Past JEE Main Entrance Papers

Hint

\(
\begin{array}{l}
\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}} \quad \text { [By sine rule] } \\
2 \mathrm{c}=8 \Rightarrow \mathrm{c}=4 \\
\mathrm{AB}=|(\mathrm{b}+1)|=4 \\
\mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 \\
\mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} \\
\mathrm{BC}:-\mathrm{y}=\frac{-1}{\sqrt{3}}(\mathrm{x}-3) \\
\sqrt{3} \mathrm{y}+\mathrm{x}=3
\end{array}
\)
Point of intersection: \(y=x+3, \sqrt{3} y+x=3\)
\(
\begin{array}{l}
(\sqrt{3+1}) y=6 \\
y=\frac{6}{\sqrt{3}+1} \\
x=\frac{6}{\sqrt{3}+1}-3 \\
=\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} \\
=3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})} \\
=\frac{-6}{(1+\sqrt{3})^2} \\
\frac{\beta^4}{\alpha^2}=36
\end{array}
\)

Hint

Co-ordinates of \(\mathrm{A}=\left(\frac{5}{3}, \frac{8}{3}\right)\)
Co-ordinates of \(\mathrm{B}=\left(\frac{10}{3}, \frac{4}{3}\right)\)
Slope of \(\mathrm{OA}=\mathrm{m}_1=\frac{8}{5}\)
Slope of \(\mathrm{OB}=\mathrm{m}_2=\frac{2}{5}\)

\(
\begin{array}{l}
\tan \theta=\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right| \\
\tan \theta=\frac{\frac{6}{5}}{1+\frac{16}{25}}=\frac{30}{41} \\
\tan \theta=\frac{30}{41}
\end{array}
\)

Hint

\(
\begin{array}{l}
\mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right) \\
\mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0
\end{array}
\)
Origin and \(\mathrm{P}\) lies same side w.r.t. \(\mathrm{L}_1\)
\(
\begin{array}{l}
\Rightarrow \mathrm{L}_1(0) \cdot \mathrm{L}_1(\mathrm{P})>0 \\
\therefore 3\left(\mathrm{a}^2\right)-(\mathrm{a}+1)+1>0
\end{array}
\)

\(
\begin{array}{l}
\Rightarrow 3 \mathrm{a}^2-\mathrm{a}>0 \\
\mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \quad \ldots(1)
\end{array}
\)
Let \(\mathrm{L}_2: \mathrm{x}+2 \mathrm{y}-5=0\)
Origin and \(\mathrm{P}\) lies same side w.r.t. \(\mathrm{L}_2\)
\(
\begin{array}{l}
\Rightarrow \mathrm{L}_2(0) \cdot \mathrm{L}_2(\mathrm{P})>0 \\
\Rightarrow \mathrm{a}^2+2(\mathrm{a}+1)-5<0 \\
\Rightarrow \mathrm{a}^2+2 \mathrm{a}-3<0 \\
\Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0 \\
\therefore \mathrm{a} \in(-3,1) \quad \ldots(2)
\end{array}
\)
Intersection of (1) and (2)
\(
\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)
\)

Hint

\(
\begin{array}{l}
2^{\mathrm{m}}-2^{\mathrm{n}}=56 \\
2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 \\
2^{\mathrm{n}}=2^3 \text { and } 2^{\mathrm{m}-\mathrm{n}}-1=7 \\
\Rightarrow \mathrm{n}=3 \text { and } 2^{\mathrm{m}-\mathrm{n}}=8 \\
\Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}-\mathrm{n}=3 \\
\Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}=6 \\
\mathrm{P}(6,3) \text { and } \mathrm{Q}(-2,-3) \\
\mathrm{PQ}=\sqrt{8^2+6^2}=\sqrt{100}=10
\end{array}
\)

Hint

\(
\begin{array}{l}
2 x-y+3=0 \\
6 x+3 y+1=0 \\
\alpha x+2 y-2=0
\end{array}
\)
Will not form a \(\Delta\) if \(\alpha x+2 y-2=0\) is concurrent with \(2 x-y+3=0\) and \(6 x+3 y+1=0\) or parallel to either of them so
Case-1: Concurrent lines
\(
\left|\begin{array}{ccc}
2 & -1 & 3 \\
6 & 3 & 1 \\
\alpha & 2 & -2
\end{array}\right|=0 \Rightarrow \alpha=\frac{4}{5}
\)
Case-2 : Parallel lines
\(
\begin{array}{l}
-\frac{\alpha}{2}=\frac{-6}{3} \text { or }-\frac{\alpha}{2}=2 \\
\Rightarrow \alpha=4 \text { or } \alpha=-4 \\
P=16+16+\frac{16}{25} \\
{[P]=\left[32+\frac{16}{25}\right]=32}
\end{array}
\)

Hint

\(
\begin{array}{l}
\mathrm{AD}: \mathrm{DC}=1: 2 \\
\frac{4-\alpha}{6-\alpha}=\frac{10}{8} \\
\alpha=\beta \\
\alpha=14 \text { and } \beta=14
\end{array}
\)

\(
\alpha+2 \beta=42
\)

Hint

\(
\begin{array}{l}
A(a,-2), B(a, 6), C\left(\frac{a}{4},-2\right), O\left(5, \frac{a}{4}\right) \\
A O=B O \\
(a-5)^2+\left(\frac{a}{4}+2\right)^2=(a-5)^2+\left(\frac{a}{4}-6\right)^2 \\
a=8 \\
A B=8, A C=6, B C=10 \\
\alpha=5, \beta=24, \gamma=24
\end{array}
\)

\(
\alpha+\beta+\gamma=53
\)

Hint

Writing \(P\) in terms of parametric co-ordinates \(2+r\)
\(
\begin{array}{l}
\cos \theta, 3+r \sin \theta \text { as } \tan \theta=\sqrt{3} \\
P\left(2+\frac{r}{2}, 3+\frac{\sqrt{3} r}{2}\right)
\end{array}
\)
\(P\) must satisfy \(2 x-3 y+28=0\)
So, \(2\left(2+\frac{r}{2}\right)-3\left(3+\frac{\sqrt{3} r}{2}\right)+28=0\)
We find \(r=4+6 \sqrt{3}\)

Hint

Solving lines \(\mathrm{L}_1(3 \mathrm{x}+2 \mathrm{y}=14)\) and \(\mathrm{L}_2(5 \mathrm{x}-\mathrm{y}=6)\) to get \(\mathrm{A}(2,4)\) and solving lines \(\mathrm{L}_3(4 \mathrm{x}+3 \mathrm{y}=8)\) and \(L_4(6 x+y=5)\) to get \(B\left(\frac{1}{2}, 2\right)\).
Finding Eqn. of \(A B: 4 x-3 y+4=0\)
Calculate distance PM
\(
\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6
\)

Hint

\(
\mathrm{Eq}^{\mathrm{n}}: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)
\)
\(
\frac{y}{\sqrt{3}-2}+x=9
\)

Hint

Cocus of point \(P(x, y)\) whose distance from Gives \(X+2 y+7=0  \&  2 x-y+8=0\) are equal is
\(
\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}(x+2 y+7)^2-(2 x-y+8)^2=0
\)
Combined equation of lines
\(
\begin{array}{l}
(x-3 y+1)(3 x+y+15)=0 \\
3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\
x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \\
x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\
h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \\
g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14
\end{array}
\)

Hint

Let \(\mathrm{E}\) is mid point of diagonals
\(
\begin{array}{ll}
\frac{\alpha+\gamma}{2}=\frac{1+1}{2} & \& \frac{\beta+\delta}{2}=\frac{2+0}{2} \\
\alpha+\gamma=2 & \beta+\delta=2 \\
2(\alpha+\beta+\gamma+\delta)=2(2+2)=8 &
\end{array}
\)

Hint

\(
\begin{array}{l}
\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8) \\
\mathrm{C} \frac{2: 1}{A} \\
(-6,-8) \quad(a, b) \quad(3,4) \\
\Rightarrow a=0, b=0 \quad \Rightarrow P(3,5)
\end{array}
\)
Distance from \(P\) measured along \(x-2 y-1=0\)
\(
\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta
\)
Where \(\tan \theta=\frac{1}{2}\)
\(
\begin{array}{l}
r(2 \cos \theta+3 \sin \theta)=-17 \\
\Rightarrow r=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7}
\end{array}
\)

Hint

\(
\begin{array}{l}
\mathrm{P} \equiv\left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) \equiv\left(\frac{\gamma+1}{2}, \frac{\delta}{2}\right) \\
\frac{\alpha-2}{2}=\frac{\gamma+1}{2} \text { and } \frac{\beta-1}{2}=\frac{\delta}{2} \\
\Rightarrow \alpha-\gamma=3 \quad \ldots(1) \\
\beta-\delta=1 \quad \ldots(2)
\end{array}
\)
Also, \((\gamma, \delta)\) lies on \(3 \mathrm{x}-2 \mathrm{y}=6\)
\(
3 \gamma-2 \delta=6 \quad \ldots(3)
\)
and \((\alpha, \beta)\) lies on \(2 x-y=5\)
\(
\Rightarrow 2 \alpha-\beta=5 \quad \ldots .(4)
\)
Solving (1), (2), (3), (4)
\(
\begin{array}{l}
\alpha=-3, \beta=-11, \gamma=-6, \delta=-12 \\
|\alpha+\beta+\gamma+\delta|=32
\end{array}
\)

Hint

Let the position vector of \(P, Q, R\) be \(\overrightarrow{0}, \vec{a}, \vec{b}\)
\(\Rightarrow\) Position vector of \(A=\frac{2 \vec{a}+\vec{b}}{3}\),
Position vector of \(B=\frac{2 \vec{b}}{3}\) and
Position vector of \(C=\frac{\vec{a}}{3}\)

\(
\begin{array}{l}
\therefore \overrightarrow{A B}=\frac{2 \vec{b}}{3}-\left(\frac{2 \vec{a}+\vec{b}}{3}\right)=\frac{\vec{b}}{3}-\frac{2 \vec{a}}{3}=\frac{\vec{b}-2 \vec{a}}{3} \\
\overrightarrow{C A}=\frac{2 \vec{a}+\vec{b}}{3}-\frac{\vec{a}}{3}=\frac{\vec{a}+\vec{b}}{3}
\end{array}
\)
Area of \(\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\frac{1}{2}|\vec{a} \times \vec{b}|\)
Area of \(\triangle A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{A B}|\)
\(
=\frac{1}{2}\left|\left(\frac{\vec{a}+\vec{b}}{3}\right) \times\left(\frac{\vec{b}-2 \vec{a}}{3}\right)\right|=\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right|
\)
\(\therefore \frac{\text { Area }(\triangle P Q R)}{\text { Area }(\triangle A B C)}=\frac{\frac{1}{2}|\vec{a} \times \vec{b}|}{\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right|}=3\) sq. units

Hint

\(
\begin{array}{ll}
\text { Assume } \mathrm{B}(\alpha,-2 \alpha) & \text { and } \mathrm{C}(\beta+3, \beta) \\
\frac{\alpha+\beta+3+1}{3}=2 & \text { also } \frac{-2 \alpha-2+\beta}{3}=\mathrm{a} \\
\Rightarrow \alpha+\beta=2 & -2 \alpha-2+\beta=3 \mathrm{a} \\
\Rightarrow \beta=2-\alpha & -2 \alpha-\not 2+\not 2-\alpha=3 \mathrm{a} \Rightarrow \alpha=-\mathrm{a}
\end{array}
\)
Now both \(B\) and \(C\) lies as given line
\(
\begin{array}{l}
\alpha-p \cdot 2 \alpha=21 a \\
\alpha(1-2 p)=21 a \\
-a(1-2 p)=21 a \Rightarrow p=11 \\
\beta+3+p \beta=21 a \\
\beta+3+11 \beta=21 a \\
21 \alpha+12 \beta+3=0
\end{array}
\)
Also \(\beta=2-\alpha\)
\(
\begin{array}{l}
21 \alpha+12(2-\alpha)+3=0 \\
21 \alpha+24-12 \alpha+3=0 \\
9 \alpha+27=0 \\
\alpha=-3, \beta=5
\end{array}
\)
So \(\mathrm{BC}=\sqrt{122}\) and \((\mathrm{BC})^2=122\)

Hint

\(
\begin{array}{l}
y^2=6 x & y^2=4 a x \\
\Rightarrow 4 a=6 \Rightarrow a=\frac{3}{2}
\end{array}
\)

\(
\begin{array}{l}
y=m x+\frac{3}{2 m} ;(m \neq 0) \\
h=\frac{6 m-3}{4 m^2}, k=\frac{6 m+3}{4 m}, \text { Now eliminating } m \text { and we get } \\
\Rightarrow 3 h=2\left(-2 k^2+9 k-9\right) \\
\Rightarrow 4 y^2-18 y+3 x+18=0
\end{array}
\)

Hint

\(
\text { If } x=1, y=\frac{57}{4}=14.25
\)
\((1,1)(1,2) \ldots(1,14) \quad \Rightarrow 14\) pts.
If \(x=2, y=\frac{27}{2}=13.5\)
\((2,2)(2,4) \ldots(2,12) \quad \Rightarrow 6\) pts.
If \(x=3, y=\frac{51}{4}=12.75\)
\((3,3)(3,6)(3,9)(3,12) \Rightarrow 4\) pts.
If \(x=4, y=12\)
\((4,4)(4,8) \quad \Rightarrow 2\) pts.
If \(x=5 . y=\frac{45}{4}=11.25\)
\((5,5),(5,10) \quad \Rightarrow 2\) pts.
If \(x=6, y=\frac{21}{2}=10.5\)
\(
\Rightarrow 1 \mathrm{pt} \text {. }
\)
If \(x=7, y=\frac{39}{4}=9.75\)
\((7,7) \quad \Rightarrow 1 \mathrm{pt}\).
If \(x=8, y=9\)
\((8,8) \quad \Rightarrow 1 \mathrm{pt}\).
If \(x=9 y=\frac{33}{4}=8.25 \Rightarrow\) no pt.
Total \(=31\) pts.

