Solution

Let the intercepts along the $X$ and $Y$-axes are $a$ and a respectively.

$\therefore$ Equation of the line is

$ \frac{x}{a}+\frac{y}{a}=1 $

Since, the point $(1,-2)$ lies on the line,

$ \therefore \frac{1}{a}-\frac{2}{a} =1 $

$ \Rightarrow \frac{1-2}{a} =1 $

$ \Rightarrow a =-1 $

On putting $a=-1$ in Eq. (i), we get

$ \Rightarrow \begin{gathered} \frac{x}{-1}+\frac{y}{-1}=1 \\ x+y=-1 \Rightarrow x+y+1=0 \end{gathered} $

Solution

Consider the given points $A(5,2), B(2,3)$ and $C(3,-1)$.

Slope of the line passing through the points $B$ and $C, m_{B C}=\frac{-1-3}{3-2}=-4$

So, the slope of required line is $\frac{1}{4}$.

Since, the equation of a line passing the point $A(5,2)$ and having slope $\frac{1}{4}$ is $y-2=\frac{1}{4}(x-5)$.

$ \begin{matrix} \Rightarrow & 4 y-8=x-5 \\ \Rightarrow & x-4 y+3=0 \end{matrix} $

Solution

$ \begin{aligned} \text{Given lines,}\quad y & =(2-\sqrt{3})(x+5) \\ \text{Slope of this line,}\quad m_1 & =(2-\sqrt{3}) \\ \text{and}\quad y & =(2+\sqrt{3})(x-7) \\ \text{Slope of this line,}\quad m_2 & =(2+\sqrt{3}) \end{aligned} $

Let $\theta$ be the angle between lines (i) and (ii), then

$\tan \theta =\Big|\frac{m_1-m_2}{1+m_1 m_2}\Big| $

$ \Rightarrow \tan \theta =\Big|\frac{(2-\sqrt{3})-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\Big| \Rightarrow \tan \theta=\Big|\frac{-2 \sqrt{3}}{1+4-3}\Big| $

$ \Rightarrow \tan \theta =\sqrt{3} $

$ \Rightarrow \tan \theta =\tan \pi / 3 $

$ \therefore \theta =\pi / 3=60^{\circ} $

For obtuse angle $=\pi-\pi / 3=2 \pi / 3=120^{\circ}$

Hence, the angle between the lines are $60^{\circ}$ or $120^{\circ}$.

Solution

Let the intercept along the axes be $a$ and $b$.

Given, $\quad a+b=14 \Rightarrow b=14-a$

Now, the equation of line is $\frac{x}{a}+\frac{y}{b}=1$.

$\Rightarrow \quad \frac{x}{a}+\frac{y}{14-a}=1$

Since, the point $(3,4)$ lies on the line.

$ \therefore \frac{3}{a}+\frac{4}{14-a} =1 $

$ \Rightarrow \frac{42-3 a+4 a}{a(14-a)} =1 \Rightarrow 42+a=14 a-a^{2} $

$ \Rightarrow a(a-7)-6(a-7) =0 \Rightarrow(a-7)(a-6)=0 $

$ \Rightarrow a-7 =0 \text { or } a-6=0 $

$ \Rightarrow a =7 \text { or } a=6 $

$\text { When } a =7, \text { then } b=7 $

$\text { When } a =6, \text { then } b=8$

$\therefore$ The equation of line, when $a=7$ and $b=7$ is

$ \frac{x}{7}+\frac{y}{7}=1 \Rightarrow x+y=7 $

So, the equation of line, when $a=6$ and $b=8$ is $\frac{x}{6}+\frac{y}{8}=1$

Solution

Let the required point be $(h, k)$ and point $(h, k)$ lies on the line $x+y=4$ i.e., $\quad h+k=4$

The distance of the point $(h, k)$ from the line $4 x+3 y=10$ is

Taking positive sign,

$ \begin{aligned} \Big|\frac{4 h+3 k-10}{\sqrt{16+9}} \Big|& =1 \\ 4 h+3 k-10 & = \pm 5 \\ 4 h+3 k & =15 \end{aligned} $

From Eq. (i) $h=4-k$ put in Eq. (ii), we get

$ \begin{aligned} & \Rightarrow 4(4-k)+3 k=15 \\ & \Rightarrow 16-4 k+3 k=15 \\ & \text { On putting } k=1 \text { in Eq. (i), we get } \end{aligned} $

So, the point is $(3,1)$.

