_1+t_2\right)=-t_3\left(x-a t_1 t_2\right) \\
\Rightarrow \quad & x t_3+y=a t_1 t_2 t_3+a\left(t_1+t_2\right) \dots(i)
\end{aligned}
\)
Similarly, by symm. equation of \(B E\) is
\(
\Rightarrow x t_1+y=a t_1 t_2 t_3+a\left(t_2+t_3\right) \dots(ii)
\)
On solving (i) and (ii), we get
\(
\left.x=-a \text { and } y=a\left(t_1+t_2+t_3\right)+a t_1 t_2 t_3\right)
\)
\(\therefore \quad\) Orthocentre is \(H\left(-a, a\left(t_1+t_2+t_3\right)+a t_1 t_2 t_3\right)\)

Hint

(c) Let \(H\) be the orthocentre.
Since \(A H \perp B C \Rightarrow m_{A H} \times m_{B C}=-1\)
\(
\Rightarrow \frac{k}{h} \times \frac{3+1}{-2-5}=-1
\)
\(
\Rightarrow \quad 4 k-7 h=0 \dots(i)
\)
Also, \(B H \perp A C\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{-1}{5} \times \frac{3-k}{-2-h}=-1 \Rightarrow 3-k=-10-5 h \\
& \Rightarrow 5 h-k+13=0 \dots(ii)
\end{aligned}
\)
On solving (i) and (ii), we get \(h=-4, k=-7\)
\(\therefore \quad\) Third vertex is \((-4,-7)\).

Hint

The given lines are \(x-2 y+4=0\) and \(4 x-3 y+2=0\)
Both the lines have constant terms of same sign.
\(\therefore\) The equation of bisectors of the angles between the given lines are
\(
\frac{x-2 y+4}{\sqrt{1+4}}= \pm \frac{4 x-3 y+2}{\sqrt{16+9}}
\)

Here \(a_1 a_2+b_1 b_2>0\), therefore taking + ve sign on RHS, we get obtuse angle bisector as
\(
(4-\sqrt{5}) x+(2 \sqrt{5}-3) y-(4 \sqrt{5}-2)=0
\)

Hint

As \(C\) lies on the line \(y=x+3\), let the co-ordinates of \(C\) be \((\lambda, \lambda+3)\). Given two points are \(A(2,1)\) and \(B(3,-2)\).
Now, area of \(\triangle B C\)
\(
=\frac{1}{2}\left|\begin{array}{ccc}
2 & 1 & 1 \\
3 & -2 & 1 \\
\lambda & \lambda+3 & 1
\end{array}\right|= \pm 5
\)
\(
\begin{aligned}
& \Rightarrow|2(-2-\lambda-3)-1(3-\lambda)(3 \lambda+9+2 \lambda)|=10 \\
& \Rightarrow|-2 \lambda-10-3+\lambda+5 \lambda+9|=10 \Rightarrow|4 \lambda-4|=10
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow 4 \lambda-4=10 & \text { or } 4 \lambda-4=-10 \\
\Rightarrow \lambda=7 / 2 & \text { or } \lambda=-3 / 2
\end{array}
\)
\(
\therefore \quad \text { Coordinates of } C \text { are }\left(\frac{7}{2}, \frac{13}{2}\right) \text { or }\left(\frac{-3}{2}, \frac{3}{2}\right)
\)

Hint

Let \(AB =\ell\) and \(P ( x , y )\) divides line segment \(AB\) in the ratio \(1: 2\).
\(
\therefore \quad AP =\ell / 3 \text { and } BP =2 \ell / 3
\)
Then \(PN = x\) and \(PM = y\)
Let \(\angle PAM =\theta=\angle BPN\)
In right \(\angle P M A, \sin \theta=\frac{y}{\ell / 3}=\frac{3 y}{\ell}\)
In right \(\angle P N B, \cos \theta=\frac{x}{2 \ell / 3}=\frac{3 x}{2 \ell}\)
Now, \(\sin ^2 \theta+\cos ^2 \theta=1\)
\(
\Rightarrow \frac{9 y^2}{\ell^2}+\frac{9 x^2}{4 \ell^2}=1 \Rightarrow 9 x ^2+36 y ^2=4 \ell^2
\)