Hint

Slope of reflected ray \(=\tan 60^{\circ}=\sqrt{3}\)
Line \(y=\frac{x}{\sqrt{3}}\) intersect \(y+x=1\) at \(\left(\frac{\sqrt{3}}{\sqrt{3}+1}, \frac{1}{\sqrt{3}+1}\right)\)
Equation of reflected ray is
\(
y-\frac{1}{\sqrt{3}+1}=\sqrt{3}\left(x-\frac{\sqrt{3}}{\sqrt{3}+1}\right)
\)
Put \(y=0 \Rightarrow x=\frac{2}{3+\sqrt{3}}\)

Hint

At A \(x=y\)
\(
\begin{array}{l}
Y-2 x=2 \\
(-2,-2)
\end{array}
\)
Height from line \(\mathrm{x}+\mathrm{y}=0\)
\(
\mathrm{h}=\frac{4}{\sqrt{2}}
\)
Area of \(\Delta=\frac{\sqrt{3}}{4} \frac{h^2}{\sin ^2 60}=\frac{8}{\sqrt{3}}\)

Hint

A triangle is formed by the tangents at the point \((2,2)\) on the curves \(y^2=2 x\) and \(x^2+\) \(y^2=4 x\), and the line \(x+y+2=0\). If \(r\) is the radius of its circumcircle, then \(r^2\) is equal to 10 .
\(
S_1: y^2=2 x \quad S_2: x^2+y^2=4 x
\)
\(\mathrm{P}(2,2)\) is common point on \(\mathrm{S}_1 \& \mathrm{~S}_2\)
\(T_1\) is tangent to \(S_1\) at \(P \quad \Rightarrow T_1: y \cdot 2=x+2\)
\(
\Rightarrow T_1: x-2 y+2=0
\)
\(\mathrm{T}_2\) is tangent to \(\mathrm{S}_2\) at \(\mathrm{P}\)
\(
\begin{array}{c}
\Rightarrow \mathrm{T}_2: \mathrm{x} \cdot 2+\mathrm{y} \cdot 2=2(\mathrm{x}+2) \\
\Rightarrow \mathrm{T}_2: \mathrm{y}=2
\end{array}
\)
& \(\mathrm{~L}_3: \mathrm{x}+\mathrm{y}+2=0\) is third line

\(
\begin{array}{l}
P Q=a=\sqrt{20} \\
Q R=b=\sqrt{8} \\
R P=c=6 \\
\text { Area }(\Delta P Q R)=\Delta=\frac{1}{2} \times 6 \times 2=6 \\
\therefore r=\frac{a b c}{4 \Delta}=\frac{\sqrt{160}}{4}=\sqrt{10} \Rightarrow r^2=10
\end{array}
\)

Hint

Equation of straight line : \(\frac{x}{a}+\frac{y}{b}=1\)
Or \(\mathrm{x} \cos \frac{\pi}{3}+\mathrm{y} \sin \frac{\pi}{3}=\mathrm{p}\)
\(
\begin{array}{l}
\frac{\mathrm{x}}{2}+\frac{\mathrm{y} \sqrt{3}}{2}=\mathrm{p} \\
\frac{\mathrm{x}}{2 \mathrm{p}}+\frac{\mathrm{y}}{\frac{2 \mathrm{p}}{\sqrt{3}}}=1
\end{array}
\)
Comparing both : \(\mathrm{a}=2 \mathrm{p}, \mathrm{b}=\frac{2 \mathrm{p}}{\sqrt{3}}\)
Now area of \(\triangle \mathrm{OAB}=\frac{1}{2} \cdot \mathrm{ab}=\frac{98}{3} \cdot \sqrt{3}\)
\(
\begin{array}{l}
\frac{1}{2} \cdot 2 \mathrm{p} \cdot \frac{2 \mathrm{p}}{\sqrt{3}}=\frac{98}{3} \cdot \sqrt{3} \\
\mathrm{p}^2=49 \\
\mathrm{a}^2-\mathrm{b}^2=4 \mathrm{p}^2-\frac{4 \mathrm{p}^2}{3}=\frac{2}{3} 4 \mathrm{p}^2 \\
=\frac{8}{3} \cdot 49=\frac{392}{3}
\end{array}
\)

Hint

Here \(\mathrm{m}_{\mathrm{BH}} \times \mathrm{m}_{\mathrm{AC}}=-1\)
\(
\begin{array}{l}
\left(\frac{\beta-3}{\alpha-2}\right)\left(\frac{1}{-2}\right)=-1 \\
\beta-3=2 \alpha-4 \\
\beta=2 \alpha-1 \\
m_{A H} \times m_{B C}=-1
\end{array}
\)
\(
\begin{array}{l}
\left(\frac{\beta-2}{\alpha-1}\right)(-2)=-1 \\
2 \beta-4=\alpha-1 \\
2(2 \alpha-1)=\alpha+3 \\
3 \alpha=5
\end{array}
\)
\(
\begin{array}{l}
\alpha=\frac{5}{3}, \beta=\frac{7}{3} \Rightarrow \mathrm{H}\left(\frac{5}{3}, \frac{7}{3}\right) \\
\alpha+4 \beta=\frac{5}{3}+\frac{28}{3}=\frac{33}{3}=11 \\
\beta+4 \alpha=\frac{7}{3}+\frac{20}{3}=\frac{27}{3}=9 \\
\mathrm{x}^2-20 \mathrm{x}+99=0
\end{array}
\)

Hint

Equation of the pair of angle bisector for the homogenous equation \(\mathrm{ax}^2+2 \mathrm{hxy}+\mathrm{by}^2=0\) is given as
\(
\frac{x^2-y^2}{a-b}=\frac{x y}{h}
\)
Here \(\mathrm{a}=2, \mathrm{~h}=1 / 2\) & \(\mathrm{b}=-3\)
Equation will become
\(
\begin{array}{l}
\frac{x^2-y^2}{2-(-3)}=\frac{x y}{1 / 2} \\
x^2-y^2=10 x y \\
x^2-y^2-10 x y=0
\end{array}
\)

Hint

The straight lines \(l_1\) and \(l_2\) pass through the origin and trisect the line segment of the line \(L: 9 x+5 y=45\) between the axes, And \(m_1\) and \(m_2\) are the slopes of the lines \(l_1\) and \(l_2\), Now on plotting the diagram we get,

Given equation of line, \(L: 9 x+5 y=45\)
\(
\Rightarrow \frac{x}{5}+\frac{y}{9}=1
\)
Now using the section formula between point \(A(5,0) \& B(0,9)\) we get the value of point \(C \& D\) \(\Rightarrow C \equiv\left(\frac{10}{3}, 3\right)\) and \(D \equiv\left(\frac{5}{3}, 6\right)\)
Now finding the slope \(m_1 \& m_2\) we get,
\(
m_1=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10} \& m_2=\frac{6 \times 3}{5}=\frac{18}{5}
\)
So, equation of line \(y=\left(m_1+m_2\right) x\) will be,
\(
y=\left(\frac{9}{10}+\frac{36}{10}\right) x=\frac{9}{2} x
\)
So, intersection point with \(L\) will be,
\(
7 y=45 \Rightarrow y=\frac{45}{7}, x=\frac{10}{7}
\)
Hence, \(y-x=\frac{45-10}{7}=5\)

Hint

Given,
\(A(0,1), B(1,1)\) and \(C(1,0)\) be the mid-points of the sides of a triangle with incentre at the point \(D\), Now finding the vertices of the triangle by using midpoint formula and plotting the diagram we get, Now finding the incentre of the above triangle using the formula,
\(
I \equiv\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}\right)
\)
We get, Incentre \(D=\left(\frac{4}{4+2 \sqrt{2}}, \frac{4}{4+2 \sqrt{2}}\right)\)
Now given parabola \(y^2=4 a x\) passes through the incentre \(D\) we get,
\(
\begin{array}{l}
\left(\frac{4}{4+2 \sqrt{2}}\right)^2=4 a\left(\frac{4}{4+2 \sqrt{2}}\right) \\
\Rightarrow a=\frac{1}{4+2 \sqrt{2}}
\end{array}
\)
Now we know that focus of parabola is given by, focus \((a, 0)\)
Hence, focus will be \(\left(\frac{1}{4+2 \sqrt{2}}, 0\right)=\left(\frac{4-2 \sqrt{2}}{8}, 0\right)\)
Now comparing with \((\alpha+\beta \sqrt{2}, 0)\) we get,
\(
\alpha=\frac{4}{8}=\frac{1}{2}, \beta=-\frac{1}{4}
\)
Hence, \(\frac{\alpha}{\beta^2}=\frac{\frac{1}{2}}{\left(\frac{-1}{4}\right)^2}=\frac{16}{2}=8\)

Hint

Given,
The quadrilateral \(A B C D\) with vertices \(A(2,1,1), B(1,2,5), C(-2,-3,5)\) and \(D(1,-6,-7)\) Now finding, \(\overrightarrow{A B}=-\hat{i}+\hat{j}+4 \hat{k}\) and \(\overrightarrow{A D}=-\hat{i}-7 \hat{j}-8 \hat{k}\)
And plotting the diagram we get,

Now we know that,
Area of \(\triangle A B D\) is given by,
\(
\begin{array}{l}
=\frac{1}{2}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & 4 \\
-1 & -7 & 8
\end{array}\right| \\
=|10 \hat{i}-6 \hat{j}+4 \hat{k}|=2 \sqrt{38}
\end{array}
\)
Now finding, \(\overrightarrow{C B}=3 \hat{i}+5 \hat{j}\) and \(\overrightarrow{C D}=3 \hat{i}-3 \hat{j}-12 \hat{k}\)
So, Area of \(\triangle C B D=\frac{1}{2}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & -3 & -12 \\ 3 & 5 & 0\end{array}\right|\)
\(
\begin{array}{l}
=|6(5 \hat{i}-3 \hat{j}-2 \hat{k})| \\
=6 \sqrt{38}
\end{array}
\)
\(\therefore\) Area of quadrilateral \(A B C D=8 \sqrt{38}\) square units

Hint

Given,
A line segment \(A B\) of length \(\lambda\) moves such that the points \(A\) and \(B\) remain on the periphery of a circle of radius \(\lambda\), Now taking the points on the circle of radius \(\lambda\) as \(B\left(\lambda \cos \theta_1, \lambda \sin \theta_1\right) \& A\left(\lambda \cos \theta_2 \lambda \sin \theta_2\right)\) and taking the point \(P(h, k)\) which divides the line segment in \(A B\) of length \(\lambda\) in \(2: 3\)
Now plotting the diagram we get,

Now, let \(O\) be the origin and radius of circle is \(\lambda\) and \(A B=\lambda\) and using distance formula we get,
\(
\begin{array}{l}
A B=\lambda=\sqrt{\left(\lambda \cos \theta_1-\lambda \cos \theta_2\right)^2+\left(\lambda \sin \theta_1-\lambda \sin \theta_2\right)^2} \\
\Rightarrow 1=2-2 \cos \left(\theta_1-\theta_2\right) \\
\Rightarrow \cos \left(\theta_1-\theta_2\right)=\frac{1}{2}
\end{array}
\)
Now using section formula we get,
\(
h=\frac{2 \lambda \cos \theta_1+3 \lambda \cos \theta_2}{5} \text { and } k=\frac{2 \lambda \sin \theta_1+3 \lambda \sin \theta_2}{5}
\)
Now squaring and adding above two value we get,
\(
\begin{array}{l}
h^2+k^2=\frac{\lambda^2}{25}\left[4+9+12\left(\cos \left(\theta_1-\theta_2\right)\right)\right] \\
\Rightarrow h^2+k^2=\frac{\lambda^2}{25} \cdot 19
\end{array}
\)
Hence, Radius \(=\frac{\lambda}{5} \sqrt{19}\)