Taking negative sign,

$ \begin{aligned} & 4 h+3 k-10 =-5 \\ \Rightarrow & 4(4-k)+3 k =5 \\ \Rightarrow & 16-4 k+3 k =5 \\ \therefore & -k=5-16 =-11 \\ \therefore & k =11 \end{aligned} $

On putting $k=11 in \mathrm{Eq.}$ (i), we get

$ h+11=4 \Rightarrow h=-7 $

Hence, the required points are $(3,1)$ and $(-7,11)$.

Solution

Given equation of lines are

$ \frac{x}{a}+\frac{y}{b} =1 $

$ \text { and } \text { Slope, } m_1 =-\frac{b}{a} $

$ \therefore \frac{x}{a}-\frac{y}{b} =1 $

$ \text { Slope, } m_2 =\frac{b}{a} $

Let $\theta$ be the angle between the given lines, then

$ \begin{aligned} & \tan \theta=\Big|\frac{m_1-m_2}{1+m_1 m_2}\Big| \Rightarrow \tan \theta=\Big|\frac{-\frac{b}{a}-\frac{b}{a}}{1+(\frac{-b}{a})(-\frac{b}{a})}\Big| \\ & \Rightarrow \quad \tan \theta=\Big|\frac{\frac{-2 b}{a}}{\frac{a^{2}-b^{2}}{a^{2}}}\Big| \Rightarrow \tan \theta=\frac{2 a b}{a^{2}-b^{2}} \end{aligned} $

Hence proved.

Solution

Given that, angle with $Y$-axis $=30^{\circ}$

and angle with $X$-axis $=60^{\circ}$

$\therefore$ Slope of the line, $m=\tan 60^{\circ}=\sqrt{3}$

So, the equation of a line passing through $(1,2)$ and having slope $\sqrt{3}$, is

$ \begin{matrix} \Rightarrow & y-2 =\sqrt{3}(x-1) \\ \Rightarrow & y-2 =\sqrt{3} x-\sqrt{3} \\ \Rightarrow & y-\sqrt{3} x-2+\sqrt{3} =0 \end{matrix} $

Solution

Given equation of lines

$ \begin{aligned} 2 x+y & =5 \\ x+3 y & =-8 \\ y & =5-2 x \end{aligned} $

From Eq. (i),

Now, put the value of $y$ in Eq. (ii), we get

$ x+3(5-2 x) =-8 $

$ \Rightarrow x+15-6 x =-8 $

$ \Rightarrow -5 x =-23 \Rightarrow x=\frac{23}{5} $

Now, $x=\frac{23}{5}$ put in Eq. (i), we get

$ y=5-\frac{46}{5}=\frac{25-46}{5}=\frac{-21}{5} $

Since, the required line is parallel to the line $3 x+4 y=7$. So, slope of the line is $m=\frac{-3}{4}$.

So, the equation of the line passing through the point $(\frac{23}{5}), (\frac{-21}{5})$ having slope $(\frac{-3}{4})$ is

$ y+\frac{21}{5} =\frac{-3}{4} x\Big(-\frac{23}{5}\Big) $

$ \Rightarrow 4 y+\frac{84}{5} =-3 x+\frac{69}{5} $

$ \Rightarrow 3 x+4 y =\frac{84-69}{5} \Rightarrow 3 x+4 y+\frac{15}{5}=0 $

$ \Rightarrow 3 x+4 y+3 =0 $

Solution

Given equation of line

$ \begin{matrix} a x+b y+8=0 \\ \\ \Rightarrow\frac{x}{\frac{-8}{a}}+\frac{y}{\frac{-8}{b}}=1 \end{matrix} $

So, the intercepts are $\frac{-8}{a}$ and $\frac{-8}{b}$.

and another given equation of line is $2 x-3 y+6=0$.

