Quiz

Hint

\(
\text { The slope of the line through the points }(3,-2) \text { and }(3,4) \text { is }
\)
\(
m=\frac{4-(-2)}{3-3}=\frac{6}{0}, \text { which is not defined. }
\)

Hint

Let \(\mathrm{ABC}\) be the given equilateral triangle with side \(2 a\).
Accordingly, \(\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=2 \mathrm{a}\)
Assume that base \(B C\) (the base of the triangle) lies along the \(y\)-axis such that the mid-point of \(\mathrm{BC}\) is at the origin. i.e., \(\mathrm{BO}=\mathrm{OC}=\mathrm{a}\), where \(\mathrm{O}\) is the origin.
Now, it is clear that the coordinates of point \(\mathrm{C}(\mathrm{O}, \mathrm{a})\), while the coordinates of point \(B(0,-a)\).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
Hence, vertex A lies on the \(y\)-axis.
On applying Pythagoras theorem to DAOC, we obtain
\(
\begin{aligned}
& \mathrm{AC}^2=O A^2+O C^2 \\
& O A^2=A C^2-O C^2 =(2 a)^2+(a)^2 \\
& O A^2=4 a^2-a^2 =3 a^2 \\
& O A=\pm \sqrt{3} a
\end{aligned}
\)
Therefore, coordinates of point \(A(\pm \sqrt{3} a, 0)\)
Thus, the vertices of the given equilateral triangle are \((0, a),(0,-a)\) and \((\sqrt{3} a, 0)\) or \((0, a),(0,-a)\) and \((-\sqrt{3} a, 0)\)

Hint

(i) When \(P Q\) is parallel to \(y\)-axis \(x_1=x_2\)
In this case distance between \(P\) and \(Q\)
\(
\begin{aligned}
& =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& =\sqrt{\left(y_2-y_1\right)^2}=\left|y_2-y_1\right|
\end{aligned}
\)

(ii) Since \(P Q\) is parallel to \(x\)-axis
\(
y_1=y_2 .
\)
\(
\text { Distance } \begin{aligned}
\mathrm{PQ} & =\sqrt{\left(x_2-x_1^2+\left(y_2-y_1\right)^2\right.} \\
& =\sqrt{\left(x_2-x_1\right)^2+\left(y_1-y_1\right)^2} \\
& =\sqrt{\left(x_2-x_1\right)^2+0} \\
& =\sqrt{\left(x_2-x_1\right)^2} \\
& =\pm\left(x_2-x_1\right) \\
& =\left|x_2-x_1\right|
\end{aligned}
\)

Hint

Let \((a, 0)\) be the point on the \(X\) – axis that is equidistant from the points \((7\),
\(6)\) and \((3,4)\).
Using distance formula,
\(
\begin{aligned}
& \sqrt{(7-a)^2+(6-0)^2}=\sqrt{(3-a)^2+(4-0)^2} \\
& \Rightarrow \sqrt{49+a^2-14 a+36}=\sqrt{9+a^2-6 a+16} \\
& \Rightarrow \sqrt{a^2-14 a+85}=\sqrt{a^2-6 a+25}
\end{aligned}
\)
On squaring on both sides, we obtain
\(
\begin{aligned}
& a^2-14 a+85=a^2-6 a+25 \\
& \Rightarrow-14 a+6 a=25-85 \\
& \Rightarrow-8 a=60 \\
& \Rightarrow a=60 / 8 \\
& \Rightarrow a=15 / 2
\end{aligned}
\)
Thus, the required point on the \(x\)-axis is \((15 / 2,0)\)

Hint

The coordinates of the mid-point of the line segment joining the points \(P(0,-4)\) and \(B(8,0)\) are \([(0+8) / 2,(-4+0) / 2]=(4,-2)\)

It is known that the slope \((\mathrm{m})\) of a non-vertical line passing through the points \(\left(x_1, y_1\right)\) and \(\left(x_2, y_2\right)\) is given by \(m=\left(y_2-y_1\right) /\left(x_2-x_1\right), x_2 \neq x_1\).
Therefore, the slope of the line passing through \((0,0)\) and \((4,-2)\) is \((-2-0) /(4-0)=-2 / 4=-1 / 2\)
Hence, the required slope of the line is \(-1 / 2\)

Hint

The vertices of the given triangle are \(A(4,4)\), B \((3,5)\), and \(C(-1,-1)\). It is known that the slope \((m)\) of a non-vertical line passing through the points \(\left(x_1, y_1\right)\) and \(\left(x_2, y_2\right)\) is given by \(m=\frac{y_2-y_1}{x_2-x_1}, x_2 \neq x_1\).
\(\therefore\) Slope of \(\mathrm{AB}\left(m_1\right)=\frac{5-4}{3-4}=-1\)
Slope of \(B C\left(m_2\right)=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of CA \(\left(m_3\right)=\frac{4+1}{4+1}=\frac{5}{5}=1\)
It is observed that \(m_1 m_3=-1\)
This shows that line segments \(A B\) and \(C A\) are perpendicular to each other i.e., the given triangle is right-angled at \(A(4,4)\).
Thus, the points \((4,4),(3,5)\), and \((-1,-1)\) are the vertices of a right-angled triangle.

Hint

The line makes angle of \(30^{\circ}\) from positive \(y\) axis Hence it makes angle of \(\left(90^{\circ}+30^{\circ}\right)\) from positive \(x\) axis i.e. \(120^{\circ}\) from positive \(x\) axis
Hence \(\theta=120^{\circ}\)
\(
\begin{aligned}
\text { Slope }=m & =\tan \theta \\
& =\tan 120^{\circ}
\end{aligned}
\)
\(
\begin{aligned}
& =\tan \left(180-60^{\circ}\right) \\
& =-\tan 60^{\circ} \quad\left(\tan \left(180^{\circ}-\theta\right)=-\tan \theta\right) \\
& =-\sqrt{3}
\end{aligned}
\)
Hence, Slope of line \(=-\sqrt{3}\)

Hint

If points \(A(x,-1), B(2,1)\), and \(C(4,5)\) are collinear, then Slope of \(A B=\) Slope of \(B C\)
\(
\begin{aligned}
& \Rightarrow \frac{1-(-1)}{2-x}=\frac{5-1}{4-2} \\
& \Rightarrow \frac{1+1}{2-x}=\frac{4}{2} \\
& \Rightarrow \frac{2}{2-x}=2 \\
& \Rightarrow 2=4-2 x \\
& \Rightarrow 2 x=2 \\
& \Rightarrow x=1
\end{aligned}
\)
Thus, the required value of \(x\) is 1.

Hint

Let the vertices be \(A(-2,-1), B(4,0), C(3,3), D(-3,2)\)
Now \(\mathrm{m}_{\mathrm{AB}}=\frac{0+1}{4+2}=\frac{1}{6}\)
\(
\begin{aligned}
& m_{B C}=\frac{3-0}{3-4}=-3 \\
& m_{A D}=\frac{2+1}{-3-3}=\frac{-1}{-6}=\frac{1}{6} \\
& m_{A D}=\frac{2+1}{-3+2}=-3
\end{aligned}
\)
Now \(m_{A B}=m_{C D} \Rightarrow A B \| C D\)
and \(\mathrm{m}_{B C}=\mathrm{m}_{\mathrm{AD}} \Rightarrow \mathrm{BC} \| \mathrm{AD}\)
Hence \(A B C D\) is a parallelogram.

Hint

The slope of the line joining the points \((3,-1)\) and \((4,-2)\) is
\(
\begin{aligned}
& m=(-2-(-1)) /(4-3) \\
& =(-2+1) / 1 \\
& =-1
\end{aligned}
\)
If the inclination of the line joining the points \((3,-1)\) and \((4,-2)\) is \(\theta\), then its slope is \(\tan \theta\). Then,
\(
\begin{aligned}
& \tan \theta=-1 \\
& \theta=\left(90^{\circ}+45^{\circ}\right) \\
& \theta=135^{\circ}
\end{aligned}
\)
Thus, the angle between the \(x\)-axis and the line joining the points \((3,-1)\) and \((4,-2)\) is \(135^{\circ}\)

Hint

Let \(m_1\) and \(m\) be the slopes of the two given lines such that \(m_1=2 m\).
We know that if \(\theta\) is the angle between the lines \(l_1\) and \(l_2\) with slopes \(m_1\) and \(m_2\), then
\(
\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right| .
\)
It is given that the tangent of the angle between the two lines is \(\frac{1}{3}\)
\(
\begin{aligned}
& \therefore \frac{1}{3}=\left|\frac{m-2 m}{1+(2 m) \cdot m}\right| \\
& \Rightarrow \frac{1}{3}=\left|\frac{-m}{1+2 m^2}\right| \\
& \Rightarrow \frac{1}{3}=\frac{-m}{1+2 m^2} \text { or } \frac{1}{3}=-\left(\frac{-m}{1+2 m^2}\right)=\frac{m}{1+2 m^2}
\end{aligned}
\)
Case I
\(
\begin{aligned}
& \Rightarrow \frac{1}{3}=\frac{-m}{1+2 m^2} \\
& \Rightarrow 2 m^2+3 m+1=0 \\
& \Rightarrow(m+1)(2 m+1)=0 \\
& \Rightarrow m=-1 \text { or } m=-\frac{1}{2}
\end{aligned}
\)
If \(m=-1\), then the slopes of the lines are \(-1\) and \(-2\).
If \(m=-\frac{1}{2}\), then the slopes of the lines are \(-\frac{1}{2}\) and \(-1\).
Case II
\(
\begin{aligned}
& \frac{1}{3}=\frac{m}{1+2 m^2} \\
& \Rightarrow 2 m^2+1=3 m \\
& \Rightarrow 2 m^2-3 m+1=0 \\
& \Rightarrow(m-1)(2 m-1)=0 \\
& \Rightarrow m=1 \text { or } m=\frac{1}{2}
\end{aligned}
\)
If \(m=1\), then the slopes of the lines are 1 and 2 .
If \(m=\frac{1}{2}\), then the slopes of the lines are \(\frac{1}{2}\) and 1
Hence, the slopes of the lines are \(-1\) and \(-2\) or \(-\frac{1}{2}\) and \(-1\) or 1 and 2 or \(\frac{1}{2}\) and 1