Hint

Given,
\(A\) be the point \((1,2)\) and \(B\) be any point on the curve \(x^2+y^2=16\),
And the centre of the locus of the point \(P\), which divides the line segment \(A B\) in the ratio \(3: 2\) is the point \(C(\alpha, \beta)\),
Now let the point on the circle \(x^2+y^2=16\) be \(B(4 \cos \theta, 4 \sin \theta)\)
Now using section formula in \(A(1,2)\) and \(B(4 \cos \theta, 4 \sin \theta)\) we get,
\(
\begin{array}{l}
P\left(\frac{12 \cos \theta+2}{5}, \frac{12 \sin \theta+4}{5}\right) \equiv(h, k) \\
\Rightarrow \cos \theta=\frac{5 h-2}{12} \& \sin \theta=\frac{5 k-4}{12}
\end{array}
\)
Now squaring and adding we get,
\(
\begin{array}{l}
\left(\frac{5 h-2}{12}\right)^2+\left(\frac{5 k-4}{12}\right)^2=1 \\
\Rightarrow\left(h-\frac{2}{5}\right)^2+\left(k-\frac{4}{5}\right)^2=\left(\frac{12}{5}\right)^2
\end{array}
\)
So, the centre of the locus is \(C\left(\frac{2}{5}, \frac{4}{5}\right)\)
Hence, by distance formula we get, \(A C=\sqrt{\left(\frac{3}{5}\right)^2+\left(\frac{6}{5}\right)^2}=\frac{3 \sqrt{5}}{5}\)

Hint

Given,
The equations of two adjacent sides of a parallelogram \(A B C D\) be \(2 x-3 y=-23\) and \(5 x+4 y=23\),
So, \(A B \equiv 2 x-3 y=-23\) and \(B C \equiv 5 x+4 y=23\)
Also given, \(A C \equiv 3 x+7 y=23\)
Solving the above lines we get, \(A(-4,5), B(-1,7), C(3,2)\)

We know that,
Diagonal of parallelogram have same midpoint,
So \(A C\) and \(B D\) have same mid-point and let point \(D\) be \((x, y)\),
So midpoint formula we get,
\(
\frac{x-1}{2}=\frac{-4+3}{2} \Rightarrow x=0 \text { and } \frac{y+7}{2}=\frac{2+5}{2} \Rightarrow y=0
\)
Hence, point \(D\) is \((0,0)\)
Now Equation of \(B D\) will be \(7 x+y=0\)
Now finding the distance of \(A(-4,5)\) from \(7 x+y=0\) we get,
\(
d=\left|\frac{7(-4)+5}{\sqrt{7^2+1^2}}\right|=\frac{23}{\sqrt{50}}
\)
Hence, \(50 d^2=23^2=529\)

Hint

Given,
\(\mathrm{R}\) be a rectangle given by the lines \(x=0, x=2, y=0\) and \(y=5\) And point \(\mathrm{A}(\alpha, 0)\) and \(\mathrm{B}(0, \beta), \alpha \in[0,2]\) and \(\beta \in[0,5]\),
Now plotting the diagram we get,

Now given that line segment \(A B\) divides the ratio of area in \(4: 1\), we get
\(
\begin{array}{l}
\frac{10-\frac{1}{2} \alpha \beta}{\frac{1}{2} \alpha \beta}=\frac{4}{1} \\
\Rightarrow 20-\alpha \beta=4 \alpha \beta \\
\Rightarrow \alpha \beta=4 \ldots \ldots(1)
\end{array}
\)
Now using midpoint formula we get,
\(
h=\frac{\alpha}{2} \& \beta=\frac{k}{2}
\)
Now using equation (1) we get,
\(
\Rightarrow 4 h k=4
\)
\(\Rightarrow x y=1\) which is a equation of rectangular hyperbola.

Hint

Given,
\(L_1: 3 y-2 x=3\) is angular bisector of \(L_2: x-y+1=0\) and \(L_3: \alpha x+\beta y+17=0\) Now finding point of intersection of \(L_1 \& L_2\) we get, \((0,1)\)
And point will lie on \(L_3\), so \(\alpha \times 0+\beta \times 1+17=0\)
\(
\Rightarrow \beta=-17
\)
Any point, say \(\left(\frac{-3}{2}, 0\right)\) on \(L_1\) should be equidistant from lines \(L_2 \& L_3\)
Now using the formula of distance of a point from the line we get,
\(
\Rightarrow\left|\frac{\frac{-3}{2}-0+1}{\sqrt{1^2+1^2}}\right|=\left|\frac{\frac{-3 \alpha}{2}+0+17}{\sqrt{\alpha^2+(-17)^2}}\right|
\)
\(
\Rightarrow(\alpha-7)(\alpha-17)=0
\)
Now, for \(\alpha=17, L_2 \& L_3\) coincides
So, \(\alpha=7\)
Now putting the value of \(\alpha \& \beta\) in given expression we get,
\(
\alpha^2+\beta^2-\alpha-\beta=(-17)^2+7^2-7+17=348
\)

Hint

The straight line is shown below with the intercepts.

\(
\begin{array}{l}
M\left(\frac{7}{2 \cos \theta}, \frac{7}{2 \sin \theta}\right) \equiv(h, k) \\
\therefore \cos \theta=\frac{7}{2 h} \quad \sin \theta=\frac{7}{2 k} \\
\because \theta \in\left(0, \frac{\pi}{2}\right) \\
\Rightarrow \frac{49}{4 h^2}+\frac{49}{4 k^2}=1 \quad \therefore x \text { and } y \text { are positive } \\
\therefore \text { Locus }=\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{49}
\end{array}
\)
Now \(\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) lies on curve
\(
\begin{array}{l}
\Rightarrow \frac{1}{\alpha^2}+\frac{9}{147}=\frac{4}{49} \\
\frac{1}{\alpha^2}=\frac{1}{49} \\
\therefore \alpha= \pm 7 \\
\therefore \alpha=7
\end{array}
\)
Hence this is the required option.

Hint

The given set of equations are:
\(
\begin{array}{l}
15 x-y=82 \\
6 x-5 y=-4 \\
9 x+4 y=17
\end{array}
\)
Now on solving the above equation by taking two at a time and plotting the diagram we get,

\(
(\alpha, \beta) \equiv\left(\frac{1+5+6}{3}, \frac{2-7+8}{3}\right) \equiv(4,1)
\)

Hence, \(\alpha+2 \beta=6\) and \(2 \alpha-\beta=7\)
The required equation is \((x-6)(x-7)=0\)
Equation \(x^2-13 x+42=0\)
Hence this is the required option.

Hint

The required diagram will be:

\(
\begin{array}{l}
\text { Equation of } A D: y+7=\frac{-5}{3}(x-3) \\
\Rightarrow 3 y+21=-5 x+15 \\
\Rightarrow 5 x+3 y+6=0 \ldots \ldots(\mathrm{i}) \\
\text { Equation of } B E: y-2=\frac{-1}{12}(x+1) \\
\Rightarrow 12 y-24=-x-1 \\
\Rightarrow x=23-12 y \ldots(i i)
\end{array}
\)
by (ii) \(115-60 y+3 y+6=0\)
\(
\begin{array}{l}
\Rightarrow 57 y=121 \\
y=\frac{121}{57}, x=23-12 \times \frac{121}{57} \\
\therefore 9 \alpha-6 \beta+60=9 \times 23-108 \times \frac{121}{57}-6 \times \frac{121}{57}+60 \\
=207-242+60=25
\end{array}
\)
Hence this is the required option.

Hint

\(
\begin{array}{l}
\mathrm{A}=\left(\frac{3}{\sqrt{\mathrm{a}}}, \sqrt{\mathrm{a}}\right) \\
\mathrm{B}=\left(\frac{-3}{\sqrt{\mathrm{a}}}, \sqrt{\mathrm{a}}\right) \\
\mathrm{C}=\left(-\frac{3}{\sqrt{\mathrm{a}}},-\sqrt{\mathrm{a}}\right)
\end{array}
\)
Area of ACD
\(
=\frac{1}{2}\left|\begin{array}{cc}
\frac{3}{\sqrt{a}} & \sqrt{a} \\
-\frac{3}{\sqrt{a}} & -\sqrt{a} \\
3 \cos \theta & a \sin \theta \\
\frac{3}{\sqrt{a}} & \sqrt{a}
\end{array}\right|
\)
\(
\begin{array}{l}
=\frac{1}{2} 6 \sqrt{\mathrm{a}}(\cos \theta-\sin \theta) \\
=3 \sqrt{\mathrm{a}}(\cos \theta-\sin \theta)
\end{array}
\)
max values of function is \(3 \sqrt{\mathrm{a}} \sqrt{2}\)
\(
\begin{array}{l}
3 \sqrt{a} \sqrt{2}=12 \\
2 a=16 \\
a=8
\end{array}
\)

Hint

\(
\frac{1}{2}\left|\begin{array}{lll}
\alpha & 0 & 1 \\
1 & \alpha & 1 \\
0 & \alpha & 1
\end{array}\right|= \pm 4
\)
\(
\alpha= \pm 8
\)

Now given points \((8,-8),(-8,8),(64, \beta)\)
\(
\text { OR }(-8,8),(8,-8),(64, \beta)
\)
are collinear \(\Rightarrow\) Slope \(=-1\).
\(
\beta=-64
\)

Hint

Altitude of equilateral triangle,
\(
\begin{array}{l}
\frac{\sqrt{3} l}{2}=\frac{5}{\sqrt{2}} \\
l=\frac{5 \sqrt{2}}{\sqrt{3}}
\end{array}
\)

Area of triangle
\(
=\frac{\sqrt{3}}{4} l^2=\frac{\sqrt{3}}{4} \cdot \frac{50}{3}=\frac{25}{2 \sqrt{3}}
\)

Hint

\(
\begin{array}{l}
2 x+y=4 \dots(1) \\
2 x+6 y=14 \dots(2)
\end{array}
\)
On solving eq(1) and eq(2)
\(
\begin{array}{l}
y=2, x=3 \\
B(1,2) \text { and } C(k, 4-2 k) \\
\text { Hence, } A B^2=A C^2 \\
52+(-1)^2=(6-k)^2+(-3+2 k)^2 \\
\Rightarrow 5 k^2-24 k+19=0 \\
(5 k-19)(k-1)=0 \\
\Rightarrow k=\frac{19}{5} \\
C\left(\frac{19}{5},-\frac{18}{5}\right) \Rightarrow \text { Centroid }(\alpha, \beta) \\
\alpha=\frac{6+1+\frac{19}{5}}{3} \\
\alpha=\frac{18}{5} \\
\beta=\frac{1+2-\frac{18}{5}}{3} \\
\beta=-\frac{1}{5}
\end{array}
\)
Now \(15(\alpha+\beta)=15\left(\frac{18}{5}+\left(-\frac{1}{5}\right)\right)\)
\(
\begin{array}{l}
=15 \times \frac{17}{5} \\
=51
\end{array}
\)

Hint

Equation of line \(A B\)
\(
\mathrm{y}=2 \mathrm{x}
\)

Slope of \(A B=2\)
Slope of given diameter \(=2\)
So the diameter is parallel to \(\mathrm{AB}\)
Distance between diameter and line \(A B\)
\(
=\left(\frac{4}{\sqrt{2^2+12}}\right)=\frac{4}{\sqrt{5}}
\)

Thus BC \(=2 \times \frac{4}{\sqrt{5}}=\frac{8}{\sqrt{5}}\)
\(
\begin{array}{l}
\qquad \mathrm{AB}=\sqrt{(1-3)^2+(2-6)^2}=\sqrt{20}=2 \sqrt{5} \\
\text { Area }=\mathrm{AB} \times \mathrm{BC}=\frac{8}{\sqrt{5}} \times 2 \sqrt{5}=16
\end{array}
\)

Hint

By observation we see that \(\mathrm{A}(\alpha, 0)\).
And \(\beta=\mathrm{y}\)-cordinate of \(\mathrm{R}\)
\(
=\frac{2 \times 4+1 \times 0}{2+1}=\frac{8}{3} \dots(1)
\)
Now \(P^{\prime}\) is image of \(P\) in \(y=0\) which will be \(\mathrm{P}^{\prime}(2,-3)\)
Equation of \(P^{\prime} Q\) is \((y+3)=\frac{4+3}{5-2}(x-2)\)
i.e. \(3 y+9=7 x-14\)
\(\mathrm{A} \equiv\left(\frac{23}{7}, 0\right)\) by solving with \(\mathrm{y}=0\)
\(
\therefore \alpha=\frac{23}{7} \dots(2)
\)
\(
\begin{array}{l}
\text { By (1), (2) } \\
\qquad 7 \alpha+3 \beta=23+8=31
\end{array}
\)