$\Rightarrow \quad \frac{x}{-3}+\frac{y}{2}=1$

So, the intercepts are -3 and 2 .

According to the question,

$ \begin{aligned} & \frac{-8}{a} =3 \text { and } \frac{-8}{b}=-2 \\ \therefore & a =-\frac{8}{3}, b=4 \end{aligned} $

Solution

Let intercept of a line are $(h, k)$.

The coordinates of $A$ and $B$ are $(h, 0)$ and $(0, k)$ respectively.

$ \begin{aligned} & -5=\frac{1 \times 0+2 \times h}{1+2} \\ & \therefore \quad-5=\frac{2 h}{3} \Rightarrow+h=-\frac{15}{2} \\ & \text { and } \quad 4=\frac{1 \cdot k+0 \cdot 2}{1+2} \\ & \Rightarrow \quad k=12 \\ & \therefore \quad A=(-\frac{15}{2}, 0) \text { and } B=(0,12) \end{aligned} $

Hence, the equation of a line $A B$ is

$ \begin{aligned} & & y-0=\frac{12-0}{0+15 / 2} \Big(x+\frac{15}{2}\Big) \\ \Rightarrow & y =\frac{12 \cdot 2}{15} \Big(x+\frac{15}{2}\Big) \\ \Rightarrow & 5 y =8 x+60 \Rightarrow 8 x-5 y+60=0 \end{aligned} $

Solution

Given that, $O C=P=4$ units

$ \begin{aligned} & \angle B A X=120^{\circ} \\ & \text { Let } \quad \angle C O A=\alpha, \angle O C A=90^{\circ} \\ & \because \quad \angle B A X=\angle C O A+\angle O C A \quad \text { [exterior angle property] } \\ & \Rightarrow \quad 120^{\circ}=\alpha+90^{\circ} \\ & \therefore \quad \alpha=30^{\circ} \\ & \text{ Now, the equation of required line is} \\ & x \cos 30^{\circ}+y \sin 30^{\circ}=4 \\ & \Rightarrow \quad x \cdot \frac{\sqrt{3}}{2}+y \cdot \frac{1}{2}=4 \\ & \Rightarrow \sqrt 3 x+y=8 \end{aligned} $

Solution

Let slope of line $A C$ be $m$ and slope of line $B C$ is $\frac{-3}{4}$ and let angle between line $A C$ and $B C$ be $\theta$.

$ \therefore \tan \theta=\Big|\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\Big| \Rightarrow \tan 45^{\circ} = \pm \Big[\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}} \Big]$

$ \text { Taking positive sign, } 1 =\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}} $

$ \Rightarrow m+\frac{3}{4} =1-\frac{3 m}{4}$

$ \Rightarrow m+\frac{3 m}{4} =1-\frac{3}{4} $

$ \Rightarrow \frac{7 m}{4} =\frac{1}{4} \Rightarrow m=\frac{1}{7} $

Taking negative sign,

$ \begin{aligned} & \quad 1=-\Big(\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\Big) \Rightarrow 1-\frac{3 m}{4}=-m-\frac{3}{4} \\ & \Rightarrow \quad m-\frac{3 m}{4}=-1-\frac{3}{4} \\ & \Rightarrow \quad \frac{m}{4}=\frac{-7}{4} \Rightarrow m=-7 \\ & \begin{matrix} \therefore \text { Equation of side } A C \text { having slope } (\frac{1}{7}) \text { is } \end{matrix} \\ & \quad y-2=\frac{1}{7}(x-2) \\ & \Rightarrow \quad 7 y-14=x-2 \\ & \text { and equation of side } A B \text { having slope }(-7) \text { is } \\ & \Rightarrow \quad y-2=-7(x-2) \\ & \Rightarrow \quad y-2=-7 x+14 \\ & \Rightarrow \quad 7 x+y-16=0 \end{aligned} $

Solution

Given that, equilateral $\triangle A B C$ having equation of base is $x+y=2$.