Hint

We know that slope of line passing through \(\left(x_1, y_1\right)\) and \(\left(x_2, y_2\right)\) is
\(
\mathrm{m}=\frac{y_2-y_1}{x_2-x_1}
\)
So, Slope of line passing through \(\left(x_1, y_1\right)\) and \((h, k)\) is
Here \(x_1=x_1, y_1=y_1\)
\(
x_2=h, y_2=k
\)
Putting values
\(
\begin{aligned}
& m=\frac{k-y_1}{h-x_1} \\
& m\left(h-x_1\right)=k-y_1 \\
& k-y_1=m\left(h-x_1\right)
\end{aligned}
\)

Hint

If the points \(\mathrm{A}(\mathrm{h}, 0), \mathrm{B}(\mathrm{a}, \mathrm{b})\) and \(\mathrm{C}(0, \mathrm{k})\) lie on line, then,
Slope of \(A B=\) slope of \(B C\)
\(
\begin{aligned}
& \Rightarrow \frac{b-0}{a-h}=\frac{k-b}{0-a} \\
& \Rightarrow \frac{b}{a-h}=\frac{k-b}{-a} \\
& \Rightarrow-a b=(k-b)(a-h) \\
& \Rightarrow-a b=k a-k h-a b+b h \\
& \Rightarrow k a+b h=k h
\end{aligned}
\)
On dividing both sides by kh, we obtain
\(
\begin{aligned}
& \frac{\mathrm{ka}}{\mathrm{kh}}+\frac{\mathrm{bh}}{\mathrm{kh}}=\frac{\mathrm{kh}}{\mathrm{kh}} \\
& \Rightarrow \frac{\mathrm{a}}{\mathrm{h}}+\frac{\mathrm{b}}{\mathrm{k}}=1
\end{aligned}
\)

Hint

Since line AB passes through points \(A(1985,92)\) and B \((1995,97)\), its slope is
\(
(97-92) /(1995-1985)=5 / 10=1 / 2
\)
Let \(y\) be the population in the year 2010.
Then, according to the given graph, line \(A B\) must pass through point \(C\) \((2010, y)\)
Therefore, Slope of \(A B=\) Slope of \(B C\)
\(
\begin{aligned}
& 1 / 2=(y-97) /(2010-1995) \\
& 1 / 2=(y-97) / 15 \\
& 15 / 2=(y-97) \\
& y=7.5+97 \\
& y=104.5
\end{aligned}
\)
Thus, the slope of line \(A B\) is \(1 / 2\), while in the year 2010 , the population will be \(104.5\) crores.

Hint

At x-axis, value of \(y\) is always 0 Hence equation for \(x\)-axis is
\(
y=0
\)
Similarly
At y-axis, value of \(x\) is always 0
Hence equation for \(y\)-axis is
\(
x=0
\)

Hint

We know that the equation of the line passing through point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\), whose slope is \(m\), is \(\left(y-y_1\right)=m\left(x-x_1\right)\)
Thus, the equation of the line passing through point \((-4,3)\), whose slope is \(1 / 2\),
\(
\begin{aligned}
& (y-3)=1 / 2(x+4) \\
& 2(y-3)=x+4 \\
& 2 y-6=x+4 \\
& x-2 y+10=0
\end{aligned}
\)
Hence the equation is \(x-2 y+10=0\)

Hint

We know that the equation of the line passing through point \(\left(\mathrm{x}_0, \mathrm{y}_0\right)\), whose slope is \(m\) is,
\(
\left(y-y_0\right)=m\left(x-x_0\right)
\)
Substituting \(\left(\mathrm{x}_0, \mathrm{y}_0\right)=(0,0)\) here,
\(
\begin{aligned}
& y-0=m(x-0) \\
& y=m x
\end{aligned}
\)
Hence the equation is \(y=m x\)

Hint

\(
\begin{aligned}
& \text { Here, } m=\tan 75^{\circ} \\
& \Rightarrow m=\tan \left(45^{\circ}+30^{\circ}\right) \\
& \Rightarrow m=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}} \\
& \Rightarrow m=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \\
& \Rightarrow m=\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=2+\sqrt{3}
\end{aligned}
\)
So, the equation of the line that passes through \((2,2 \sqrt{3})\) and has slope \(2+\sqrt{3}\)
\(
\begin{aligned}
& \text { is } y-2 \sqrt{3}=(2+\sqrt{3})(x-2) \\
& \Rightarrow y-2 \sqrt{3}=(2+\sqrt{3}) x-4-2 \sqrt{3} \\
& \Rightarrow(2+\sqrt{3}) x-y-4=0
\end{aligned}
\)

Hint

It is known that if a line with slope \(m\) makes \(x\)-intercept \(d\), then the equation of the time is given as \(y=m(x-d)\).
For the line intersecting the \(x\)-axis at a distance of 3 units to the left of the origin, \(d=-3\).
The slope of the line is given as \(m=-2\)
Thus, the required equation of the given line is
\(
\begin{aligned}
& y=-2[x-(-3)] \\
& y=-2 x-6 \\
& 2 x+y+6=0
\end{aligned}
\)
Hence the equation is \(2 x+y+6=0\)

Hint

It is known that if a line with slope \(m\) makes \(y\)-intercept \(c\), then the equation of the line is given as \(y=m x+c\).
Here, \(c=2\) and \(m=\tan 30^{\circ}=1 / \sqrt{3}\)
Thus, the required equations of the given line is
\(
\begin{aligned}
& y=(1 / \sqrt{3}) x+2 \\
& y=(x+2 \sqrt{3}) / \sqrt{3} \\
& \sqrt{3} y=x+2 \sqrt{3} \\
& x-\sqrt{3} y+2 \sqrt{3}=0
\end{aligned}
\)
Hence, the equation of the line is \(x-\sqrt{3} y+2 \sqrt{3}=0\)

Hint

It is known that the equation of the line passes through points \(\left(x_1, y_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\) is
\(
y-y_1=\left[\left(y_2-y_1 /\left(x_2-x_1\right]\left(x-x_1\right)\right.\right.
\)
Therefore, the equation of the line passing through the points \((-1,1)\) and \((2,-4)\) is
\(
\begin{aligned}
& (y-1)=[(-4-1) /(2+1)](x+1) \\
& (y-1)=(-5 / 3)(x+1) \\
& 3(y-1)=-5(x+1) \\
& 3 y-3=-5 x-5 \\
& 5 x+3 y+2=0
\end{aligned}
\)
Hence, the equation of the line is \(5 x+3 y+2=0\)

Hint

If \(p\) is the length of the normal from the origin to a line and \(w\) is the angle made by the normal with the positive direction of the \(x\)-axis, then the equation of the line given by
\(
x \cos \omega+y \sin \omega=p
\)
Here, \(p=5\) units and \(\omega=30^{\circ}\)
Thus, the required equation of the given line is
\(
\begin{aligned}
& x \cos 30^{\circ}+y \sin 30^{\circ}=5 \\
& x \cdot \sqrt{ } 3 / 2+y \cdot 1 / 2=5 \\
& \sqrt{ } 3 x+y=10
\end{aligned}
\)
Hence, the equation of the line is \(\sqrt{3} x+y=10\)

Hint

Since, median bisects the opposite side i.e., \(S\) is the mid-point of \(P Q\).
\(
\begin{aligned}
& S=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\
& =\left(\frac{2-2}{2}, \frac{1+3}{2}\right)=\left(0, \frac{4}{2}\right)=(0,2) \\
& \left(\because x_1=2, y_1=1, x_2=-2, y_1=3\right)
\end{aligned}
\)
\(\therefore\) Equation of line \(R S\) by using
\(
y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)
\)
\(
\begin{aligned}
& \Rightarrow y-5=\frac{2-5}{0-4}(x-4) \\
& \Rightarrow\left(\because x_1=4, y_1=5, x_2=0, y_2=2\right) \\
& \Rightarrow y-5=\frac{-3}{-4}(x-4) \\
& \Rightarrow 4 y-20=3 x-12 \\
& \Rightarrow 3 x-4 y+8=0
\end{aligned}
\)

Hint

The slope of the line joining the points \((2,5)\) and \((-3,6)\) is \(m=(6-5) /(-3-2)=-1 / 5\)
We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, the slope of the line perpendicular to the line through the points \((2,5)\) and \((-3,6)\) is
\(
-1 / m=-1 /(-1 / 5)=5
\)
Now, the equation of the line passing through point \((-3,5)\), whose slope is 5 , is
\(
\begin{aligned}
& (y-5)=5(x+3) \\
& y-5=5 x+15 \\
& 5 x-y+20=0
\end{aligned}
\)
Hence, the equation of the line is \(5 x-y+20=0\)

Hint

According to the section formula, the coordinates of the point that divides the line segment joining the points \((1,0)\) and \((2,3)\) in the ratio \(1: n\) is given by
\(
[n(1)+1(2) /(1+n), n(0)+1(3) /(1+n)]=[(n+2) /(n+1), 3 / /(n+1)]
\)
The slope of the line joining the points \((1,0)\) and \((2,3)\) is \(m=(3-0) /(2-\) 1) \(=3\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points \((1,0)\) and \((2,3)\) is \(-1 / m=-1 / 3\)