Hint

Points \(A\) lies on \(\mathrm{L}_2\)
\(
\mathrm{A}\left(\alpha, 4+\frac{4}{3} \alpha\right)
\)
Points B lies on \(\mathrm{L}_1\)
\(
\mathrm{B}\left(\beta, 2-\frac{2}{5} \beta\right)
\)
Points \(\mathrm{P}\) divides \(\mathrm{AB}\) internally in the ratio \(1: 3\)
\(
\begin{array}{l}
\Rightarrow \mathrm{P}(2,3)=\mathrm{P}\left(\frac{3 \alpha+\beta}{4}, \frac{3\left(4+\frac{4}{3} \alpha\right)+1\left(2-\frac{2}{5} \beta\right)}{4}\right) \\
\Rightarrow \alpha=\frac{3}{13}, \beta=\frac{95}{13}
\end{array}
\)
Point \(\mathrm{A}\left(\frac{3}{13}, \frac{56}{13}\right), \mathrm{B}\left(\frac{95}{13},-\frac{12}{13}\right)\)
Vertex \(\mathrm{C}\) of triangle is the point of intersection \(\mathrm{L}_1 \& \mathrm{~L}_2\)
\(
\Rightarrow \mathrm{C}\left(-\frac{15}{13}, \frac{32}{13}\right)
\)
\(
\begin{array}{l}
\text { area } \triangle \mathrm{ABC}=\frac{1}{2}\left\|\begin{array}{ccc}
\frac{3}{13} & \frac{56}{13} & 1 \\
\frac{95}{13} & -\frac{12}{13} & 1 \\
-\frac{15}{13} & \frac{32}{13} & 1
\end{array}\right\| \\
=\frac{1}{2 \times 13^3}\left\|\begin{array}{ccc}
3 & 56 & 13 \\
95 & -12 & 13 \\
-15 & 32 & 13
\end{array}\right\| \\
\text { area } \triangle \mathrm{ABC}=\frac{132}{13} \text { sq. units. }
\end{array}
\)

Hint

\(
\begin{array}{l}
\mathrm{M}_1=3 / 2 \quad \mathrm{M}_2=2 / \mathrm{x} \\
\tan \pi / 4=\left|\frac{3 / 2-2 / \mathrm{x}}{1+6 / 2 \mathrm{x}}\right|=1 \\
\Rightarrow \mathrm{x}_1=10, \quad \mathrm{x}_2=-2 / 5 \\
\Rightarrow \mathrm{AA}^{\prime}=52 / 5
\end{array}
\)

Hint

\(
\begin{array}{l}
\because B D \perp A C \\
\therefore \quad(m)_{B D}(m)_{A C}=-1 \\
\Rightarrow\left(\frac{1}{2}\right)\left(\frac{\frac{7}{3}-2 k+1}{\frac{7}{3}-k}\right)=-1
\end{array}
\)
\(
\begin{aligned}
k & =2 \\
\text { Also, }(m)_{C E}(m)_{A B} & =-1 \\
m & =2
\end{aligned}
\)
\(
\therefore A(1,1), B(2,3), C(3,2)
\)
So, coordinates of centroid is
\(
\begin{aligned}
G & =\left(\frac{1+2+3}{3}, \frac{1+3+2}{3}\right)=(2,2) \\
\therefore \quad O G & =\sqrt{2^2+2^2}=2 \sqrt{2}
\end{aligned}
\)

Hint

Let \(\mathrm{B}\left(\mathrm{x}_1, \mathrm{x}_1-2\right)\)
\(
\sqrt{\left(\mathrm{x}_1-4\right)^2+\left(\mathrm{x}_1-2-3\right)^2}=\frac{\sqrt{29}}{3}
\)
Squaring on both side
\(
\begin{array}{lll}
18 x_1^2-162 x_1+340 & =0 \\
x_1=\frac{51}{9} & \text { or } & x_1=\frac{10}{3} \\
y_1=\frac{33}{9} & \text { or } & y_1=\frac{4}{3}
\end{array}
\)
Option (c) will satisfy \(\left(\frac{10}{3}, \frac{4}{3}\right)\)

Hint

By observing origin and \(P\) lies in same region.
\(
\begin{array}{l}
L_1(0,0)>0 ; L_1(\alpha, \beta)>0 \Rightarrow 3 \alpha-4 \beta+12>01=\left|\frac{3 \alpha-4 \beta+12}{5}\right| \\
3 \alpha-4 \beta+12=5 \ldots \ldots(1)
\end{array}
\)
Similarly for \(L_2\)
\(
\begin{array}{l}
L_2(0,0)>0 ; L_2(\alpha, \beta)>0 \\
1=\left|\frac{8 \alpha+6 \beta+11}{10}\right| \Rightarrow 8 \alpha+6 \beta+11=10 \ldots \ldots .(2)
\end{array}
\)
Solving (1) and (2)
\(
\begin{array}{l}
\alpha=-\frac{23}{25} ; \beta=\frac{106}{100} \\
100(\alpha+\beta)=100\left(\frac{-92}{100}+\frac{106}{100}\right)=14
\end{array}
\)

Hint

\(
\begin{array}{l}
(x-1)^2+(y-2)^2+(x+2)^2+(y-1)^2=14 \\
\Rightarrow x^2+y^2+x-3 y-2=0 \\
\text { Put } x=0 \\
\Rightarrow y^2-3 y-2=0 \\
\Rightarrow y=\frac{3 \pm \sqrt{17}}{2}
\end{array}
\)
Put \(y=0\)
\(
\begin{array}{l}
\Rightarrow \mathrm{x}^2+\mathrm{x}-2=0 \\
(\mathrm{x}+2)(\mathrm{x}-1)=0 \\
\therefore \mathrm{A}(-2,0), \mathrm{B}(1,0), \mathrm{C}\left(0, \frac{3+\sqrt{17}}{2}\right), \mathrm{D}\left(0, \frac{3-\sqrt{17}}{2}\right) \\
\text { Area }=\frac{1}{2} \cdot 3 \cdot \sqrt{17}=\frac{3 \sqrt{17}}{2}
\end{array}
\)

Hint

Coordinates of \(\mathrm{A}(1,-2), \mathrm{B}\left(\frac{15 \mathrm{a}}{1-2 \mathrm{p}}, \frac{-30 \mathrm{a}}{1-2 \mathrm{p}}\right)\) and orthocentre \(\mathrm{H}(2, \mathrm{a})\)
Slope of \(\mathrm{AH}=\frac{a+2}{1}\)
Slope of BC \(=-\frac{1}{p}\)
\(
\text { Slope of } \mathrm{AH}=\mathrm{p}
\)
\(
a+2=p \dots(1)
\)
Slope of BH \(=-1\)
\(
31 a-2 a b=15 a+4 p-2 \dots(2)
\)
From (1) and (2)
\(
a=1 \& p=3
\)

Hint

Given \(\Delta_1=\frac{1}{2}\left|\begin{array}{lll}x & y & 1 \\ 1 & 1 & 1 \\ -4 & 3 & 1\end{array}\right|\)
\(\& \Delta_2=\frac{1}{2}\left|\begin{array}{ccc}1 & 1 & 1 \\ -4 & 3 & 1 \\ -2 & -5 & 1\end{array}\right|\)
Given \(\frac{\Delta_1}{\Delta_2}=\frac{4}{7} \Rightarrow \frac{-2 x-5 y+7}{36}=\frac{4}{7}\)
\(
\Rightarrow 14 x+35 y=-95 \dots(1)
\)

Equation of \(\mathrm{BC}\) is \(4 \mathrm{x}+\mathrm{y}=-13 \dots(2)\)
Solve equation (1) & (2)
Point \(P\left(\frac{-20}{7}, \frac{-11}{7}\right)\)
Here point \(Q\left(\frac{-1}{2}, 0\right) \& R\left(\frac{1}{2}, 0\right)\)
So Area of triangle \(\mathrm{AQR}=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\)

Hint

Equation of line
\(R P\) is \(x+y=5\)
By solving \(x-y=3\)
and \(x+y=5\), we get \(R(4,1)\)
The point \(R\) is mid point of \(A\) and \(C\).
So, point \(C\) is \((7,4)\).
Since line \(B C\) passes through point \(C(7,4)\). So, it must satisfy the equation of line \(B C\).
\(
\Rightarrow 7+4 p=39 \Rightarrow p=8
\)
Coordinates of point \(B\) are \(\left(\frac{-39}{15}, \frac{78}{15}\right)\)
\(
\text { Area of } \triangle A B C=\frac{1}{2}\left|\begin{array}{lll}
1 & -2 & 1 \\
7 & 4 & 1 \\
\frac{-39}{15} & \frac{78}{15} & 1
\end{array}\right|=32.4
\)

Hint

Let \(P(h, k)\) be the orthocentre of \(\triangle \mathrm{ABC}\)
Then
\(
h=\frac{\sin t+\cos t+a}{3}, k=\frac{-\cos t+\sin t+b}{3}
\)
(Orthocentre coincide with centroid)
\(
\begin{array}{l}
\therefore(3 h-a)^2+(3 k-b)^2=2 \\
\therefore\left(h-\frac{a}{3}\right)^2+\left(k-\frac{b}{3}\right)^2=\frac{2}{9}
\end{array}
\)
\(\because\) Orthocentre lies on circle with centre \(\left(1, \frac{1}{3}\right)\)
\(
\begin{array}{l}
\therefore a=3, b=1 \\
\therefore a^2-b^2=8
\end{array}
\)

Hint

\(
\begin{array}{l}
\mathrm{m}_{\mathrm{AC}} \longrightarrow \infty \\
\mathrm{m}_{\mathrm{PD}}=0 \\
\mathrm{D}\left(\frac{\mathrm{a}+\mathrm{a}}{2}, \frac{\mathrm{b}+3}{2}\right)
\end{array}
\)
\(
\begin{array}{l}
\mathrm{m}_{\mathrm{PD}}=0 \\
\frac{\mathrm{b}+3}{2}-1=0; \mathrm{~b}+3-2=0; \mathrm{~b}=-1
\end{array}
\)
\(
\begin{array}{l}
\mathrm{E}\left(\frac{\mathrm{b}+\mathrm{a}}{2}, \frac{5+\mathrm{b}}{2}\right)=\left(\frac{\mathrm{af}}{2}, 2\right) \\
\mathrm{m}_{\mathrm{CB}} \cdot \mathrm{m}_{\mathrm{EP}}=-1
\end{array}
\)
\(
\left(\frac{5-b}{b-a}\right)=\left(\frac{2-1}{\frac{a-1}{2}-1}\right)=-1
\)
\(
\begin{array}{l}
(a-5)(a+3)=0 \\
a=5 \text { or } a=-3; \text { Given } a b>0 \\
a(-1)>0
\end{array}
\)
\(
\begin{array}{l}
a<0 \\
a=-3 \text { Accept }
\end{array}
\)
AP line \(A(-3,3), \mathrm{P}(1,1)\)
\(
\begin{array}{l}
y-1=\left(\frac{3-1}{-3-1}\right)(x-1) ; -2 y+2=x-1
\end{array}
\)
\(
x+2 y=3 \dots(1)
\)
\(
\begin{array}{l}
\text { Line BC B }(-1,5), \quad C(-3,-1) \\
(y-5)=\frac{6}{2}(x+1) \\
y-5=3 x+3; y=3 x+8 \dots(2)
\end{array}
\)
\(
\text { Solving (1) & (2) }
\)
\(
x+y=\frac{-13+17}{7}=\frac{4}{7}
\)

Hint

\(
\begin{array}{l}
m_1 m_2=-1 \\
a^2+11 a+3\left(m_1^2+\frac{1}{m_1^2}\right)=220
\end{array}
\)
Eq. of \(A C\)
\(
\begin{array}{l}
\mathrm{AC}=(\cos \alpha-\sin \alpha)+(\sin \alpha+\cos \alpha) y=10 \\
\mathrm{BD}=(\sin \alpha-\cos \alpha) x+(\sin \alpha-\cos \alpha) y=0
\end{array}
\)
So, the point of intersection of the diagonal will be
\(
(5(\cos \alpha-\sin \alpha), 5(\cos \alpha+\sin \alpha)) \text {. }
\)
Therefore, the vertex opposite to the given vertex is \((0,0)\).
So, the diagonal length \(=10 \sqrt{2}\)
Side length \((a)=10\)
It is given that
\(
\begin{array}{l}
a^2+11 a+3\left(m_1^2+m_2^2\right)=220 \\
m_1^2+m_2^2=\frac{220-100-110}{3}=\frac{10}{3} \\
\text { and } m_1 m_2=-1
\end{array}
\)
Slopes of the sides are \(\tan \alpha\) and \(-\cot \alpha\)
\(
\begin{array}{l}
\tan ^2 \alpha=3 \text { or } \frac{1}{3} \\
72\left(\sin ^4 \alpha+\cos ^4 \alpha\right)+a^2-3 a+13 \\
=72 \cdot \frac{\tan ^4 \alpha+1}{\left(1+\tan ^2 \alpha\right)^2}+a^2-3 a+13=128
\end{array}
\)

Hint

\(\mathrm{A}(\alpha,-2): \mathrm{B}(\alpha, 6): \mathrm{C}\left(\frac{\alpha}{4},-2\right)\)
since \(A C\) is perpendicular to \(A B\).
So, \(\triangle \mathrm{ABC}\) is right angled at \(\mathrm{A}\).
\(
\begin{array}{l}
\text { Circumcentre }=\text { mid point of } \mathrm{BC} .=\left(\frac{5 \alpha}{8}, 2\right) \\
\therefore \frac{5 \alpha}{8}=5 \& \frac{\alpha}{4}=2 \\
\alpha=8
\end{array}
\)

\(
\text { Area }=\frac{1}{2}(6)(8)=24
\)