Hint

(d) \(S\) is the midpoint of \(Q\) and \(R\)
\(
\therefore S \equiv\left(\frac{7+6}{2}, \frac{3-1}{2}\right)=\left(\frac{13}{2}, 1\right)
\)
Now slope of \(P S=\frac{2-1}{2-13 / 2}=-\frac{2}{9}\)
Now equation of the line passing through \((1,-1)\) and parallel to \(P S\) is
\(
y+1=-\frac{2}{9}(x-1) \Rightarrow 2 x+9 y+7=0
\)

Hint

\(
\begin{aligned}
& \text { (c) } x^2-5 x+6=0 \\
& \Rightarrow \quad(x-2)(x-3)=0 \\
& \therefore \quad x=2 \text { and } x=3 \\
& \text { And } y^2-6 y+5=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad(y-1)(y-5)=0 \\
& \therefore \quad y=1 \text { and } y=5
\end{aligned}
\)
The sides of parallelogram are
\(
x=2, x=3, y=1, y=5 \text {. }
\)
\(\therefore \quad\) Diagonal \(A C\) is \(\frac{y-1}{5-1}=\frac{x-2}{3-2} \Rightarrow y=4 x-7\)
Equation of diagonal \(B D\) is \(\frac{x-2}{3-2}=\frac{y-5}{1-5} \Rightarrow 4 x+y=13\)

Hint

(b, c) We know that length of intercept made by a circle on a line is given by \(=2 \sqrt{r^2-p^2}\), where
\(p=\) perpendicular distance of the line from the centre of the circle.
Here, circle is \(x^2+y^2-x+3 y=0\) with centre \(\left(\frac{1}{2}, \frac{-3}{2}\right)\) and radius
\(
=\frac{\sqrt{10}}{2}
\)
Let \(L_1: y=m x\) (any line through origin)
Now, \(L_z: x+y-1=0\) (given line)
ATQ circle makes equal intercepts on \(L_1\) and \(L_2\)
\(
\begin{aligned}
& \Rightarrow 2 \sqrt{\frac{10}{4}-\frac{\left(\frac{m}{2}+\frac{3}{2}\right)^2}{m^2+1}}=2 \sqrt{\frac{10}{4}-\frac{\left(\frac{1}{2}-\frac{3}{2}-1\right)^2}{2}} \\
& \Rightarrow \frac{\left(\frac{m+3}{2}\right)^2}{m^2+1}=2 \\
& \Rightarrow m^2+6 m+9=8 m^2+8 \Rightarrow 7 m^2-6 m-1=0 \\
& \Rightarrow(7 m+1)(m-1)=0 \Rightarrow m=1,-1 / 7
\end{aligned}
\)
\(\therefore \quad\) The required line \(L_1\) is \(y=x\) or \(y=-\frac{x}{7}\), i.e., \(x-y=0 \quad\) or \(\quad x+7 y=0\).

Hint

The given straight lines are \(3 x+4 y=5\) and \(4 x-3 y=15\). Clearly these straight lines are perpendicular to each other as \(m_1 m_2=-1\) i.e., product of their slopes is -1 . The given two lines intersect at \(A\).
\(
\begin{array}{r}
A B=A C \\
\angle B=\angle C=45^{\circ}
\end{array}
\)
\(
\text { Let slope of } B C \text { be } m \text {. Then }
\)
\(
\begin{aligned}
& \tan 45^{\circ}=\left|\frac{m+3 / 4}{1-\frac{3}{4} m}\right| \\
\Rightarrow & 4 m+3= \pm(4-3 m) \\
\Rightarrow & 4 m+3=4-3 m \text { or } \quad 4 m+3=-4+3 m \\
\Rightarrow & m=1 / 7 \text { or } m=-7 \\
\therefore & \text { Equation of } B C \text { is, } y-2=\frac{1}{7}(x-1) \\
& \quad \text { or } y-2=-7(x-1) \\
\Rightarrow & 7 y-14=x-1 \text { or } y-2=-7 x+7 \\
\Rightarrow & x-7 y+13=0 \text { or } 7 x+y-9=0
\end{aligned}
\)