Now the equation of the line passing through \([(n+2) /(n+1), 3 / /(n+1)]\) and whose slope is \(-1 / 3\), given by
\(
\begin{aligned}
& {[y-3 / /(n+1)]=-1 / 3[x-(n+2) /(n+1)]} \\
& 3[(n+1) y-3]=-[x(n+1)-(n-2)] \\
& 3(n+1) y-9=-(n+1) x+n+2 \\
& (1+n) x+3(1+n) y=n+11
\end{aligned}
\)
Hence, the equation of the line is \((1+n) x+3(1+n) y=n+11\)

Hint

Equation of line in intercept form is
\(
\frac{x}{a}+\frac{y}{b}=1 \dots(i)
\)
Since, the line makes equal intercepts on the axes i.e.,
\(
a=b
\)
\(\therefore\) From Eq. (i), \(\frac{x}{a}+\frac{y}{a}=1\)
\(
\Rightarrow x+y=a \ldots \text {…(ii) }
\)
Line (ii) passes through the point \((2,3)\), it will satisfy the line (ii).
\(
2+3=a \Rightarrow a=5
\)
Put the value of \(a\) in Eq. (ii), we get
\(
x+y=5
\)

Hint

Equation of line in intercept form is
\(
\frac{x}{a}+\frac{y}{b}=1 \ldots \ldots \text { (i) }
\)
Given, \(a+b=9 \dots(ii)\)
and line (i) passes through the point \((2,2)\), it will satisfy the line (i) i.e., put \(x=2, y=2\) in Eq. (i)
\(
\Rightarrow \frac{2}{a}+\frac{2}{b}=1 \text {. } \dots(iii)
\)
Solve Eqs. (ii) and (iii), to find the values of \(a\) and \(b\).
From Eq. (ii), \(b=9-a\) put in Eq. (iii), we get
\(
\begin{aligned}
& \frac{2}{a}+\frac{2}{9-a}=1 \\
& \Rightarrow 2(9-a)+2 a=a(9-a) \\
& \Rightarrow 18-2 a+2 a=9 a-a^2 \\
& \Rightarrow a^2-9 a+18=0
\end{aligned}
\)
Factorize it by splitting the middle term,
\(
\begin{aligned}
& \Rightarrow a^2-6 a-3 a+18=0 \\
& \Rightarrow a(a-6)-3(a-6)=0 \\
& \Rightarrow(a-6)(a-3)=0 \\
& \Rightarrow a=6 \text { or } 3 \Rightarrow b=3 \text { or } 6
\end{aligned}
\)
Hence, Eq. (i) becomes
\(
\begin{aligned}
& \frac{x}{6}+\frac{y}{3}=1 \text { or } \frac{x}{3}+\frac{y}{6}=1 \\
& \Rightarrow 3 x+6 y=18 \text { or } 6 x+3 y=18 \\
& \Rightarrow x+2 y=6 \text { or } 2 x+y=6 \text { (divide by } 3 \text { ) }
\end{aligned}
\)

Hint

The slope of the line making an angle \(2 \pi / 3\) with the positive \(x\)-axis is \(m=\) \(\tan (2 \pi / 3)=-\sqrt{3}\)

Now, the equation of the line passing through points \((0,2)\) and having a slope \(-\sqrt{3}\) is
\(
\begin{aligned}
& (y-2)=-\sqrt{3}(x-0) \\
& \sqrt{3} x+y-2=0
\end{aligned}
\)
The slope of line parallel to line \(\sqrt{3} x+y-2=0\) is \(\sqrt{3}\).
It is given that the line parallel to line \(\sqrt{3} x+y-2=0\) crosses the \(y\)-axis 2 units below the origin, i.e., it passes through point \((0,2)\).
Hence, the equation of the line passing through points \((0,2)\) and having a slope \(-\sqrt{3}\) is
\(
\begin{aligned}
& y-(-2)=-\sqrt{3}(x-0) \\
& y+2=-\sqrt{3} \\
& \sqrt{3} x+y+2=0
\end{aligned}
\)
Thus, the equation of the line is \(\sqrt{3} x+y+2=0\).
Now we will find the equation of line parallel to it and crossing the \(y\)-axis at a distance of 2 units below the origin
The equation of the line parallel to \(\sqrt{3} x+y+2=0\) is of the form \(\sqrt{3} x+y+\) \(k=0 \ldots(1)\)
Since this line crosses the \(y\)-axis at a distance of 2 units below the origin, its \(y\)-intercept is \((0,2)\). Substituting \(x=0\) and \(y=2\) in the above equation,
\(
\begin{aligned}
& \sqrt{ } 3(0)+2+k=0 \\
& k=-2
\end{aligned}
\)
Substituting this in (1),
\(
\sqrt{3} x+y-2=0
\)
Thus, the equations of the required lines are \(\sqrt{ } 3 x+y+2=0\) and \(\sqrt{3} x+y-2=0\)

Hint

The slope of joining the origin \((0,0)\) and point \((-2,9)\) is,
\(
\mathrm{m}_1=\frac{9-0}{-2-0}=-\frac{9}{2}
\)
Accordingly the slope of the line perpendicular to the line joining the origin and point \((-2,9)\) is,
\(
\mathrm{m}_2=-\frac{1}{\mathrm{~m}_1}=-\frac{1}{\left\{-\frac{9}{2}\right\}}=\frac{2}{9}
\)
Thus the equation of the line passing through point \((-2,9)\) and having a slope \(\mathrm{m}_2\) is:
\(
(y-9)=\frac{2}{9}(x+2)
\)
\(
\Rightarrow 9 \mathrm{y}-81=2 \mathrm{x}+4 \Rightarrow 2 \mathrm{x}-9 \mathrm{y}+85=0
\)

Hint

Assuming \(C\) is along \(x\)-axis \& \(L\) is along \(y\)-axis
\(
(x, y)=(C, L)
\)
So, we are given two points
\((20,124.942)\) and \((110,125.134)\)
So, from two point form
\(
\mathrm{y}-\mathrm{y}_1=\frac{y_2-y_1}{x_2-x_1}\left(\mathrm{x}-\mathrm{x}_1\right)
\)
\(
\begin{aligned}
& L-124.942=\frac{125.134-124.942}{110-20}(C-20) \\
& L-124.942=\frac{125.134-124.942}{110-20}(C-20) \\
& L-124.942=\frac{0.192}{90}(C-20) \\
& L=\frac{\mathbf{0 . 1 9 2}}{90}(C-20)+124.942
\end{aligned}
\)

Hint

Let \(x\) denote the price per litre and \(y\) denote the quantity of the milk sold at this price.
Since there is a linear relationship between the price and the quantity, the line representing this relationship passes through \((14,980)\) and \((16,1220)\).
So, the equation of the line passing through these points is
\(
\begin{aligned}
& y-980=\frac{1220-980}{16-14}(x-14) \\
& \Rightarrow y-980=120(x-14) \\
& \Rightarrow 120 x-y-700=0
\end{aligned}
\)
When \(x=17\) then we have,
\(
\begin{aligned}
& 120 \times 17-y-700=0 \\
& \Rightarrow y=1340
\end{aligned}
\)
Hence, the owner of the milk store can sell 1340 litres of milk at Rs 17 per litre.

Hint

Let \(A(x, 0)\) and \(B(0, y)\) be two points where the line intersect \(x\) and \(y\)-axes respectively and \(P(a, b)\) is mid point of \(A B\).
Then
\(
\begin{aligned}
& \frac{0+x}{2}=a \Rightarrow x=2 a \\
& \text { and } \frac{0+y}{2}=b \Rightarrow y=2 b
\end{aligned}
\)
Now equation of the required line is
\(
\frac{x}{2 a}+\frac{y}{2 b}=1 \quad \Rightarrow \quad \frac{x}{a}+\frac{y}{b}=2
\)

Hint

Let \(A B\) be the line segment between the axes such that point \(R(h, k)\) divides \(A B\) in the ratio \(1: 2\).
Let the respective coordinates of \(A\) and \(B\) be \((x, 0)\) and \((0, y)\).
Since point \(R(h, k)\) divides \(A B\) in the ratio \(1: 2\), according to the section formula,
\(
\begin{aligned}
& (h, k)=[(1(0)+2(x)) /(1+2),((1(y)+2(0)) /(1+2) \\
& (h, k)=(2 x / 3, y / 3) \\
& h=2 x / 3, k=y / 3 \\
& x=3 h / 2, y=3 k
\end{aligned}
\)
Therefore, the respective coordinates of \(A\) and \(B\) are \((3 h / 2,0)\) and \((0,3 k)\)
Now, the equation of the line AB passing through points \((3 \mathrm{~h} / 2,0)\) and \((0\), \(3 \mathrm{k})\) is
\(
\begin{aligned}
& (y-0)=[(3 k-0) /(0-3 h / 2)][x-3 h / 2] \\
& y=-2 k / h(x-3 h / 2) \\
& h y=-2 k(2 x-3 h) / 2 \\
& h y=-k(2 x-3 h) \\
& h y=-2 k x+3 k h \\
& 2 k x+h y=3 k h
\end{aligned}
\)
Thus, the required equation of a line is \(2 k x+h y=3 k h\)

Hint

In order to show that points \((3,0),(-2,-2)\), and \((8,2)\) are collinear, it suffices to show that the line passing through points \((3,0)\) and \((-2,-2)\) also passes through point \((8,2)\).
The equation of the line passing through points \((3,0)\) and \((-2,-2)\) is
\(
\begin{aligned}
& (y-0)=\frac{(-2-0)}{(-2-3)}(x-3) \\
& y=\frac{-2}{-5}(x-3) \\
& 5 y=2 x-6 \\
& \text { i.e., } 2 x-5 y=6
\end{aligned}
\)
It is observed that at \(x=8\) and \(y=2\),
L.H.S. \(=2 \times 8-5 \times 2=16-10=6=\) R.H.S.
Therefore, the line passing through points \((3,0)\) and \((-2,-2)\) also passes through point \((8,2)\). Hence, points \((3,0),(-2,-2)\), and \((8,2)\) are collinear.