$\ln \triangle A B D$,

$ \begin{aligned} \sin 60^{\circ} & =\frac{A D}{A B} \\ A D & =A B \sin 60^{\circ}=A B \frac{\sqrt{3}}{2} \\ \Rightarrow A D & =A B \frac{\sqrt{3}}{2} \end{aligned} $

Now, the length of perpendicular from $(2,-1)$ to the line $x+y=2$ is given by

$AD = \Big| \frac{2+(-1)-2}{\sqrt {1^{2}+1^{2}}}\Big| = \frac{1}{\sqrt 2}$

From Eq. (i),

$ \begin{aligned} & \frac{1}{\sqrt{2}}=A B \frac{\sqrt{3}}{2} \\ & A B=\sqrt{\frac{2}{3}} \end{aligned} $

Solution

Let slope of the line be $m$ and the coordinates of fixed point $P$ are $(x_1, y_1)$.

$\therefore \quad$ Equation of line is $y-y_1=m(x-x_1)$

Since, the given points are $A(2,0), B(0,2)$ and $C(1,1)$.

Now, perpendicular distance from $A$, is

$ \frac{0-y_1-m(2-x_1)}{\sqrt{1+m^{2}}} $

Perpendicular distance from $B$, is

$ \frac{2-y_1-m(0-x_1)}{\sqrt{1+m^{2}}} $

Perpendicular distance from $C$, is

$ \frac{1-y_1-m(1-x_1)}{\sqrt{1+m^{2}}}$

$ \text { Now, } \frac{-y_1-2 m+m x_1+2-y_1+m x_1+1-y_1-m+m x_1}{\sqrt{1+m^{2}}}=0 $

$ \Rightarrow -3 y_1-3 m+3 m x_1+3=0 $

$ \Rightarrow -y_1-m+m x_1+1=0 $

Since, $(1,1)$ lies on this line. So, the point $P$ is $(1,1)$.

Solution

Let slope of the line be $m$. As, the line passes through the point $A(1,2)$.

$\therefore \quad$ Equation of line is $y-2=m(x-1)$

$ \begin{aligned} m x-y+2-m & =0 \\ x+v-4 & =0 \end{aligned} $

and

$ \frac{x}{(4-2+m)}=\frac{y}{2-m+4 m}=\frac{1}{1+m} $

$ \Rightarrow \quad \frac{x}{2+m}=\frac{y}{3 m+2}=\frac{1}{1+m} $

$\Rightarrow \quad x=\frac{2+m}{1+m}$

$ y=\frac{3 m+2}{1+m} $

So, the point of intersection is $B \Big(\frac{m+2}{m+1}, \frac{3 m+2}{m+1}\Big)$.

$ \begin{aligned} & \text { Now, } \\ & A B^{2}=\Big(\frac{m+2}{m+1}-1\Big)^{2}+\Big(\frac{3 m+2}{m+1}-2\Big)^{2} \\ & \because \quad A B=\frac{\sqrt{6}}{3} \\ & \therefore \quad \Big(\frac{m+2-m-1}{m+1}\Big)^{2}+\Big(\frac{3 m+2-2 m-2}{m+1}\Big)^{2}=\frac{6}{9} \\ & \Rightarrow \quad \Big(\frac{1}{m+1}\Big)^{2}+\Big(\frac{m}{m+1}\Big)^{2}=\frac{6}{9} \\ & \Rightarrow \quad \frac{1+m^{2}}{(1+m)^{2}}=\frac{6}{9} \\ & \Rightarrow \quad \frac{1+m^{2}}{1+m^{2}+2 m}=\frac{6}{9} \\ & \Rightarrow \quad 9+9 m^{2}=6+6 m^{2}+12 m \\ & \Rightarrow \quad 3 m^{2}-12 m+3=0 \\ & \Rightarrow \quad m^{2}-4 m+1=0 \\ & \therefore \quad m=\frac{4 \pm \sqrt{16-4}}{2} \\ & =2 \pm \sqrt{3} \\ & =2+\sqrt{3} \text { or } 2-\sqrt{3} \\ & \theta=75^{\circ} \text { or } 15^{\circ} \end{aligned} $

Solution

Since, the intercept form of a line is $\frac{x}{a}+\frac{y}{b}=1$.