Perimeter \(=24\)
Circumradius \(=5\)
Inradius \(=\frac{\Delta}{\mathrm{s}}=\frac{24}{12}=2\)

Hint

\(
\begin{array}{l}
\frac{x}{a}+\frac{y}{b}=1 \\
\frac{h}{a}+\frac{k}{b}=1 \ldots(1) \\
\text { and } \frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{1}{4} \\
\therefore \frac{1}{a}+\frac{1}{b}=\frac{1}{2} \ldots(2)
\end{array}
\)
From (1) and (2)
Line passes through fixed point \(B(2,2)\)

Hint

Considering the points according to the given data:
\(
P(h, k), A(5,0) \text { and } B(-5,0)
\)
Given, \(P A=3 P B\)
Distance between the points \(P(h, k) \& A(5,0)\) is:
\(
P A=\sqrt{(h-5)^2+k^2}
\)
Similarly, the distance between the points \(P(h, k) \& B(-5,0)\)
\(
P B=\sqrt{(h+5)^2+k^2}
\)
Given
\(
\begin{array}{l}
P A=3 P B\\
P A^2=9 P B^2
\end{array}
\)
Euqating both sides we get,
\(
\begin{aligned}
(h-5)^2+k^2 & =9\left[(h+5)^2+k^2\right] \\
8 h^2+8 k^2+100 h+200 & =0 \\
x^2+y^2+\frac{25}{2} x+25 & =0
\end{aligned}
\)
Therefore, the locus of point \(P\) is \(x^2+y^2+\frac{25}{2} x+25=0 \ldots \ldots(i)\)
Finding the radius of the circle:
We know that a circle of the form \(x^2+y^2+2 g x+2 f y+c=0\) has a center \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\).
Therefore, from equation \((i)\) we have,
\(
\begin{array}{l}
g=\frac{25}{4}, f=0 \& c=25 \\
\text { Thus, Centre }=(-g,-f)=\left(-\frac{25}{4}, 0\right)
\end{array}
\)
\(
r=\sqrt{\left(\frac{25}{4}\right)^2-25}
\)
\(
4 r^2=56.25
\)

Hint

Step 1.Find the image of the point
We know that if \(B\left({ }^x, y\right)\) is the image of a point \(P(p, q)\) with respect to a line
\(a x+b y+c=0\) then
\(
\frac{x-p}{a}=\frac{y-q}{b}=\frac{-2(a p+b q+c)}{a^2+b^2}
\)
Image of the point \((3,5)\) in the line \(x-y+1=0\), is
\(
\begin{array}{l}
\frac{x-3}{1}=\frac{y-5}{-1}=\frac{-2(3-5+1)}{2}=1 \\
\therefore x=4, y=4
\end{array}
\)
Step2: Finding answer
Image of the point \((3,5)\) is \((4,4)\)
If the image lies on the circle \((x-2)^2+(y-4)^2=4\), then the image must satisfy the equation of the circle
Since,
\(
\begin{array}{l}
(4-2)^2+(4-4)^2 \\
=2^2+0 \\
=4
\end{array}
\)

Hint

Given equations of the straight lines are as follows,
\(
\begin{array}{l}
x-y=0 \dots(i) \\
x+2 y=3 \dots(ii) \\
2 x+y=6 \dots(iii)
\end{array}
\)
Step 1: Solving Equation (ii) – Equation (i).
\(
\begin{array}{l}
(x+2 y)-(x-y)=3-0 \Rightarrow 3 y=3 \Rightarrow y=1
\end{array}
\)
Substitute \(y\) with 1 in equation (i).
\(
\begin{array}{l}
x-y=0 \Rightarrow \quad x-1=0 \Rightarrow \quad x=1
\end{array}
\)
Thus, the point is \(A(1,1)\).
Step 2: Solve the equation (iii) -equation (ii).
Equation (iii) \(\times 2-\) Equation (ii).
\(
\begin{array}{l}
2 \times(2 x+y)-(x+2 y)=(2 \times 6)-3 \\
\quad \Rightarrow 3 x=9 \Rightarrow x=3
\end{array}
\)
Substitute \(x\) with 3 in equation (ii).
\(
\begin{array}{l}
x+2 y=3 \Rightarrow 3+2 y=3 \Rightarrow \quad y=0
\end{array}
\)
Thus, the point is \(B(3,0)\).
Step 3: Solve the equation (iii) -equation \((i)\)
Equation (iii) + Equation (i).
\(
\begin{array}{l}
(2 x+y)+(x-y)=6+0 \\
\Rightarrow 3 x=6 \\
\Rightarrow x=2
\end{array}
\)
Substitute \(x\) with 2 in equation (i).
\(
\begin{array}{l}
x-y=0 \\
\Rightarrow 2-y=0 \\
\Rightarrow \quad y=2
\end{array}
\)
Thus, the point is \(C(2,2)\).
Step 4: Check which type of triangle formed.
Apply the distance formula, \(\sqrt{\left(\mathbf{x}_1-\mathbf{x}_2\right)^2+\left(\mathbf{y}_1-\mathbf{y}_2\right)^2}\). \(A B=\sqrt{(1-3)^2+(1-0)^2}\) units \(=\sqrt{4+1}\) units \(=\sqrt{5}\) units. \(B C=\sqrt{(3-2)^2+(0-2)^2}\) units \(=\sqrt{1+4}\) units \(=\sqrt{5}\) units. \(C A=\sqrt{(2-1)^2+(2-1)^2}\) units \(=\sqrt{1+1}\) units \(=\sqrt{2}\) units.
So, the given triangle has \(A B=B C \neq C A\), thus the triangle is an isosceles triangle.

Hint

Step 1: Derive a relation between the general point on the given circle and the mid-point.
In the question, an equation of the circle \(x^2+y^2=1\) and a point \((3,2)\) is given.
We know that the standard equation of a circle is given by \(\left(x-x_1\right)^2+\left(y-y_1\right)^2=r^2\), where \((x, y)\) is the general point on the circle, the centre of the circle is at \(\left(x_1, y_1\right)\) and \(r\) is the radius of the circle.
Assume that, the mid-point of the line segment from the point \((3,2)\) to a point on the circle, \(x^2+y^2=1\) is \((h, k)\).
So, \(\left(\frac{x+3}{2}, \frac{y+2}{2}\right)=(h, k)\)
Now, compute \(x\) and \(y\) is terms of \(h\) and \(k\) respectively.
Therefore,
\(
\begin{array}{l}
\frac{x+3}{2}=h \\
\Rightarrow \quad x=2 h-3
\end{array}
\)
\(
\begin{array}{l}
\frac{y+2}{2}=k \\
\Rightarrow \quad y=2 k-2
\end{array}
\)
Therefore, \((x, y)=(2 h-3,2 k-2)\).
Step 2: Drive an equation for the locus of the mid-point.
Since, \((x, y)\) lies on the given circle. So, it satisfy the given equation of the circle.
So,
\(
\begin{aligned}
& (2 h-3)^2+(2 k-2)^2=1 \\
\Rightarrow & \frac{(2 h-3)^2}{4}+\frac{(2 k-2)^2}{4}=\frac{1}{4} \\
\Rightarrow \quad & \left(h-\frac{3}{2}\right)^2+(k-1)^2=\left(\frac{1}{2}\right)^2
\end{aligned}
\)
Which represents the locus of the mid-point of the line segment from the point \((3,2)\) to a point on the circle, \(x^2+y^2=1\).
Since, the radius of the locus of the mid-point of the line segment from the point \((3,2)\) to a point on the circle, \(x^2+y^2=1\) is \(\frac{1}{2}\).
Therefore, the value of \(r\) is \(\frac{1}{2}\).

Hint

\(
\begin{array}{l}
\mathrm{P} \equiv\left(\mathrm{x}_1, \mathrm{mx}_1\right) \\
\mathrm{Q} \equiv\left(\mathrm{x}_2, \mathrm{mx}_2\right) \\
A_1=\frac{1}{2}\left|\begin{array}{ccc}
3 & 4 & 1 \\
2 & 0 & 1 \\
-1 & 1 & 1
\end{array}\right|=\frac{13}{2} \\
A_2=\frac{1}{2}\left|\begin{array}{ccc}
x_1 & m x_1 & 1 \\
x_2 & m x_2 & 1 \\
2 & 0 & 1
\end{array}\right| \\
\mathrm{A}_2=\frac{1}{2}\left|2\left(\mathrm{mx}_1-\mathrm{mx}_2\right)\right|=\mathrm{m}\left|\mathrm{x}_1-\mathrm{x}_2\right| \\
A_1=3 A_2 \Rightarrow \frac{13}{2}=3 m\left|x_1-x_2\right|
\end{array}
\)
\(
\begin{array}{l}
A C: x+3 y=2 \\
B C: y=4 x-8 \\
P: x+3 y=2 \& y=m x \Rightarrow x_1=\frac{2}{1+3 m} \\
Q: y=4 x-8 \& y=m x \Rightarrow x_2=\frac{8}{4-m} \\
\left|x_1-x_2\right|=\left|\frac{2}{1+3 m}-\frac{8}{4-m}\right| \\
=\left|\frac{-26 m}{(1+3 m)(4-m)}\right|=\frac{26 m}{(3 m+1)|m-4|} \\
=\frac{26 m}{(3 m+1)(4-m)} \\
\left|x_1-x_2\right|=\frac{13}{6 m} \\
\frac{26 m}{(3 m+1)(4-m)}=\frac{13}{6 m}
\end{array}
\)
\(
\begin{array}{l}
\Rightarrow 12 \mathrm{~m}^2=-(3 \mathrm{~m}+1)(\mathrm{m}-4) \\
\Rightarrow 12 \mathrm{~m}^2=-\left(3 \mathrm{~m}^2-11 \mathrm{~m}-4\right) \\
\Rightarrow 15 \mathrm{~m}^2-11 \mathrm{~m}-4=0 \\
\Rightarrow 15 \mathrm{~m}^2-15 \mathrm{~m}+4 \mathrm{~m}-4=0 \\
\Rightarrow(15 \mathrm{~m}+4)(\mathrm{m}-1)=0 \\
\Rightarrow \mathrm{m}=1
\end{array}
\)

Hint

Perpendicular bisector of \(\mathrm{PR}\) is
\(
2 x-y+2=0 \dots(i)
\)

Mid-points of \(P Q\) is \(M \equiv(1,1)\)
Equation of perpendicular bisector of \(P Q\) is
\(
\begin{array}{l}
y-1=-\left(\frac{4+2}{-2-4}\right)(x-1) \\
\Rightarrow x=y \dots(ii)
\end{array}
\)
Solving eqn (i) and eqn (ii), therefore, circumcentre is point of intersection of the two perpendicular bisectors i.e., \((-2,-2)\)

Hint

Tangent to circle \(3 x+4 y=25\)

\(
\begin{aligned}
\mathrm{OP}+\mathrm{OQ} & +\mathrm{OR}=25 \\
\text { Incentre } & =\left(\frac{\frac{25}{4} \times \frac{25}{3}}{25}, \frac{\frac{25}{4} \times \frac{25}{3}}{25}\right) \\
& =\left(\frac{25}{12}, \frac{25}{12}\right)
\end{aligned}
\)
\(
\therefore \mathrm{r}^2=2\left(\frac{25}{12}\right)^2=2 \times \frac{625}{144}=\frac{625}{72}
\)

Hint

Since orthocentre and circumcentre both lies on y-axis.
\(\Rightarrow\) Centroid also lies on \(y\)-axis.
\(
\begin{array}{l}
\Rightarrow \sum \cos \alpha=0 \\
\cos \alpha+\cos \beta+\cos \gamma=0 \\
\Rightarrow \cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma=3 \cos \alpha \cos \beta \cos \gamma \\
\therefore \frac{\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma}{\cos \alpha \cos \beta \cos \gamma} \\
=\frac{4\left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)-3(\cos \alpha+\cos \beta+\cos \gamma)}{\cos \alpha \cos \beta \cos \gamma}=12
\end{array}
\)
then, \(\left(\frac{\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma}{\cos \alpha \cos \beta \cos \gamma}\right)^2=144\)

Hint

Given equation of the lines are \(3 x+4 y=9\) and \(y=m x+1\)
\(
\begin{array}{l}
\therefore \quad 3 x+4 y=9 \\
\Rightarrow 3 x+4(m x+1)=9 \\
\Rightarrow \quad x(3+4 m)=9-4 \\
\Rightarrow \quad x=\frac{5}{3+4 m} \\
\end{array}
\)
Clearly \(x\) will attain integer values when \(3+4 m=5,-5,1,-1\)
\(
\Rightarrow \mathrm{m}=\frac{1}{2},-2,-\frac{1}{2},-1
\)
so, number of integral values of \(m\) is 2.