Hint

The equation is \(3 x+2 y-12=0\)
It can be written as
\(
\begin{aligned}
& 3 x+2 y=12 \\
& \frac{3 x}{12}+\frac{2 y}{12}=1 \\
& \text { i.e; } \frac{x}{4}+\frac{y}{6}=1 \dots(1)
\end{aligned}
\)
This equation is of the form \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\), where \(\mathrm{a}=4\) and \(\mathrm{b}=6\).
Therefore equation (1) is in the intercept form, where the intercepts on the \(\mathrm{x}\) and \(y\) axes are 4 and 6 respectively.

Hint

The given equation is
\(
x-\sqrt{3} y+8=0
\)
which can be written as
\(
\begin{aligned}
& x-\sqrt{3} y=-8 \\
& \Rightarrow-x+\sqrt{3} y=8
\end{aligned}
\)
On dividing both sides by \(\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{4}=2\), we get
\(
\begin{aligned}
& -\frac{x}{2}+\frac{\sqrt{3}}{2} y=\frac{8}{2} \\
& \Rightarrow\left(-\frac{1}{2}\right) x+\left(\frac{\sqrt{3}}{2}\right) y=4 \\
& \Rightarrow x \cos 120^{\circ}+y \sin 120^{\circ}=4
\end{aligned}
\)
which is the normal form.
On comparing it with the normal form of equation of line \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\), we get
\(
\alpha=120^{\circ} \text { and } p=4
\)
So, the perpendicular distance of the line from the origin is 4 and the angle between the perpendicular and the positive \(\mathrm{x}\)-axis is \(120^{\circ}\)

Hint

The given equation of the line is \(12(x+6)=5(y-2)\).
\(
\begin{aligned}
& \Rightarrow 12 x+72=5 y-10 \\
& \Rightarrow 12 x-5 y+82=0 \ldots(1)
\end{aligned}
\)
On comparing equation (1) with general equation of line \(A x+B y+C=0\), we obtain \(A=12, B=-5\), and \(C=\) 82.
It is known that the perpendicular distance \((d)\) of a line \(A x+B y+C=0\) from a point \(\left(x_1, y_1\right)\) is given by \(d=\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}}\).
The given point is \(\left(x_1, y_1\right)=(-1,1)\). Therefore, the distance of point \((-1,1)\) from the given line
\(
=\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^2+(-5)^2}} \text { units }=\frac{|-12-5+82|}{\sqrt{169}} \text { units }=\frac{|65|}{13} \text { units }=5 \text { units }
\)

Hint

The given equation of line is
\(
\frac{x}{3}+\frac{y}{4}=1
\)
or, \(4 x+3 y-12=0 \dots(1)\)
On comparing equation (1) with general equation of line \(A x+B y+C=0\),
we obtain \(A=4, B=3\), and \(C=-12\).
Let \((a, 0)\) be the point on the \(x\)-axis whose distance from the given line is 4 units.
It is known that the perpendicular distance (d) of a line \(A x+B y+C=0\) from a point
\(
\left(x_1, y_1\right) \text { is given by } d=\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}}
\)
Therefore,
\(
\begin{aligned}
& 4=\frac{|4 a+3 \times 0-12|}{\sqrt{4^2+3^2}} \\
& \Rightarrow 4=\frac{|4 a-12|}{5} \\
& \Rightarrow|4 a-12|=20 \\
& \Rightarrow \pm(4 a-12)=20 \\
& \Rightarrow(4 a-12)=20 \text { or }-(4 a-12)=20 \\
& \Rightarrow 4 a=20+12 \text { or } 4 a=-20+12 \\
& \Rightarrow a=8 \text { or }-2
\end{aligned}
\)
Thus, the required points on the \(x\)-axis are \((-2,0)\) and \((8,0)\).

Hint

It is known that the distance \((d)\) between parallel lines \(A x+B y+C_1=0\) and \(A x+B y+C_2=0\)
given by \(d=\frac{\left|C_1-C_2\right|}{\sqrt{A^2+B^2}}\).
The given parallel lines are \(15 x+8 y-34=0\) and \(15 x+8 y+31=0\)
Here, \(A=15, B=8, C_1=-34\), and \(C_2=31\)
Therefore, the distance between the parallel lines is
\(d=\frac{\left|C_1-C_2\right|}{\sqrt{A^2+B^2}}=\frac{|-34-31|}{\sqrt{(15)^2+(8)^2}}\) units \(=\frac{|-65|}{17}\) units \(=\frac{65}{17}\) units

Hint

Given equation line, \(3 x-4 y+2=0\)
\(
\begin{aligned}
& \therefore 4 y=3 x+2 \\
& y=\left(\frac{3}{4}\right) x+\frac{1}{2}
\end{aligned}
\)
Slope of given line \(=\frac{3}{4}\)
As slope of parallel lines are equal
\(\therefore\) Slope of line parallel to given line \(=\frac{3}{4}\)
Equations of line passes through \((-2,3)\) with slope \(\left(\frac{3}{4}\right)\) is
\(
\begin{aligned}
& (y-3)=\frac{3}{4}(x+2) \\
& 4 y-12=3 x+6
\end{aligned}
\)
So, \(3 x-4 y+18=0\) be the equation of line parallel to \(3 x-4 y+2=0\) and passing through \((-2,3)\)

Hint

The given equation of line is \(x-7 y+5=0\).
Or \(y=\frac{1}{7} x+\frac{5}{7}\)
Therefore, slope of the given line \(=\frac{1}{7}\)
The slope of the line perpendicular to the line having slope \(\frac{1}{7}\) is : \(\mathrm{m}=-7\)
The equation of the line with slope \(-7\) and \(\mathrm{x}\)-intercept 3 is given by:
\(
y=m(x-d)
\)
\(
y=-7(x-3)
\)
\(
7 x+y=21
\)

Hint

The given lines are \(\sqrt{3} x+y=1\) and \(x+\sqrt{3} y=1\).
\(
y=-\sqrt{3} x+1 \dots(1)
\)
and \(y=-\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} \dots(2)\)
The slope of line (1) is \(m_1=-\sqrt{3}\), while the slope of line (2) is \(m_2=-\frac{1}{\sqrt{3}}\)
The acute angle i.e., \(\theta\) between the two lines is given by
\(
\begin{aligned}
& \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
& \tan \theta=\left|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\right|=\left|\frac{-2}{2 \times \sqrt{3}}\right| \\
& \tan \theta=\frac{1}{\sqrt{3}} \\
& \theta=30^{\circ} \\
&
\end{aligned}
\)
Thus, the angle between the given lines is either \(30^{\circ}\) or \(180^{\circ}-30^{\circ}=150^{\circ}\).

Hint

The slope of the line passing through points \((h, 3)\) and \((4,1)\) is \(m_1=\frac{1-3}{4-h}=\frac{-2}{4-h}\)
The slope of line \(7 x-9 y-19=0\) or \(y=\frac{7}{9} x-\frac{19}{9}\) is \(m_2=\frac{7}{9}\). It is given that the two lines are perpendicular.
\(
\begin{aligned}
& \therefore m_1 \times m_2=-1 \\
& \Rightarrow\left(\frac{-2}{4-h}\right) \times\left(\frac{7}{9}\right)=-1 \\
& \Rightarrow \frac{-14}{36-9 h}=-1 \\
& \Rightarrow 14=36-9 h \\
& \Rightarrow 9 h=36-14 \\
& \Rightarrow h=\frac{22}{9}
\end{aligned}
\)
Thus, the value of \(h\) is \(\frac{22}{9}\).