Given that,

$\because \frac{1}{a}+\frac{1}{b}=\text { constant }$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{k}$

$\Rightarrow \frac{k}{a}+\frac{k}{b}=1$

So, $(k, k)$ lies on $\frac{x}{a}+\frac{y}{b}=1$.

Hence, the line passes through the fixed point.

Solution

Since, the line intersects $X$ and $Y$-axes respectively at $A(x, 0)$ and $B(0, y)$.

$\begin{aligned}-4 & =\frac{5 \times 0+3 x}{5+3} \\ \Rightarrow -4 & =\frac{3 x}{8} \Rightarrow x=\frac{-32}{3} \\ \text{and}\quad 3 & =\frac{5 \cdot y+3 \cdot 0}{5+3} \\ \Rightarrow 3 & =\frac{5 y}{8} \Rightarrow y=\frac{24}{5}\end{aligned}$

Since, the intercept on the $X$ and $Y$-axes respectively are $a=\frac{-32}{3}$ and $b=\frac{24}{5}$.

$\therefore$ Equation of required line is

$ \begin{aligned} & \frac{x}{-32 / 3}+\frac{y}{24 / 5} & =1 \\ \Rightarrow & \frac{-3 x}{32}+\frac{5 y}{24} & =1 \\ \Rightarrow & 9 x-20 y+96 & =0 \end{aligned} $

Solution

Given equation of lines

$ \begin{aligned} x-y+1 & =0 \\ 2 x-3 y+5 & =0 \\ x & =y-1 \end{aligned} $

Now, put the value of $x$ in Eq. (ii), we get

$ \begin{matrix} \Rightarrow & 2(y-1)-3 y+5 =0 \\ \Rightarrow & 2 y-2-3 y+5 =0 \\ y=3 \text { put in Eq. (i), we get } & 3-y=0 \Rightarrow y =3 \\ & x =2 \end{matrix} $

Since, the point of intersection is $(2,3)$.

Let slope of the required line be $m$.

$\therefore$ Equation of line is

$ y-3=m(x-2) $

$\Rightarrow$ $ m x-y+3-2 m=0 $

Since, the distance from $(3,2)$ to line (iii) is $\frac{7}{5}$.

$ \begin{aligned} & \therefore \quad \frac{7}{5}=\Big|\frac{3 m-2+3-2 m}{\sqrt{1+m^{2}}}\Big| \\ & \Rightarrow \quad \frac{49}{25}=\frac{(m+1)^{2}}{1+m^{2}} \\ & \Rightarrow \quad 49+49 m^{2}=25(m^{2}+2 m+1) \\ & \Rightarrow \quad 49+49 m^{2}=25 m^{2}+50 m+25 \\ & \Rightarrow \quad 24 m^{2}-50 m+24=0 \\ & \Rightarrow \quad 12 m^{2}-25 m+12=0 \\ & \therefore \quad m=\frac{25 \pm \sqrt{625-4 \cdot 12 \cdot 12}}{24} \\ & =\frac{25 \pm \sqrt{49}}{24}=\frac{25 \pm 7}{24}=\frac{32}{24} \text { or } \frac{18}{24}=\frac{4}{3} \text { or } \frac{3}{4} \\ & \therefore \text { First equation of a line is } \quad y-3=\frac{4}{3}(x-2) \\ & \begin{matrix} \Rightarrow & 3 y-9=4 x-8 \end{matrix} \\ & \Rightarrow \quad 4 x-3 y+1=0 \\ & \text { and second equation of line is } \quad y-3=\frac{3}{4}(x-2) \\ & \Rightarrow \\ & 4 y-12=3 x-6 \\ & \Rightarrow \\ & 3 x-4 y+6=0 \end{aligned} $

Solution

Let the coordinates of moving point $P$ be $(x, y)$. Given that, the sum of distances of this point in a plane from the axes is 1 .