Hint

\(
\begin{array}{l}
y=m x+c \\
3=m+c \\
\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right| \\
=6 m+\sqrt{2}=m-3 \sqrt{2}
\end{array}
\)
\(
\begin{array}{l}
=\sin =-4 \sqrt{2} \rightarrow \mathrm{m}=\frac{-4 \sqrt{2}}{5} \\
=6 m-\sqrt{2}=m-3 \sqrt{2} \\
=7 \mathrm{~m}-2 \sqrt{2} \rightarrow \mathrm{m}=\frac{2 \sqrt{2}}{7}
\end{array}
\)
According to options take \(\mathrm{m}=\frac{-4 \sqrt{2}}{5}\)
So \(y=\frac{-4 \sqrt{2} x}{5}+\frac{3+4 \sqrt{2}}{5}\)
\(
4 \sqrt{2} \mathrm{x}+5 \mathrm{y}-(15+4 \sqrt{2})=0
\)

Hint

\(
\begin{array}{l}
r=O M=\frac{3}{\sqrt{2}} \\
\& \sin 30^{\circ}=\frac{1}{2}=\frac{r}{R} \Rightarrow R=\frac{6}{\sqrt{2}} \\
\therefore r+R=\frac{9}{\sqrt{2}}
\end{array}
\)

Hint

\(
\begin{array}{l}
(\sqrt{50})^2=(\sqrt{45})^2+(\sqrt{5})^2 \\
\angle \mathrm{B}=90^{\circ}
\end{array}
\)
\(
\angle \mathrm{B}=90^{\circ}
\)
Circum-center \(=\left(\frac{1}{2}, \frac{11}{2}\right)\)
Mid point of \(\mathrm{BC}=\left(2, \frac{17}{2}\right)\)
Line : \(\left(y-\frac{11}{2}\right)=2\left(x-\frac{1}{2}\right) \Rightarrow y=2 x+\frac{9}{2}\)
Passing though \(\left(0, \frac{\alpha}{2}\right)\)
\(
\frac{\alpha}{2}=\frac{9}{2} \Rightarrow \alpha=9
\)

Hint

\(
\tan \theta=-\frac{2}{3}
\)
Using symmetric from of line
\(
\begin{array}{l}
P, Q:(2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta) \\
\left(2 \pm \sqrt{13} \cdot\left(-\frac{3}{\sqrt{13}}\right), 3 \pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right) \\
(-1,5) \&(5,1)
\end{array}
\)

Hint

Image of \(\mathrm{A}(\mathrm{a}, \mathrm{b})\) along \(\mathrm{y}=\mathrm{x}\) is \(\mathrm{B}(\mathrm{b}, \mathrm{a})\).
Translating it 2 units it becomes \(\mathrm{C}(\mathrm{b}+2, \mathrm{a})\).

Now, applying rotation theorem
\(
\begin{array}{l}
-\frac{1}{\sqrt{2}}+\frac{7}{\sqrt{2}} i=((b+2)+a i)\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \\
-\frac{1}{\sqrt{2}}+\frac{7}{\sqrt{2}} i=\left(\frac{b+2}{\sqrt{2}}-\frac{a}{\sqrt{2}}\right)+i\left(\frac{b+2}{\sqrt{2}}+\frac{a}{\sqrt{2}}\right) \\
\Rightarrow \mathrm{b}-\mathrm{a}+2=-1 \ldots \ldots \text { (i) }
\end{array}
\)
and \(b+2+a=7 \dots(ii)\)
\(
\Rightarrow a=4 ; b=1
\)
\(
\Rightarrow 2 \mathrm{a}+\mathrm{b}=9
\)

Hint

The sides are \(4 x+5 y=0 \dots(1)\)
and \(7 \mathrm{x}+2 \mathrm{y}=0 \dots(2)\)

Equation of the diagonal,
\(
11 x+7 y=9 \dots(3)
\)
Solving (1) and (3) co-ordinates of D are (5/3, -4/3)
Solving (2) and (3) co-ordinates of B are \((-2 / 3,7 / 3)\)
diagonals of parallelogram intersect at middle
let middle point of \(M=B, D\)
\(
\Rightarrow\left(\frac{\frac{5}{3}-\frac{2}{3}}{2}, \frac{\frac{-4}{3}+\frac{7}{3}}{2}\right)=\left(\frac{1}{2}, \frac{1}{2}\right)
\)
equation of diagonal \(\mathrm{AC}\)
\(
\begin{array}{l}
\Rightarrow(y-0)=\frac{\frac{1}{\alpha}-0}{\frac{1}{\alpha}-0}(\pi-0)\\
y=x
\end{array}
\)
dlagonal AC passes through \((2,2)\).

Hint

Intersection point of give lines are \((1,2),(7,5),(2,3)\)

\(
\begin{array}{l}
\Delta=\frac{1}{2}\left|\begin{array}{lll}
1 & 2 & 1 \\
7 & 5 & 1 \\
2 & 3 & 1
\end{array}\right| \\
=\frac{1}{2}[1(5-3)-2(7-2)+1(21-10)] \\
=\frac{1}{2}[2-10+11] \\
\Delta \mathrm{DEF}=\frac{1}{2}(3)=\frac{3}{2} \\
\Delta \mathrm{ABC}=4 \Delta \mathrm{DEF}=4\left(\frac{3}{2}\right)=6
\end{array}
\)

Hint

\(
\frac{1}{2}\left|\begin{array}{lll}
5 & 6 & 1 \\
3 & 2 & 1 \\
\alpha & \beta & 1
\end{array}\right|=12
\)
\(
\begin{array}{l}
4 \alpha-2 \beta= \pm 24+8 \\
\Rightarrow 4 \alpha-2 \beta=+24+8 \Rightarrow 2 \alpha-\beta=16 \\
2 x-y-16=0 \ldots(1) \\
\Rightarrow 4 \alpha-2 \beta=-24+8 \Rightarrow 2 \alpha-\beta=-8 \\
2 x-y+8=0 \ldots(2)
\end{array}
\)
perpendicular distance of (1) from \((0,0)\)
\(
\left|\frac{0-0-16}{\sqrt{5}}\right|=\frac{16}{\sqrt{5}}
\)
perpendicular distance of \((2)\) from \((0,0)\) is
\(
\left|\frac{0-0+8}{\sqrt{5}}\right|=\frac{8}{\sqrt{5}}
\)
Required length \(=\) minimum perpendicular distance from origin
\(
=\min \left\{\frac{16}{\sqrt{5}}, \frac{8}{\sqrt{5}}\right\}=\frac{8}{\sqrt{5}}
\)

Hint

Let \(L_1=x \operatorname{cosec} \alpha-y \sec \alpha=k \cot 2 \alpha\).
Then \(\frac{x}{\sin \alpha}-\frac{y}{\cos \alpha}=\frac{k \cos 2 \alpha}{\sin 2 \alpha}\)
\(\Rightarrow x \cos \alpha-y \sin \alpha=\frac{k}{2} \cos 2 \alpha\)
Perpendicular distance from \((00)\) is
\(
\begin{array}{l}
\mathrm{p}=\left|\frac{0-0-\frac{k}{2} \cos 2 \alpha}{\sqrt{\cos ^2 \alpha+\sin ^2 \alpha}}\right| \\
\Rightarrow \mathrm{p}=\left|\frac{k}{2} \cos 2 \alpha\right|
\end{array}
\)
\(
\Rightarrow 2 p=|k \cos 2 \alpha| \ldots \text { (i) }
\)
\(
\mathrm{L}_2=\mathrm{x} \sin \alpha+\mathrm{y} \cos \alpha=\mathrm{k} \sin 2 \alpha
\)
Perpendicular distance from \((0,0)\) is
\(
\begin{array}{l}
\mathrm{q}=\left|\frac{0+0-k \sin 2 \alpha}{\sqrt{\sin ^2 \alpha+\cos ^2 \alpha}}\right| \\
\Rightarrow \mathrm{q}=|\mathrm{k} \sin 2 \alpha| \ldots \text { (ii) }
\end{array}
\)
Hence \(4 p^2+q^2=k^3 \ldots(\) From (i) and (ii))

Hint

\(
\frac{1}{2}\left|\begin{array}{ccc}
a & 0 & 1 \\
b & 2 b+1 & 1 \\
0 & b & 1
\end{array}\right|=1
\)
\(
\begin{array}{l}
\Rightarrow\left|\begin{array}{ccc}
\mathrm{a} & 0 & 1 \\
\mathrm{~b} & 2 \mathrm{~b}+1 & 1 \\
0 & \mathrm{~b} & 1
\end{array}\right|= \pm 2 \\
\Rightarrow a(2 b+1-b)-0+1\left(b^2-0\right)= \pm 2 \\
\Rightarrow \mathrm{a}=\frac{ \pm 2-\mathrm{b}^2}{\mathrm{~b}+1} \\
\therefore \mathrm{a}=\frac{2-\mathrm{b}^2}{\mathrm{~b}+1} \text { and } \mathrm{a}=\frac{-2-\mathrm{b}^2}{\mathrm{~b}+1}
\end{array}
\)
sum of possible values of ‘ \(a\) ‘ is
\(
=\frac{-2 b^2}{a+1} \text { Ans. }
\)

Hint

Let \(A(0,6)\) and \(B(2 t, 0)\)
Then, mid point of \(A B\) is \(M=(t, 3)\) and slope of \(A B\) is
\(
m_{A B}=\frac{-6}{2 t}=-\frac{3}{t}
\)

Perpendicular bisector of \(A B\) is
\(
(y-3)=\frac{t}{3}(x-t)
\)
So, \(C=\left(0,3-\frac{t^2}{3}\right)\)
Let \(\mathrm{P}\) be \((\mathrm{h}, \mathrm{k})\)
\(
\begin{array}{l}
\mathrm{h}=\frac{\mathrm{t}}{2} ; \mathrm{k}=\left(3-\frac{\mathrm{t}^2}{6}\right) \\
\Rightarrow \mathrm{k}=3-\frac{4 \mathrm{~h}^2}{6} \Rightarrow 2 \mathrm{x}^2+3 \mathrm{y}-9=0
\end{array}
\)

Hint

\(
\therefore \quad M\left(\frac{a-3}{2}, \frac{b+1}{2}\right) \text { lies on } 2 x+y-3=0
\)
\(
\Rightarrow 2 \mathrm{a}+\mathrm{b}=11 \dots(i)
\)
\(\because\) C lies on \(7 x-4 y=1\)
\(
\Rightarrow 7 \mathrm{a}-4 \mathrm{~b}=1 \dots(ii)
\)
\(\therefore\) by (i) and (ii) : \(\mathrm{a}=3, \mathrm{~b}=5\)
\(
\begin{array}{l}
\Rightarrow \mathrm{C}(3,5) \\
\therefore \mathrm{m}_{A C}=2 / 3
\end{array}
\)
Also, \(\mathrm{m}_{\mathrm{CD}}=7 / 4\)
\(
\begin{array}{l}
\Rightarrow \tan \frac{\theta}{2}=\left|\frac{\frac{2}{3}-\frac{4}{4}}{1+\frac{14}{12}}\right| \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \\
\Rightarrow \tan \theta=\frac{2 \cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}
\end{array}
\)

Hint

\(
\begin{array}{l}
P \text { is the centroid which is } \equiv\left(\frac{1+6+\frac{3}{2}}{3}, \frac{0+2+6}{3}\right) \\
P \equiv\left(\frac{17}{6}, \frac{8}{3}\right) \\
Q \equiv\left(-\frac{7}{6},-\frac{1}{3}\right) \\
\therefore P Q=\sqrt{(4)^2+(3)^2}=5
\end{array}
\)

Hint

\(
\begin{array}{l}
\text { slope of } P Q=\frac{k-\alpha}{h-2 \alpha}=-1 \\
\Rightarrow k-\alpha=-h+2 \alpha \\
\Rightarrow \alpha=\frac{h+k}{3} \quad \ldots \text { (1) }
\end{array}
\)
Also \(2 h=2 \alpha+\beta\)
\(
2 k=\alpha+\beta
\)
\(
\begin{array}{l}
\Rightarrow 2 h=\alpha+2 k \\
\Rightarrow \alpha=2 h-2 k \dots(2)
\end{array}
\)
from (1) and (2)
\(
\begin{array}{l}
\frac{h+k}{3}=2(h-k) \\
\text { so locus } 6 x-6 y=x+y \Rightarrow 5 x=7 y
\end{array}
\)

Hint

\(
D=\frac{1}{2}\left|\begin{array}{ccc}
0 & 2 & 1 \\
1 & -1 & 1 \\
x^{\prime} & y^{\prime} & 1
\end{array}\right|
\)
\(
\begin{array}{l}
-2\left(1-x^{\prime}\right)+\left(y^{\prime}+x^{\prime}\right)= \pm 10 \\
-2+2 x^{\prime}+y^{\prime}+x^{\prime}= \pm 10 \\
3 x^{\prime}+y^{\prime}=12 \text { or } 3 x^{\prime}+y^{\prime}=-8 \\
\lambda=3,-2
\end{array}
\)

Hint

\(
\begin{array}{l}
x^2+2 x y-3 y^2=0 \\
x^2+3 x y-x y-3 y^2=0 \\
(x-y)(x+3 y)=0 \\
x-y=0 \quad x+3 y=0 \\
(2,2) \text { satisfy } x-y=0 \\
\text { Normal } \\
x+y=\lambda \\
\lambda=4 \\
\text { Hence } x+y=0
\end{array}
\)
perpendicular distance from origin \(=\left|\frac{0+0-4}{\sqrt{2}}\right|=2 \sqrt{2}\)

Hint

\(
\text { Centroid of } \Delta=(2,2)
\)
line passing through intersection of \(x+3 y-1=0\) and \(3 x-y+1=0\), be given by \((x+3 y-1)+\lambda(3 x-y+1)=0\)
If passes through \((2,2)\) \(\Rightarrow 7+5 \lambda=0 \Rightarrow \lambda=-7 / 5\)
\(\therefore\) Required line is \(8 x-11 y+6=0\)
\(\because(-9,-6)\) satisfies this equation.