Hint

The slope of line \(A x+B y+C=0\) or \(y=(-A / B) x+(-C / B)\) is \(m=-A / B\) It is known that parallel lines have the same slope.
Therefore, slope of the other line \(=m=-A / B\)
The equation of the line passing through point \(\left(x_1, y_1\right)\) and having slope \(m\) \(=-\mathrm{A} / \mathrm{B}\) is
\(
\begin{aligned}
& y-y_1=m\left(x-x_1\right) \\
& y-y_1=-A / B\left(x-x_1\right) \\
& B\left(y-y_1=-A\left(x-x_1\right)\right. \\
& A\left(x-x_1\right)+B\left(y-y_1\right)=0
\end{aligned}
\)
Hence, the line through the point \(\left(x_1, y_1\right)\) and parallel to the line \(A x+B y+\) \(C=0\) is \(A\left(x-x_1\right)+B\left(y-y_1\right)=0\)

Hint

It is given that the slope of the first line, \(\mathrm{m}_1=2\).
Let the slope of the other line be \(m_2\).
The angle between the two lines is \(60^{\circ}\).
\(
\begin{aligned}
& \tan \theta=\left|\left(m_1-m_2\right) /\left(1+m_1 m_2\right)\right| \\
& \tan 60^{\circ}=\left|\left(2-m_2\right) /\left(1+2 m_2\right)\right|
\end{aligned}
\)
\(
m_2=(2-\sqrt{3}) /(2 \sqrt{3}+1)
\)
\(
m_2=-(2+\sqrt{3}) /(2 \sqrt{3}-1)
\)
Case-I:
\(
m_2=(2-\sqrt{3}) /(2 \sqrt{3}+1)
\)
The equation of the line passing through the point \((2,3)\) and having a slope of \((2-\sqrt{3}) /(2 \sqrt{3}+1)\) is
\(
\begin{aligned}
& (y-3)=[(2-\sqrt{3}) /(2 \sqrt{3}+1)](x-2) \\
& (2 \sqrt{3}+1) y-3(2 \sqrt{3}+1)=(2-\sqrt{3}) x-2(2-\sqrt{3}) \\
& (\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}
\end{aligned}
\)
In this case, the equation of the other line is \((\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+\) \(8 \sqrt{3}\)
\(
m_2=-(2+\sqrt{3}) /(2 \sqrt{3}-1)
\)
The equation of the line passing through the point \((2,3)\) and having a slope of \(-(2+\sqrt{3}) /(2 \sqrt{3}-1)\) is
\(
\begin{aligned}
& (y-3)=[-(2+\sqrt{3}) /(2 \sqrt{3}-1)](x-2) \\
& (2 \sqrt{3}-1) y-3(2 \sqrt{3}-1)=-(2+\sqrt{3}) x+2(2-\sqrt{3})
\end{aligned}
\)
\(
\begin{aligned}
& (\sqrt{3}+2) x+(2 \sqrt{3}-1) y=4-2 \sqrt{3}+6 \sqrt{3}-3 \\
& (\sqrt{3}+2) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}
\end{aligned}
\)
In this case, the equation of the other line is \((\sqrt{3}+2) x+(2 \sqrt{3}-1) y=1+\) \(8 \sqrt{3}\)

Hint

Let the given points be \(A(3,4)\) and \(B(-1,2)\).
Let \(M\) be the midpoint of \(A B\).
\(\therefore\) Coordinates of \(M=\left(\frac{3-1}{2}, \frac{4+2}{2}\right)=(1,3)\)
And, slope of \(A B=\frac{2-4}{-1-3}=\frac{1}{2}\)
Let \(m\) be the slope of the right bisector of the line joining the points \((3,4)\) and \((-1,2)\).
\(
\begin{aligned}
& \therefore m \times \text { Slope of } A B=-1 \\
& \Rightarrow m \times \frac{1}{2}=-1 \\
& \Rightarrow m=-2
\end{aligned}
\)
So, the equation of the line that passes through \(M(1,3)\) and has slope \(-2\) is
\(
\begin{aligned}
& y-3=-2(x-1) \\
& \Rightarrow 2 x+y-5=0
\end{aligned}
\)
Hence, the equation of the right bisector of the line segment joining the points \((3,4)\) and \((-1,2)\) is \(2 x+y-5=0\).

Hint

Let \(A(-1,3)\) be the given point.
Also, let \(M(h, k)\) be the foot of the perpendicular drawn from \(A(-1,3)\) to the line \(3 x-4 y-16=0\)
Point \(M(h, k)\) lies on the line \(3 x-4 y-16=0\)
\(3 h-4 k-16=0 \dots(1)\)
Lines \(3 x-4 y-16=0\) and \(A M\) are perpendicular.
\(
\begin{aligned}
& \therefore \frac{k-3}{h+1} \times \frac{3}{4}=-1 \\
& \Rightarrow 4 h+3 k-5=0 \dots(2)
\end{aligned}
\)
Solving eq (1) and eq (2) by cross multiplication, we get:
\(
\begin{aligned}
& \frac{h}{20+48}=\frac{k}{-64+15}=\frac{1}{9+16} \\
& \Rightarrow a=\frac{68}{25}, b=-\frac{49}{25}
\end{aligned}
\)
Hence, the coordinates of the foot of perpendicular are \(\left(\frac{68}{25},-\frac{49}{25}\right)\).

Hint

The given line is \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) which can be written as \(m x-y+c=0 \dots(1)\)
The slope of the line perpendicular to \(y=m x+c\) is \(-\frac{1}{m}\)
So, the equation of the line with slope \(-\frac{1}{m}\) and passing through the origin is \(y=-\frac{1}{m} x\) \(x+m y=0 \dots(2)\)
Solving eq (1) and eq (2) by cross multiplication, we get
\(
\begin{aligned}
& \frac{x}{m c-0}=\frac{y}{0-c}=\frac{1}{-1-m^2} \\
& \Rightarrow x=-\frac{m c}{m^2+1}, y=\frac{c}{m^2+1}
\end{aligned}
\)
Thus, the point of intersection of the perpendicular from the origin to the line \(y=m x+c\) is
\(
\left(-\frac{m c}{m^2+1}, \frac{c}{m^2+1}\right)
\)
It is given that the perpendicular from the origin to the line \(y=m x+c\) meets it at the point \((-1,2)\)
\(
\begin{aligned}
& -\frac{m c}{m^2+1}=-1 \text { and } \frac{c}{m^2+1}=2 \\
& \Rightarrow m^2+1=m c \text { and } m^2+1=\frac{c}{2} \\
& \Rightarrow m c=\frac{c}{2} \\
& \Rightarrow m=\frac{1}{2}
\end{aligned}
\)
Now, substituting the value of \(m\) in \(m^2+1=m c\) we get
\(
\begin{aligned}
& \frac{1}{4}+1=\frac{1}{2} c \\
& \Rightarrow c=\frac{5}{2}
\end{aligned}
\)
Hence,
\(m=\frac{1}{2}\) and \(c=\frac{5}{2}\).

Hint

Given lines are
\(
\begin{aligned}
& x \cos \theta-y \sin \theta=k \cos 2 \theta \dots(1)\\
& x \sec \theta+y \operatorname{cosec} \theta=k \dots(2)\\
& =\left|\frac{A x_1+B y_1+C}{\sqrt{A^2+B^2}}\right| \\
& =\left|\frac{(0) \cos \theta-(0) \sin \theta-k \cos 2 \theta}{\sqrt{(\cos \theta)^2+(-\sin \theta)^2}}\right| \\
& \because\left(x_1, y_1\right)=(0,0) \\
& =\left|\frac{0-0-k \cos 2 \theta}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}\right| \\
& =k \cos 2 \theta \quad \because \cos ^2 \theta+\sin ^2 \theta=1 \\
& \text { Similarly, } \\
& q=\left|\frac{(0) \sec \theta+(0) \operatorname{cosec} \theta-k}{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}\right| \\
& =\left|\frac{-k}{\sqrt{\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}}}\right| \\
& =\left|\frac{-k}{\sqrt{\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \cdot \sin ^2 \theta}}}\right| \\
&
\end{aligned}
\)
\(
\therefore \quad q=\left|\frac{-k}{\sqrt{\frac{1}{\cos ^2 \theta \cdot \sin ^2 \theta}}}\right|=k \cos \theta \cdot \sin \theta
\)
Thus \(p=k \cdot \cos 2 \theta, q=k(\cos \theta \cdot \sin \theta)\)
\(
\text { i.e., } \begin{aligned}
p=k \cdot \cos 2 \theta, q & =k \frac{\sin 2 \theta}{2} \\
\because \sin 2 \theta & =2 \sin \theta \cos \theta
\end{aligned}
\)
i.e., \(\frac{p}{k}=\cos 2 \theta, \frac{2 q}{k}=\sin 2 \theta\)
squaring and adding, we get
\(
\frac{p^2}{k^2}+\frac{4 q^2}{k^2}=\cos ^2(2 \theta)+\sin ^2(2 \theta)=1
\)
i.e., \(p^2+4 q^2=k^2\) as required.

Hint

Let \(A D\) be the altitude of triangle \(A B C\) from vertex \(A\). Accordingly, \(A D \perp\) \(\mathrm{BC}\)
The slope of \(\mathrm{BC}=(2+1) /(1-4)=3 /(-3)=-1\).
Since \(A D\) is perpendicular to \(B C\), the slope of \(A D=-1 /(\) slope of \(B C)=\) \(-1 /(-1)=1\)
The equation of the line passing through point \((2,3)\) and having a slope of 1 (equation of \(A D\) ) is
\(
\begin{aligned}
& \Rightarrow(y-3)=1(x-2) \\
& \Rightarrow y-3=x-2 \\
& \Rightarrow y-x=1
\end{aligned}
\)
Therefore, equation of the altitude from vertex \(A=y-x=1\)
Length of \(A D=\) Length of the perpendicular from \(A(2,3)\) to \(B C\)
The equation of \(B C\) is
The equation of \(B C\) is
\(
\begin{aligned}
& \Rightarrow(y+1)=(-1)(x-4) \\
& \Rightarrow \mathrm{y}+1=-\mathrm{x}+4 \\
& \Rightarrow \mathrm{x}+\mathrm{y}-3=0 \ldots(1)
\end{aligned}
\)
The perpendicular distance \((d)\) of a line \(A x+B y+C=O\) from a point \((x_1, y_1)\) ) is given by
\(
d=\left|A x_1+B y_1+C\right| / \sqrt{A^2+B^2}
\)
On comparing equation (1) to the general equation of line \(A x+B y+C=0\) , we obtain \(A=1, B=1\) and \(C=-3\).
Length of \(A D=|1 \times 2+1 \times 3-3| / \sqrt{1^2+1^2}=2 / \sqrt{2}=\sqrt{2}\) units
Thus, the equation and length of the altitude from vertex \(A\) are \(y-x=1\) and \(\sqrt{ } 2\) units respectively

Hint

The equation of the line having intercepts on the \(\mathrm{x}\)-axis and \(\mathrm{y}\)-axis as \(a\) and \(b\) respectively, is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
i.e. \(b x+a y-a b=0\)
The length of the perpendicular from origin onto this line is given by \(p=\)
\(
\begin{aligned}
& \frac{|0+0-a b|}{\sqrt{a^2+b^2}}=\frac{a b}{\sqrt{a^2+b^2}} \\
& \Rightarrow \mathrm{p}^2=\frac{\mathrm{a}^2 \mathrm{~b}^2}{\mathrm{a}^2+\mathrm{b}^2} \\
& \Rightarrow \frac{1}{\mathrm{p}^2}=\frac{1}{\mathrm{~b}^2}+\frac{1}{\mathrm{a}^2}
\end{aligned}
\)

Hint

The given equation of line is
\(
(k-3) x-\left(4-k^2\right) y+k^2-7 k+6=0
\)
(a) If the given line is parallel to the \(x\)-axis, then slope of the given line= slope of the \(\mathrm{x}\)-axis \(=0\)
\(
\begin{aligned}
& \Rightarrow \frac{(\mathrm{k}-3)}{\left(4-\mathrm{k}^2\right)}=0 \\
& \Rightarrow \mathrm{k}-3=0 \\
& \Rightarrow \mathrm{k}=3
\end{aligned}
\)
Thus, if the given line is parallel to the \(x\)-axis, then the value of \(k\) is 3 .
(b) If the given line is parallel to the \(y\)-axis, it is vertical, hence, its slope will be undefined.
The slope of the given line is \(\frac{(k-3)}{\left(4-k^2\right)}\)
Now, \(\frac{(\mathrm{k}-3)}{\left(4-\mathrm{k}^2\right)}\) is undefined at \(\mathrm{k}^2=4\)
\(
\Rightarrow \mathrm{k}=\pm 2
\)
Thus, if the given line is parallel to the \(y\)-axis, then the value of \(k\) is \(\pm 2\).
(c), If the given line is passing through the origin, then point \((0,0)\) satisfies the given equation of line
\(
\begin{aligned}
& (\mathrm{k}-3)(0)-\left(4-\mathrm{k}^2\right)(0)+\mathrm{k}^2-7 \mathrm{k}+6=0 \\
& \Rightarrow \mathrm{k}^2-7 \mathrm{k}+6=0 \Rightarrow \mathrm{k}^2-6 \mathrm{k}-\mathrm{k}+6=0 \\
& \Rightarrow(\mathrm{k}-6)(\mathrm{k}-1)=0 \Rightarrow \mathrm{k}=1 \text { or } \mathrm{k}=6
\end{aligned}
\)
Thus, if the given line is passing through the origin, then the value of \(\mathrm{k}\) is either 1 or 6.

Hint

The equation of the given line is \(\sqrt{3} x+y+2=0\) this equation can be reduced as \(-\sqrt{3} x-y=2\) on dividing both sides by \(\sqrt{(-\sqrt{3})^2+(-1)^2}=2\), we obtain \(-\frac{\sqrt{3}}{2} x-\frac{1}{2} y=\frac{2}{2}\)
\(
\Rightarrow\left\{-\frac{\sqrt{3}}{2}\right\} x+\left\{-\frac{1}{2}\right\} y=1 \ldots .(1)
\)
On comparing equation (1) to \(x \cos \theta+y \sin \theta=p\),
we obtain \(\cos \theta=-\frac{\sqrt{3}}{2}, \sin \theta=-\frac{1}{2}\), and \(p=1\)
Since the value of \(\sin \theta\) and \(\cos \theta\) are both negative, \(\theta\) is in the third quadrant
\(
\therefore \theta=\pi+\frac{\pi}{6}=\frac{7 \pi}{6}
\)
Thus, the respective values of \(\theta\) and \(p\) are \(\frac{7 \pi}{6}\) and 1

Hint

Let the intercepts cut by the given lines on the axes be \(a\) and \(b\).
It is given that
\(
a+b=1 \ldots(1)
\)
\(
a b=-6 \ldots \text { (2) }
\)
On solving equations ( 1 ) and (2), we obtain
\(a=3\) and \(b=-2\) or \(a=-2\) and \(b=3\)
It is known that the equation of the line whose intercepts on the axes are \(a\) and \(b\) is \(\frac{x}{a}+\frac{y}{b}=1\) or \(b x+a y-a b=0\)
Case I: \(a=3\) and \(b=-2\)

In this case, the equation of the line is \(-2 x+3 y+6=0\), i.e., \(2 x-3 y=6\).

Case II: \(a=-2\) and \(b=3\)
In this case, the equation of the line is \(3 x-2 y+6=0\), i.e., \(-3 x+2 y=6\).
Thus, the required equation of the lines are \(2 x-3 y=6\) and \(-3 x+2 y=6\).

Hint

Let \((0, b)\) be the point on the \(y\)-axis whose distance from line \(x / 3+y / 4\) is 4 units. The given line can be written as
\(
4 x+3 y-12=0 \dots(1)
\)
On comparing equation (1) to the general equation of line \(A x+B y+C=0\), we obtain
\(
A=4, B=3 \text {, and } C=-12
\)
It is known that the perpendicular distance (d) of a line \(A x+B y+C=0\) from a point \(\left(x_1, y_1\right)\) is given by \(d=\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}}\)

Therefore, if \((0, b)\) is the point on the \(y\)-axis whose distance from line \(x / 3+y / 4=1\) is 4 units, then:
\(
\begin{aligned}
& 4=\frac{|4(0)+3(b)-12|}{\sqrt{4^2+3^2}} \\
& \Rightarrow 4=\frac{|3 b-12|}{5} \\
& \Rightarrow 20=|3 b-12| \\
& \Rightarrow 20=\pm(3 b-12) \\
& \Rightarrow 20=(3 b-12) \text { or } 20=-(3 b-12) \\
& \Rightarrow 3 b=20+12 \text { or } 3 b=-20+12 \\
& \Rightarrow b=\frac{32}{3} \text { or } b=-\frac{8}{3}
\end{aligned}
\)
Thus, the required points are \(\left(0, \frac{32}{3}\right)\) and \(\left(0,-\frac{8}{3}\right)\)

Hint

The two points A and B can be plotted as above, where \(\mathrm{A}(\cos \phi, \sin \phi)\) and \(\mathrm{B}(\cos \theta, \sin \theta)\). Using distance formula we can calculate the distance between the two points. So the distance between two points \(\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right),\left(\mathrm{x}_2, \mathrm{y}_2\right)\) can be written as
\(
\mathrm{d}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}
\)
So we can write \(\mathrm{AB}=\sqrt{(\cos \theta-\cos \phi)^2+(\sin \theta-\sin \phi)^2}\)
Finding the perpendicular distance between the centre \(O\) and the line \(A B\) We can see that \(\triangle \mathrm{AOD}\) is a right-angled triangle
According to the Pythagoras theorem, \(\mathrm{OD}^2+\mathrm{DA}^2=\mathrm{AO}^2\)
Also we can write \(\mathrm{AO}=\sqrt{\cos ^2 \phi+\sin ^2 \phi}\)
\(
\begin{aligned}
& \mathrm{OD}=\sqrt{\mathrm{AO}^2-\mathrm{AD}^2} \\
& \mathrm{OA}=\sqrt{\mathrm{AO}^2-\left(\frac{\mathrm{AB}}{2}\right)^2}
\end{aligned}
\)
\(A B=2 A D\)..as any perpendicular from the centre of the circle bisects the chord
\(\mathrm{OD}=\sqrt{\left(\sqrt{\cos ^2 \phi+\sin ^2 \phi}\right)^2-\left(\frac{\sqrt{(\cos \theta-\cos \phi)^2+(\sin \theta-\sin \phi)^2}}{2}\right)^2}\)
\(
\begin{aligned}
& \mathrm{OD}=\sqrt{\cos ^2 \phi+\sin ^2 \phi-\frac{(\cos \theta-\cos \phi)^2+(\sin \theta-\sin \phi)^2}{4}} \\
& \mathrm{OD}=\sqrt{\frac{4-\left(\cos ^2 \theta+\cos ^2 \phi-2 \cos \theta \cos \phi+\sin ^2 \theta+\sin ^2 \phi-2 \sin \theta \sin \phi\right)}{4}}
\end{aligned}
\)
\(
\begin{aligned}
\left(\text { As } \sin ^2 \mathbf{x}+\cos ^2 \mathbf{x}\right. & =\mathbf{1}) \\
\mathrm{OD} & =\frac{\sqrt{4-2+2 \sin \theta \sin \phi+2 \cos \theta \cos \phi}}{2} \\
\mathrm{OD} & =\sqrt{\frac{1+\cos (\theta-\phi)}{2}}(\text { As } \cos (\mathbf{a}-\mathbf{b})=\cos \mathbf{a} \cos \mathbf{b}+\sin \mathbf{a} \sin \mathbf{b}) \\
\mathrm{OD} & =\cos \left(\frac{\theta-\phi}{2}\right)\left(\mathrm{As} \cos 2 \theta=2 \cos ^2 \theta-\mathbf{1}\right)
\end{aligned}
\)

Hint

First we calculate point of intersection of lines
\(
\begin{aligned}
& x-7 y+5=0 \\
& \& 3 x+y=0
\end{aligned}
\)
Solving (1)
\(
\begin{aligned}
& x-7 y+5=0 \\
& x=7 y-5
\end{aligned}
\)
Putting value of \(x\) in (2)
\(
3 x+y=0
\)
\(
3(7 y-5)+y=0
\)
\(
y=\frac{15}{22}
\)
Putting value of \(y=\frac{15}{22}\) in \((1)\)
\(
\begin{aligned}
& x-7 y+5=0 \\
& x-7\left(\frac{15}{22}\right)+5=0
\end{aligned}
\)
\(
x=\frac{-5}{22}
\)