Hint

Given that both points \((1,2) \&(\sin \theta, \cos \theta)\) lie on same side of the line \(x+y-1=0\)
\(
\begin{array}{l}
\text { Let } f(x, y)=x+y-1 \\
\because \quad f(1,2) \cdot f(\sin \theta, \cos \theta)>0 \\
\Rightarrow \quad 2[\sin \theta+\cos \theta-1]>0 \\
\Rightarrow \quad \sin \theta+\cos \theta>1 \\
\Rightarrow \quad \sin \left(\theta+\frac{\pi}{4}\right)>\frac{1}{\sqrt{2}} \\
\Rightarrow \quad \theta+\frac{\pi}{4} \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) \\
\Rightarrow \quad \theta \in\left(0, \frac{\pi}{2}\right)
\end{array}
\)

Hint

Let \(H(x, y)\) be ortho center of the triangle and let \(A D\) and \(B E\) be lines passing through the orthocenter.
Thus
\(
\begin{aligned}
& m_{A H} \cdot m_{B C}=-1 \\
\Rightarrow & \left(\frac{y-7}{x+1}\right)\left(\frac{-5-1}{5+7}\right)=-1 \\
\Rightarrow & 2 x-y+9=0 \ldots \ldots \ldots(i)
\end{aligned}
\)
and
\(
\begin{array}{c}
m_{B H} \cdot m_{A C}=-1 \\
\Rightarrow\left(\frac{y-1}{x+7}\right)\left(\frac{7-5}{-1-5}\right)=-1 \\
\Rightarrow \quad x-2 y+9=0 \ldots \ldots(i i)
\end{array}
\)
Solving equations (i) and (ii) we get
\(
(x, y)=(-3,3)
\)

Hint

\(
\begin{array}{l}
\left(\frac{K-2}{h-1}\right)\left(\frac{1-2}{3-1}\right)=-1 \\
\Rightarrow K=2 h \ldots(1) \\
\sqrt{5}|h-1|=10 \\
\because[\Delta A B C]=5 \sqrt{5} \\
\Rightarrow \frac{1}{2}(\sqrt{5}) \sqrt{(h-1)^2+(K-2)^2}=5 \sqrt{5} \ldots \text { (2) } \\
\Rightarrow h=2 \sqrt{5}+1(h>0)
\end{array}
\)

Hint

Mid point of line segment \(\mathrm{PQ}\) is \(\left(\frac{\mathrm{k}+1}{2}, \frac{7}{2}\right)\)
Slope of PQ is \(\frac{1}{1-\mathrm{k}}\)
So equation of perpendicular bisector of \(P Q\) is
\(
y-\frac{7}{2}=(k-1)\left[x-\frac{k+1}{2}\right]
\)
\(\therefore\) Line passes through \((0,-4)\); then
\(
-\frac{15}{2}=-\frac{\left(k^2-1\right)}{2} \Rightarrow k= \pm 4
\)

Hint

\(
\begin{array}{l}
L_1: 2 x-y+3=0 \\
L_1: 4 x-2 y+\alpha=0 \\
L_1: 6 x-3 y+\beta=0
\end{array}
\)
Distance between \(L_1\) and \(L_2\)
\(
\begin{array}{l}
\left|\frac{\alpha-6}{2 \sqrt{5}}\right|=\frac{1}{\sqrt{5}} \Rightarrow|\alpha-6|=2 \\
\Rightarrow \alpha=4,8
\end{array}
\)
Distance between \(L_1\) and \(L_3\left|\frac{\beta-9}{3 \sqrt{5}}\right|=\frac{2}{\sqrt{5}} \Rightarrow|\beta-9|=6\)
\(
\Rightarrow \beta=15,3
\)
Sum of all values \(=4+8+15+3=30\)

Hint

Equation of incident line AP is
\(
\begin{array}{c}
y-2 \sqrt{3}=\sqrt{3}(x-2) \\
\sqrt{3} x-y=0
\end{array}
\)

Image of \(P\) w.r.t. line \(x=1\)
is point \(Q=(0,2 \sqrt{3})\)
Equation of reflected Ray \(A B\) :
\(
\begin{array}{l}
y-\sqrt{3}=\frac{2 \sqrt{3}-\sqrt{3}}{0-1}(x-1) \\
\sqrt{3} x+y=2 \sqrt{3}
\end{array}
\)
\(\therefore\) Point \((3,-\sqrt{3})\) lies on line \(A B\)

Hint

\(
\begin{array}{l}
\text { Line is } \frac{x}{3}+y=1 \\
x+3 y-3=0
\end{array}
\)

Let image be \((\alpha, \beta)\)
Hence, \(\frac{\alpha+1}{1}=\frac{\beta+y}{3}=-\frac{2(-1-12-3)}{10}\)
\(
\begin{array}{l}
\alpha+1=\frac{\beta+4}{3}=\frac{16}{5} \\
\Rightarrow \alpha=\frac{11}{5}, \beta=\frac{28}{5}
\end{array}
\)

Hint

Given, the equation of the set of lines is,
\(
p x+q y+r=0 \dots(i)
\)
such that
\(
3 p+2 q+4 r=0 \dots(ii)
\)
Dividing (ii) by 4 and subtrating from (i)
\(
\Rightarrow\left(x-\frac{3}{4}\right) p+\left(y-\frac{1}{2}\right) q=0
\)
The lines pass through the point \(\left(\frac{3}{4^{\prime}} \frac{1}{2}\right)\).
Hence, the lines are concurrent at the point \(\left(\frac{3}{4}, \frac{1}{2}\right)\).

Hint

One of the possible \(\triangle O A B\) is \(A(a, 0)\) and \(B(0, b)\). Area of \(\triangle O A B=\frac{1}{2}|a b|\).
\(
\begin{aligned}
\therefore \quad & |a b|=100 \\
& |a||b|=100
\end{aligned}
\)
So the possible pairs \((|a|,|b|)\) are \((1,100),(100,1),(2,50),(50,2),(4,25),(25,4)\),
\(
(5,20),(20,5),(10,10)
\)

No. of possible pairs \(=9\)

As we know, one pair can draw 4 triangles,
so the total number of possible triangles \(=9 \times 4=36\)

Hint

Let equation of side \(A B: 3 x-2 y+6=0 \dots(1)\)
and equation of side \(B C\) : \(4 x+5 y-20=0 \dots(2)\)
\(\mathrm{O}\) is the orthocentre.

Line \(C E\) is perpendicular to \(A B\)
So, slope of line CE is \(-\frac{2}{3}\)
\(\therefore\) Equation of line \(C E\) is
\(
\begin{array}{l}
(y-1)=-\frac{2}{3}(x-1) \\
\Rightarrow 2 x+3 y-5=0 \dots(3)
\end{array}
\)
From Equations (2) and (3), we obtain ordinate C,
\(
x=\frac{35}{2}, y=-10
\)
Now, slope of line \(B C\) is \(-\frac{4}{5}\)
\(\therefore\) Slope of line AF is \(\frac{5}{4}\)
Thus, Equation of line \(A F\) is
\(
\begin{array}{l}
y-1=\frac{5}{4}(x-1) \\
\Rightarrow 5 x-4 y-1=0 \dots(4)
\end{array}
\)
From Equations (1) and (4), we obtain ordinate A, \(x=-13, y=-\frac{66}{4}\)
\(
\therefore \text { Equation of } A B \text { is } \frac{y+10}{x-\frac{35}{2}}=\frac{-10+\frac{66}{4}}{\frac{35}{2}+13}
\)
\(
26 x-122 y-1675=0
\)

Hint

Consider the lines
\(
\begin{array}{l}
x=a y+b, z=c y+d \\
\Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c} \\
\therefore \vec{b}_1=a \hat{i}+\hat{j}+c \hat{k}
\end{array}
\)
Now,
\(
\begin{array}{l}
x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime} \\
\Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=\frac{y-d^{\prime}}{c^{\prime}}=\frac{z}{1} \\
\therefore \vec{b}_2=a^{\prime} \hat{i}+c^{\prime} \hat{j}+\hat{k}
\end{array}
\)
The two lines are perpendicular.
\(
\begin{array}{l}
\text { So, } \vec{b}_1 \cdot \vec{b}_2=0 \\
\Rightarrow a a^{\prime}+c^{\prime}+c=0
\end{array}
\)

Hint

Let the centroid of the \(\triangle P Q R\) is \(C(\alpha, \beta)\)
\(
\begin{array}{l}
\therefore a=\frac{h+1+3}{3} \Rightarrow h=3 a-4 \\
\beta=\frac{4+\mathrm{k}-2}{3} \Rightarrow k=3 \beta-2
\end{array}
\)
\(P(h, k)\) passes through the line \(2 x-3 y+4=0\)
\(
\begin{array}{l}
\therefore 2(3 a-4)-3(3 \beta-2)+4=0 \\
\Rightarrow 6 a-9 \beta+2=0
\end{array}
\)
Hence, the locus is \(6 x-9 y+2=0\)
\(
\therefore \text { slope }=\frac{6}{9}=\frac{2}{3}
\)

Hint

Given equation of line is
\(
3 x+4 y-24=0
\)
For intersection with \(X\)-axis put \(y=0\)
\(
\Rightarrow 3 x-24=0 \Rightarrow x=8
\)

For intersection with \(Y\)-axis, put \(x=0\)
\(
\Rightarrow \quad 4 y-24=0 \Rightarrow y=6
\)
Let \(A B=c=8^2+6^2=10\)
\(O B=a=6\) and \(O A=b=8\)
Also, let incentre is \((h k)\), then
\(
\begin{array}{l}
h=\frac{a x_1+b x_2+c x_3}{a+b+c} \\
=\frac{6 \times 8+8 \times 0+10 \times 0}{6+8+10}=\frac{48}{24}=2 \\
\text { and } k=\frac{a y_1+b y_2+c y_3}{a+b+c} \\
\left(\text { here, } y_1=0, y_2=6, y_3=0\right) \\
=\frac{6 \times 0+8 \times 6+10 \times 0}{6+8+10}=\frac{48}{24}=2 \\
\therefore \text { incentre is }(2,2)
\end{array}
\)

Hint

Let the point \(C\) is \((x, y)\)
\(
\begin{array}{l}
x+y=3 \quad \cdots(1) \\
\text { and } x-y+3=0 \quad \cdots(2)
\end{array}
\)
The point of intersection from the equation (1) and (2) is point \(A:(0,3)\) So, if point \(C\) is \((x, y)\) then, \(\frac{x+0}{2}=2 \Rightarrow x=4\) and \(\frac{y+3}{2}=4 \Rightarrow y=5\)
As line \(B C\) and \(A D\) are parrallel
\(\therefore\) slope of line \(B C=1\)
\(\therefore\) equation of \(\mathrm{BC}\) is
\(
\begin{array}{l}
y-5=1(x-4) \\
\Rightarrow x-y+1=0 \dots(3)
\end{array}
\)
As line \(C D\) and \(A B\) are parrallel
\(\therefore\) slope of line \(C D=-1\)
\(\therefore\) equation of \(C D\) is
\(
\begin{array}{l}
y-5=-1(x-4) \\
\Rightarrow x+y=9 \dots(4)
\end{array}
\)
\(
\text { Hence, from (1) and ( } 3 \text { ), point } B \text { is }(1,2) \text { and from ( } 2 \text { ) and (4), point } D \text { is }(3,6 \text { ). }
\)

Hint

\(
\begin{array}{l}
m_{A D} \cdot m_{B C}=-1 \Rightarrow m_{A H} \cdot m_{B C}=-1 \\
\Rightarrow\left(\frac{y-0}{x-0}\right) \cdot\left(\frac{3-2}{4-0}\right)=-1 \\
\Rightarrow 4 x+y=0 \quad \cdots(1)
\end{array}
\)

Now,
\(
\begin{array}{l}
m_{B E} \cdot m_{A C}=-1 \Rightarrow m_{B H} \cdot m_{A C}=-1 \\
\Rightarrow\left(\frac{3-0}{4-0}\right) \cdot\left(\frac{y-2}{x-0}\right)=-1 \\
\Rightarrow 4 x+3 y=6 \quad \cdots(2)
\end{array}
\)
from the equations (1) and (2), we get \(x=-\frac{3}{4}, y=3\)
Hence, Point \((x, y)\) lies in second quadrant.