Hint

The equation of the given line is \(\frac{x}{4}+\frac{y}{6}=1\).
This equation can also be written as \(3 x+2 y-12=0\) \(\Rightarrow \mathrm{y}=\frac{-3}{2} \mathrm{x}+6\), which is of the form \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) \(\therefore\) slope of given line \(=-\frac{3}{2}\)
\(
\therefore \text { slope of line perpendicular to given line }=-\frac{1}{\left\{-\frac{3}{2}\right\}}=\frac{2}{3}
\)
On substituting \(x=0\) in the given equation of line,
we obtain \(\frac{\mathrm{y}}{6}=1 \Rightarrow \mathrm{y}=6\)
\(\therefore\) the given line intersect the \(y\)-axis at \((0,6)\)
Hence, the equation of line that has slope \(\frac{2}{3}\) and passes through point \((0,6)\) is
Hence, the equation of line that has slope \(\frac{2}{3}\) and passes through point \((0,6)\) is
\(
\begin{aligned}
& (\mathrm{y}-6)=\frac{2}{3}(\mathrm{x}-0) \\
& \Rightarrow 3 \mathrm{y}-18=2 \mathrm{x} \Rightarrow 2 \mathrm{x}-3 \mathrm{y}+18=0
\end{aligned}
\)
Thus, the required equation of the line is \(2 x-3 y+18=0\).

Hint

The equations of the given lines are:
\(
\begin{aligned}
& y-x=0 \ldots(1) \\
& x+y=0 \ldots(2) \\
& x-k=0 \ldots(3)
\end{aligned}
\)
The point of intersection of lines (1) and (2) is obtained by solving them. \(x=0\) and \(y=0\)
The point of intersection of lines (2) and (3) is obtained by solving them. \(x=k\) and \(y=-k\)
The point of intersection of lines (3) and (1) is obtained by solving them. \(x=k\) and \(y=k\).
Thus, the vertices of the triangle formed by the three given lines are \((0\), \(0),(k,-k)\) and \((k, k)\).
We know that the area of a triangle whose vertices are \(\left(x_1, y_1\right),\left(x_2, y_2\right)\) and \(\left(x_3, y_3\right)\) is
\(
1 / 2\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|
\)
Therefore, area of the triangle formed by three given lines
\(
\begin{aligned}
& =1 / 2|0(-k+k)+k(k-0)+k(0+k)| \\
& =1 / 2\left|k^2+k^2\right| \\
& =1 / 2\left(2 k^2\right) \\
& =k^2
\end{aligned}
\)
Hence, the area of the triangle is \(k^2\) square units

Hint

The equation of the given lines are
\(
\begin{aligned}
& 3 x+y-2=0 \ldots(1) \\
& p x+2 y-3=0 \ldots .(2) \\
& 2 x-y-3=0 \ldots(3)
\end{aligned}
\)
On solving equation (1) and (3) we obtain \(x=1\) and \(y=-1\).
Since these three lines may intersect at one point, the point of intersection of lines(1) and (3) will also satisfy lines(2)
\(
\begin{aligned}
& p(1)+2(-1)-3=0 \\
& p-2-3=0 \Rightarrow p=5
\end{aligned}
\)

Hint

Let the slope of the required line be \(\mathrm{m}_1\).
The given line can be written as \(y=\frac{1}{2} x-\frac{3}{2}\), which is of the form \(y=m x+c\)
\(\therefore\) slope of the given line \(=m_2=\frac{1}{2}\)
It is given that the angle between the required line and line \(x-2 y=3\) is \(45^{\circ}\)
We know that if \(\theta\) is the acute angle between lines \(l_1\) and \(l_2\) with slopes \(m_1\) and \(m_2\) respectively, then \(\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|\)
\(
\begin{aligned}
& \therefore \tan 45^{\circ}=\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right| \\
& \Rightarrow 1=\left|\frac{\frac{1}{2}-\mathrm{m}_1}{1+\frac{\mathrm{m}_1}{2}}\right| \\
& \Rightarrow 1=\left|\frac{\left(\frac{1-2 \mathrm{~m}_{\mathrm{l}}}{2}\right)}{\frac{2+\mathrm{m}_1}{2}}\right| \\
& \Rightarrow 1=\pm\left|\frac{1-2 \mathrm{~m}_1}{2+\mathrm{m}_1}\right| \\
& \Rightarrow \mathrm{m}_1=-\frac{1}{3} \text { or } {m}_1=3
\end{aligned}
\)
case \(1 ; \mathrm{m}_1=3\)
The equation of the line passing through \((3,2)\) and having a slope of 3 is \(y-2=3(x-3) \Rightarrow 3 x-y=7\)
case \(2: \mathrm{m}_1=-\frac{1}{3}\)
The equation of the line passing through \((3,2)\) and having a slope of \(-\frac{1}{3}\) is.
\(
\begin{aligned}
& y-2=-\frac{1}{3}(x-3) \\
& \Rightarrow 3 y-6=x+3 \Rightarrow x+3 y=9
\end{aligned}
\)

Hint

Let the equation of the line having equal intercepts on the axes be
\(
\begin{aligned}
& x / a+y / b=1 \\
& \Rightarrow x+y=a b .
\end{aligned}
\)
On solving equations \(4 x+7 y-3=0\) and \(2 x-3 y+1=0\), we obtain \(x=\) \(1 / 13\) and \(y=5 / 13\)
Therefore,
\((1 / 13,5 / 13)\) is the point of the intersection of the two given lines.
Since equation (1) passes through point \((1 / 13,5 / 13)\)
\(
\begin{aligned}
& 1 / 13+5 / 13=a \\
& \Rightarrow a=6 / 13
\end{aligned}
\)
Equation (1) becomes
\(
\begin{aligned}
& \Rightarrow x+y=6 / 13 \\
& \Rightarrow 13 x+13 y=6
\end{aligned}
\)
Thus, the required equation of the line \(13 x+13 y=6\)

Hint

\(
\text { Let the point } P \text { divides the line joining the points } A(-1,1) \text { and } B(5,7) \text { in the ratio is } K: 1 \text {. }
\)
Then,
\(
\begin{aligned}
& P=\left(\frac{K \times x_2+1 \times x_1}{K+1}, \frac{K \times y_2+1 \times y_1}{K+1}\right) \\
& \left(\because x_1=-1, y_1=1, x_2=5, y_2=7\right) \\
& =\left(\frac{K \times 5+1 \times(-1)}{K+1}, \frac{K \times 7+1 \times 1}{K+1}\right) \\
& =\left(\frac{5 K-1}{K+1}, \frac{7 K+1}{K+1}\right)
\end{aligned}
\)
Point \(P\) will satisfy the line \(x+y=4\)
\(
\begin{aligned}
& \text { i.e. } \frac{5 K-1}{K+1}+\frac{7 K+1}{K+1}=\frac{4}{1} \\
& \Rightarrow \frac{5 K-1+7 K+1}{K+1}=4 \\
& \Rightarrow 12 K=4 K+4 \\
& \Rightarrow 8 K=4 \\
& \Rightarrow K=\frac{4}{8}=\frac{1}{2}
\end{aligned}
\)
Hence, the required ratio is \(K: 1=1: 2\) (internally)

Hint

The given lines are
\(
2 x-y=0 \dots(1)
\)
\(
4 x+7 y+5=0 \dots(2)
\)
Let \(\mathrm{A}(1,2)\) is a point on the line \((1)\) and \(B\) be the point intersection of lines
(1) and (2).
On solving equations (1) and (2), we obtain \(x=-5 / 18\) and \(y=-5 / 9\)
Coordinates of point \(B\) are \((-5 / 18,-5 / 9)\)
By using distance formula, the distance between points \(A\) and \(B\) can be obtained as
\(
\begin{aligned}
& \mathrm{AB}=\sqrt{(1+5 / 18)^2+(2+5 / 9)^2} \\
& =\sqrt{(23 / 18)^2+(23 / 9)^2} \\
& =\sqrt{(23 / 9 \times 2)^2+(23 / 9)^2} \\
& =\sqrt{(23 / 9)^2(1 / 2)^2+(23 / 9)^2} \\
& =\sqrt{(23 / 9)^2(1 / 4+1)} \\
& =23 / 9 \sqrt{5 / 4} \\
& =23 / 9 \times \sqrt{5 / 2} \\
& =23 \sqrt{5} / 18
\end{aligned}
\)
Thus, the required distance is \(23 \sqrt{5} / 18\) units

Hint

Required equation of the line through \(\left(x_1, y_1\right)\), making an angle \(\theta\) with the positive direction \(x\)-axis and at a distance \(r\) from \((x, y)\) is of the form
\(
\begin{aligned}
& \frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r \\
& \text { i.e., } \frac{x+1}{\cos \theta}=\frac{y-2}{\sin \theta}=3 \\
& \left(\because\left(x_1, y_1\right)=(-1,2)|r|=3\right) \\
& \Rightarrow x+1=3 \cos \theta, y-2=3 \sin \theta \\
& \Rightarrow \quad x=3 \cos \theta-1, y=3 \sin \theta+2 \\
& \therefore \quad(x, y)=(3 \cos \theta-1, y=3 \sin \theta+2)
\end{aligned}
\)
If this point lies on the line \(x+y=4\), then
\(
\begin{aligned}
& 3 \cos \theta-1+3 \sin \theta+2=4 \\
\Rightarrow & 3 \sin \theta+3 \cos \theta=3 \\
\Rightarrow & \sin \theta+\cos \theta=1 \\
\Rightarrow & \theta=0^{\circ} \text { or } 90^{\circ}
\end{aligned}
\)
\(\Rightarrow\) Line is parallel to \(x\)-axis or parallel to \(y\)-axis.