Hint

(1)Given line \(A B\) is \(x+2 y=1\)
Slope of line \(A B\)
\(
=-\frac{1}{2}
\)

Coordinates of \(A\) and \(B\) are
\(
A=(1,0), B=\left(0, \frac{1}{2}\right)
\)
Co-ordinates of \(\mathrm{C}\)
\(
=\left(\frac{1+0}{2}, \frac{0+\frac{1}{2}}{2}\right)=\left(\frac{1}{2}, \frac{1}{4}\right)
\)

Slope of OC
\(
=\frac{0-\frac{1}{4}}{0-\frac{1}{2}}=\frac{1}{2}
\)
\(\therefore\) Slope of tangent \(=-2\)
So, the equation of tangent is
\(
2 \mathrm{x}+\mathrm{y}=0
\)

Sum of distance from \(A(1,0)\) and \(B\left(0, \frac{1}{2}\right)\) of tangent
\(
\begin{array}{l}
=\left|\frac{2}{\sqrt{2^2+1^2}}\right|+\left|\frac{\frac{1}{2}}{\sqrt{2^2+1^2}}\right| \\
=\frac{2}{\sqrt{5}}+\frac{\frac{1}{2}}{\sqrt{5}}=\frac{5}{2 \sqrt{5}}=\frac{\sqrt{5}}{2}
\end{array}
\)

Hint

Diagonals of a parallelogram bisect each other.
\(
\begin{array}{l}
\therefore\left(\frac{x_1+1}{2}, \frac{y_1+2}{2}\right)=\left(\frac{3+2}{2}, \frac{5+4}{2}\right) \\
\Rightarrow \frac{x_1+1}{2}=\frac{5}{2} \Rightarrow x_1=4 \\
\text { and } \frac{y_1+2}{2}=\frac{9}{2} \Rightarrow y_1=7
\end{array}
\)
\(\therefore\) Coordinates of \(D\) is \((4,7)\)
\(\therefore\) Equation of \(A D\) is
\(
\frac{y-2}{x-1}=\frac{7-2}{4-1}
\)
\(
\Rightarrow \frac{y-2}{x-1}=\frac{5}{3}
\)
\(
\begin{array}{l}
\Rightarrow 3 y-6=5 x-5 \\
\Rightarrow 5 x-3 y+1=0
\end{array}
\)

Hint

Let \(\vec{v}_1\) and \(\vec{v}_2\) be the vectors perpendicular to the plane \(O P Q\) and \(P Q R\) respectively,
\(
\begin{array}{l}
\vec{v}_1=\overrightarrow{P Q} \times \overrightarrow{O Q}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
2 & 1 & 3
\end{array}\right| \\
=5 \hat{i}-\hat{j}-3 \hat{k}
\end{array}
\)
\(
\begin{array}{l}
\vec{v}_2=\overrightarrow{P Q} \times \overrightarrow{P R}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 2 \\
-2 & -1 & 1
\end{array}\right| \\
=\hat{i}-5 \hat{j}-3 \hat{k}
\end{array}
\)
\(
\begin{array}{l}
\because \quad \cos \theta=\frac{\vec{v}_1 \cdot \vec{v}_2}{\left|\vec{v}_1\right|\left|\vec{v}_2\right|}=\frac{5+5+9}{25+1+9}=\frac{19}{35} \\
\therefore \quad \theta=\cos ^{-1}\left(\frac{19}{35}\right)
\end{array}
\)

Hint

Slope of straight line. \(2 x-3 y+17=0\) is \(m_1=\frac{2}{3}\)
Let slope of line passing through \((7,17)\) and \((15, \beta)\) be \(m_2\).
\(
\Rightarrow m_2=\frac{\beta-17}{15-7}
\)

Since, both the lines are perpendicular
\(
\begin{array}{l}
\text { So, } m_1 \cdot m_2=-1 \\
\Rightarrow \frac{2}{3} \cdot \frac{\beta-17}{8}=-1 \\
\Rightarrow \beta-17=-12 \\
\Rightarrow \beta=5
\end{array}
\)

Hint

Let the line be \(\frac{x}{a}+\frac{y}{b}=1\)
\(
\begin{array}{l}
(-3,4)=\left(\frac{a}{2}, \frac{b}{2}\right) \\
a=-6, b=8
\end{array}
\)
equation of line is \(4 x-3 y+24=0\)

Hint

A point which is equidistant from both the axes lies on either \(y=x\) and \(y=-x\).
Since, point lies on the line \(3 x+5 y=15\)
Then the required point
\(
\begin{array}{l}
3 x+5 y=15 \\
x+y=0 \\
\hline x=-\frac{15}{2}
\end{array}
\)
\(
y=\frac{15}{2} \Rightarrow(x, y)=\left(-\frac{15}{2}, \frac{15}{2}\right)\left\{2^{\text {nd }} \text { quadrant }\right\}
\)
or
\(
\begin{array}{c}
3 x+5 y=15 \\
x-y=0 \\
\hline x=\frac{15}{8}
\end{array}
\)
\(
y=\frac{15}{8} \Rightarrow(x, y)=\left(\frac{15}{8}, \frac{15}{8}\right)\left\{1^{\text {st }} \text { quadrant }\right\}
\)
Hence, the required point lies in \(1^{\text {st }}\) and \(2^{\text {nd }}\) quadrant.

Hint

\(
O A+A P+O P=4
\)
Let point \(P(h, k)\)
\(
\begin{array}{l}
\because \\
\text { So, } O P+A P=3 \quad O A=1 \\
\Rightarrow \sqrt{h^2+k^2}+\sqrt{h^2+(k-1)^2}=3
\end{array}
\)
\(
\begin{array}{l}
\Rightarrow h^2+(k-1)^2=9+h^2+k^2-6 \sqrt{h^2+k^2} \\
\Rightarrow 6 \sqrt{h^2+k^2}=2 k+8 \\
\Rightarrow 9 h^2+8 k^2-8 k-16=0
\end{array}
\)
Hence, locus of point \(P\) is
\(
9 x^2+8 y^2-8 y-16=0
\)

Hint

\((h, k),(1,2)\) and \((-3,4)\) are collinear
\(
\begin{array}{l}
\therefore\left|\begin{array}{lll}
h & k & 1 \\
1 & 2 & 1 \\
-3 & 4 & 1
\end{array}\right|=0 \Rightarrow-2 h-4 k+10=0 \\
\Rightarrow h+2 k=5 \dots(i) \\
\text { Now, } m_{L_1}=\frac{4-2}{-3-1}=-\frac{1}{2} \Rightarrow m_{L_2}=2 \quad\left[\because L_1 \perp L_2\right]
\end{array}
\)

By the given points \((h, k)\) and \((4,3)\),
\(
\begin{array}{l}
m_{L_2}=\frac{k-3}{h-4} \Rightarrow \frac{k-3}{h-4}=2 \Rightarrow k-3=2 h-8 \\
2 h-k=5 \dots(ii)
\end{array}
\)

From (i) and (ii)
\(
h=3, k=1 \Rightarrow \frac{k}{h}=\frac{1}{3}
\)

Hint

Since point at 4 units from \(P(2,3)\) will be \(\mathrm{A}(4 \cos \theta+2,4 \sin (\theta+3)\) and this point will satisfy the equation of line \(x+y=7\)
\(
\Rightarrow \cos \theta+\sin \theta=\frac{1}{2}
\)

On squaring
\(
\begin{array}{l}
\Rightarrow \sin 2 \theta-\frac{3}{4} \Rightarrow \frac{2 \tan \theta}{1+\tan ^2 \theta}=-\frac{3}{4} \\
\Rightarrow 3 \tan ^2 \theta+8 \tan \theta+3=0 \\
\Rightarrow \tan \theta=\frac{-8 \pm 2 \sqrt{7}}{6} \quad \text { (ignoring – ve sign) } \\
\Rightarrow \tan \theta=\frac{-8+2 \sqrt{7}}{6}=\frac{1-\sqrt{7}}{1+\sqrt{7}}
\end{array}
\)

Hint

\(\because\) two lines are perpendicular \(\Rightarrow m_1 m_2=-1\)
\(
\begin{array}{l}
\Rightarrow\left(\frac{-1}{a-1}\right)\left(\frac{-2}{a^2}\right)=-1 \\
\Rightarrow 2=a^2(1-a) \Rightarrow a^3-a^2+2=0 \\
\Rightarrow(a+1)\left(a^2+2 a+2\right)=0 \Rightarrow a=-1
\end{array}
\)

Hence equations of lines are \(x-2 y=1\) and \(2 x+y=1\)
\(\therefore\) intersection point is \(\left(\frac{3}{5}, \frac{-1}{5}\right)\)
Now, distance from origin \(=\sqrt{\frac{9}{25}+\frac{1}{25}}=\sqrt{\frac{10}{25}}=\sqrt{\frac{2}{5}}\)

Hint

Let the vertices of the rectangle are \(\mathrm{A}(-8,5), \mathrm{B}(6,5), \mathrm{C}(6, \alpha)\) and \(\mathrm{D}(-8, \beta)\). So, coordinates of mid point of diameter \(\mathrm{AC}\) are \(\left(\frac{-8+6}{2}, \frac{\alpha+5}{2}\right)=\left(-1, \frac{\alpha+5}{2}\right)\)
It lies on the line, \(3 y=x+7\)
\(
\begin{array}{l}
\Rightarrow 3\left(\frac{\alpha+5}{2}\right)=-1+7 \\
\Rightarrow \alpha+5=\frac{6 \times 2}{3} \Rightarrow \alpha=-1 \\
\text { Now, } \mathrm{AB}=\sqrt{(-8-6)^2+0}=14 \\
\mathrm{BC}=\sqrt{0+(5+1)^2}=6
\end{array}
\)

Sides of rectangle are 14 and 6 .
Hence, area of rectangle \(=14 \times 6=84\) sq. units.

Hint

Equation of line parallel to the line
\(
4 x-3 y+2=0 \text { is } 4 x-3 y+\lambda=0 \dots(1)
\)

Distance of line from the origin is
\(
\begin{array}{l}
\left|\frac{4(0)-3(0)+\mid \lambda}{\sqrt{4^2+(-3)^3}}\right|=\frac{3}{5} \\
\Rightarrow \lambda= \pm 3
\end{array}
\)
putting the value of \(\lambda\) in equation (1)
\(\therefore\) the required lines are \(4 x-3 y+3=0\) and \(4 x-3 y-3=0\)
Point \(\left(-\frac{1}{4}, \frac{2}{3}\right)\) satisfies the line \(4 x-3 y+3=0\)

Hint

Consider the equation,
\(
\begin{array}{l}
y=\sin x \cdot \sin (x+2)-\sin ^2(x+1) \\
=\frac{1}{2} \cos (-2)-\frac{\cos (2 x+2)}{2}-\left[\frac{1-\cos (2 x+2)}{2}\right] \\
=\frac{(\cos 2)-1}{2}=-\sin ^2 1
\end{array}
\)

\(
\text { By the graph } y \text { lies in III and IV quadrant. }
\)

Hint

As \(E\) is the midpoint of \(A B\), we get
\(
\begin{array}{l}
\frac{\mathrm{x}_2+1}{2}=-1, \quad \frac{\mathrm{y}_2+2}{2}=1 \\
\Rightarrow \mathrm{x}_2=-3, \mathrm{y}_2=0
\end{array}
\)

Point \(B\) is \((-3,0)\)

Similarly, point \(C\) is \((3,4)\)
So, centroid of the triangle is
\(
\left(\frac{1-3+3}{3}, \frac{2+0+4}{3}\right)=\left(\frac{1}{3}, 2\right)
\)

Hint

Given \(O A=p=4\)
\(x+y=0\) makes an angle of \(135^{\circ}\) with the positive \(x\)-axis because \(\tan \theta=\) slope of line \(x+y=0 \Rightarrow y=-x\) comparing with \(y=m x\), we get slope \(m=-1\)
\(
\begin{array}{l}
\Rightarrow \tan \theta_1=-1 \\
\Rightarrow \tan \theta_1=\tan 135^{\circ} \\
\therefore \theta_1=135^{\circ}
\end{array}
\)

Thus, line OA makes an angle of \(135-60=75^{\circ}\) with the \(x\)-axis.
Equation of the line \(L\) in perpendicular form is \(x \cos \theta+y \sin \theta=p\)
\(
\begin{array}{l}
\Rightarrow \mathrm{x} \cos 75^{\circ}+\mathrm{y} \sin 75^{\circ}=\mathrm{p} \\
\Rightarrow {x}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)+{y}\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)=4
\end{array}
\)
\(\therefore x(\sqrt{3}-1)+y(\sqrt{3}+1)=8 \sqrt{2}\) is the required equation.

Hint

Finding the locus of \(R\)
The general equation of straight line through point is \(\frac{x}{h}+\frac{y}{k}=1\)
where \(h\) is the \(x\)-intercept and \(k\) is the \(y\)-intercept.
The line passes through \((2,3)\), so we get
\(
\begin{aligned}
\Rightarrow & \frac{2}{h}+\frac{3}{k}=1 \\
& 2 k+3 h=h k
\end{aligned}
\)

Now substitute \(h\) as \(x\) and \(k\) as \(y\), we get the locus of \(R\) as
\(
\Rightarrow 3 x+2 y=x y
\)