Hint

Let \(P Q R\) be the right-angled triangle, where \(\angle R=90^{\circ}\)
Assume that \(P=(1,3)\) and \(Q=(-4,1)\). Then \(R\) will be \((1,1)\) which you can get from the following figure.

Now we will find the equations of the two legs \(\mathrm{PR}\) and \(\mathrm{QR}\).
Equation of PR:
\(P R\) is a vertical line passing through \((1,1)\). Hence its equation is \(y=1\).
Equation of \(\mathrm{QR}\) :
\(\mathrm{QR}\) is a horizontal line passing through \((1,1)\). Hence its equation is \(x=1\).
Thus, the equations of legs of the triangle are \(x=1\) and \(y=1\)

Hint

The equation of the given line is
\(
x+3 y=7 \dots(1)
\)
Let point \(B(a, b)\) be the image of point \(A(3,8)\)
Accordingly, line (1) is the perpendicular bisector of \(A B\)
Slope of \(A B=(b-8) /(a-3)\), while the slope of the line \((1)\) is \(-1 / 3\)
Since line (1) is perpendicular to \(A B\)
\(
\begin{aligned}
& ((b-8) /(a-3))(-1 / 3)=-1 \\
& \Rightarrow(b-8) /(3 a-9)=1 \\
& \Rightarrow b-8=3 a-9 \\
& \Rightarrow 3 a-b=1 \ldots(2)
\end{aligned}
\)
Mid-point of \(\mathrm{AB}=[(\mathrm{a}+3) / 2,(b+8) / 2\)
The mid-point of the line segments AB will also satisfy line (1).
Hence, from equation (1), we have
\(
\begin{aligned}
& (a+3) / 2+3(b+8) / 2=7 \\
& \Rightarrow a+3+3 b+24=14 \\
& \Rightarrow a+3 b=-13 \ldots \text {.(3) }
\end{aligned}
\)
On solving equations (2) and (3), we obtain
\(a=-1\) and \(b=-4\).
Thus, the image of the given point with respect to the given line is \((-1,-\) 4)

Hint

The equations of the given lines are
\(
\begin{aligned}
& y=3 x+1 \ldots \text { (1) } \\
& 2 y=x+3 \ldots \text { (2) } \\
& y=m x+4 \ldots \text { (3) }
\end{aligned}
\)
Slope of line (1), \(m_1=3\), slope of line (2), \(m_2=\frac{1}{2}\), Slope of line (3), \(m_3=m\)
It is given that lines (1) and (2) are equally inclined to line (3). This means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).
\(
\begin{aligned}
& \therefore\left|\frac{m_1-m_3}{1+m_1 m_3}\right|=\left|\frac{m_2-m_3}{1+m_2 m_3}\right| \\
& \Rightarrow\left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{1}{2} m}\right| \\
& \Rightarrow\left|\frac{3-m}{1+3 m}\right|=\left|\frac{1-2 m}{m+2}\right| \\
& \Rightarrow \frac{3-m}{1+3 m}=\pm\left(\frac{1-2 m}{m+2}\right) \\
& \Rightarrow \frac{3-m}{1+3 m}=\frac{1-2 m}{m+2} \text { or } \frac{3-m}{1+3 m}=-\left(\frac{1-2 m}{m+2}\right)
\end{aligned}
\)
If \(\frac{3-m}{1+3 m}=\frac{1-2 m}{m+2}\), then
\(
(3-m)(m+2)=(1-2 m)(1+3 m)
\)
\(
\begin{aligned}
& \Rightarrow-m^2+m+6=1+m-6 m^2 \\
& \Rightarrow 5 m^2+5=0 \\
& \Rightarrow\left(m^2+1\right)=0
\end{aligned}
\)
\(\Rightarrow m=\sqrt{-1}\), which is not real
Hence, this case is not posible.
If \(\frac{3-m}{1+3 m}=-\left(\frac{1-2 m}{m+2}\right)\), then
\(
\begin{aligned}
& \Rightarrow 7 m^2-2 m-7=0 \\
& \Rightarrow m=\frac{2 \pm \sqrt{4-4(7)(-7)}}{2(7)} \\
& \Rightarrow m=\frac{1 \pm 5 \sqrt{2}}{7}
\end{aligned}
\)
Thus, the required value of \(m\) is \(\frac{1 \pm 5 \sqrt{2}}{7}\).

Hint

The equations of the given lines are
\(
\begin{aligned}
& 9 x+6 y-7=0 \dots(1)\\
& 3 x+2 y+6=0 \dots(2)
\end{aligned}
\)
Let \(P(h, k)\) be the arbitrary point which is equidistant from lines \((1)\) and (2).
Then the perpendicular distance of \(P(h, k)\) from the line (1) is given by
\(
\begin{aligned}
& d_1=|9 h+6 k-7| / \sqrt{(9)^2+(6)^2} \\
& =|9 h+6 k-7| / 3 \sqrt{13}
\end{aligned}
\)
And the perpendicular distance of \(P(h, k)\) from the line (2) is given by
\(
\begin{aligned}
& d_2=|3 h+2 k+6| / \sqrt{(3)^2+(2)^2} \\
& =|3 h+2 k+6| / \sqrt{13}
\end{aligned}
\)
Since \(P(h, k)\) is equidistant from lines (1) and (2), \(d_1=d_2\)
Therefore,
\(
\begin{aligned}
& |9 h+6 k-7| / 3 \sqrt{13}=|3 h+2 k+6| / \sqrt{13} \\
& \Rightarrow|9 h+6 k-7|=3|3 h+2 k+6| \\
& \Rightarrow 9 h+6 k-7=\pm 3(3 h+2 k+6)
\end{aligned}
\)
Case I:
\(
\begin{aligned}
& 9 \mathrm{~h}+6 \mathrm{k}-7=3(3 \mathrm{~h}+2 \mathrm{k}+6) \\
& \Rightarrow 9 \mathrm{~h}+6 \mathrm{k}-7-9 \mathrm{~h}-6 \mathrm{k}-18=0
\end{aligned}
\)
\(\Rightarrow 25=0\), which is absurd, hence this case is not possible.
Case II:
\(
\begin{aligned}
& 9 h+6 k-7=-3(3 h+2 k+6) \\
& \Rightarrow 18 h+12 k+11=0
\end{aligned}
\)
Thus, the required equation of the line is \(18 x+12 y+11=0\)

Hint

We can see the scenario of the given problem in the following figure.

Let the coordinates of point \(\mathrm{A}\) be \((\mathrm{a}, \mathrm{O})\).
Draw a line, \(\mathrm{AL}\) perpendicular to the \(x\)-axis
We know that angle of incidence is equal to angle of reflection.
Hence, let \(\angle \mathrm{BAL}=\angle \mathrm{CAL}=\Phi\) and \(\angle \mathrm{CAX}=\theta\)
Now,
\(
\begin{aligned}
& \angle O A B=180^{\circ}-(\theta+2 \Phi) \\
& =180^{\circ}-\left[\theta+2\left(90^{\circ}-\theta\right)\right] \\
& =180^{\circ}-\left[\theta+180^{\circ}-2 \theta\right] \\
& =180^{\circ}-180^{\circ}+\theta =\theta
\end{aligned}
\)
Therefore, \(\angle B A X=180^{\circ}-\theta\)
Now, Slope of line \(A C=(3-0) /(5-a)\)
\(
\begin{aligned}
& \Rightarrow \tan \theta=3 /(5-a) \ldots(1) \\
& \Rightarrow \text { Slope of line } A B=(2-0) /(1-a) \\
& \Rightarrow \tan \left(180^{\circ}-\theta\right)=2 /(1-a) \\
& \Rightarrow \tan \theta=2 /(1-a) \ldots(2)
\end{aligned}
\)
From equations (1) and (2), we obtain
\(
\begin{aligned}
& \Rightarrow 3 /(5-a)=2 /(1-a) \\
& \Rightarrow 3 a-3=10-2 a \\
& \Rightarrow a=13 / 5
\end{aligned}
\)
Thus, the coordinates of point \(\mathrm{A}\) are \((13 / 5,0)\)

Hint

The equations of the given lines are
\(
\begin{aligned}
& 2 x-3 y+4=0 \ldots(1) \\
& 3 x+4 y-5=0 \ldots(2) \\
& 6 x-7 y+8=0 \ldots(3)
\end{aligned}
\)
The person is standing at the junction of the paths represented by lines (1) and (2).
On solving equations (1) and (2), we obtain \(x=-1 / 17\) and \(y=22 / 17\)
Thus, the person is standing at point \((-1 / 17,22 / 17)\)
The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point \((-1 / 17,22 / 17)\)
Now,
Slope of the line \((3)=6 / 7\)
Slope of the line perpendicular to line \((3)=-1 /(6 / 7)=-7 / 6\)
The equation of the line passing through \((-1 / 17,22 / 17)\) and having a slope of \(-7 / 6\) is given by
\(
\begin{aligned}
& (y-22 / 17)=-7 / 6(x+1 / 17) \\
& \Rightarrow 6(17 y-22)=-7(17 x+1) \\
& \Rightarrow 102 y-132=-119 x-7 \\
& \Rightarrow 119 x+102 y=125
\end{aligned}
\)
Hence, the path that the person should follow is \(119 x+102 y=125\)