$ \begin{matrix} \therefore & |x|+|y|=1 \\ \Rightarrow & \pm x \pm y=1 \\ \Rightarrow & x+y=1 \\ \Rightarrow & -x-y=1 \\ \Rightarrow & -x+y=1 \\ \Rightarrow & x-y=1 \end{matrix} $

So, these equations give us locus of the point which is a square.

Solution

Given equation of lines are

$ \begin{aligned} y-\sqrt{3} x & =2 \\ y+\sqrt{3} x & =2 \\ y & =\sqrt{3} x+2 \\ y & =-\sqrt{3} x+2 \\ \sqrt{3} x+2 & =-\sqrt{3} x+2 \end{aligned} $

$ [\because x \geq 0] $

$ \Rightarrow \quad 2 \sqrt{3} x=0 \Rightarrow x=0 $

On putting $x=0$ in Eq. (i), we get

So, the point of intersection of line (i) and (ii) is $(0,2)$.

Here,

$ \begin{aligned} O C & =2 \\ \text{In $\triangle D E C$}\quad \frac{C D}{C E} & =\cos 30^{\circ} \\ C D =5 \cos 30 \\ & =5 \cdot \frac{\sqrt{3}}{2} \end{aligned} $

$ \begin{aligned} \therefore & C D =5 \cos 30^{\circ} \\ & =5 \cdot \frac{\sqrt{3}}{2} \\ \Rightarrow & O D =O C+C D=2+5 \frac{\sqrt{3}}{2} \end{aligned} $

So, the coordinates of the foot of perpendiculars are $\Big(0,2+\frac{5 \sqrt{3}}{2}\Big)$.

Solution

Given equation of line is,

$ \frac{x}{a}+\frac{y}{b}=1 $

Perpendicular length from the origin on the line (i) is given by $p$

$ \begin{matrix} \text { i.e., } & p=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{a b}{\sqrt{a^{2}+b^{2}}} \\ \therefore \quad p^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \end{matrix} $

Given that, $a^{2}, p^{2}$ and $b^{2}$ are in AP.

$ \begin{matrix} \therefore & 2 p^{2}=a^{2}+b^{2} \\ \Rightarrow & \frac{2 a^{2} b^{2}}{a^{2}+b^{2}}=a^{2}+b^{2} \\ \Rightarrow & 2 a^{2} b^{2}=(a^{2}+b^{2})^{2} \\ \Rightarrow & 2 a^{2}+b^{2}=a^{4}+b^{4}+2 a^{2} b^{2} \\ \Rightarrow & a^{4}+b^{4}=0 \end{matrix} $

Solution

(a) Given that,

$ c=-3 \text { and } m=\frac{3}{5} $

$\therefore$ Equation of the line is $y=m x+c$

$ y =\frac{3}{5} x-3$

$ \Rightarrow 5 y =3 x-15 $

$ \Rightarrow 5 y-3 x+15 =0 $

Solution

(a) Let equation of line be

$ \begin{aligned} & \frac{x}{a}+\frac{y}{a}=1 \\ & x+y=a \end{aligned} $

$ \begin{aligned} \Rightarrow & & x+y & =a \\ \Rightarrow & & y & =-x+a \\ \therefore & & \text { Required slope } & =-1 \end{aligned} $

Solution

(b) Since, line passes through the point $(3,2)$ and perpendicular to the line $y=x$.

$\because$ Slope $(m)=-1 \quad$ [since, line is perpendicular to the line $y=x]$

$\therefore \quad$ Equation of line which passes through $(3,2)$ is

$ \begin{aligned} & y-2=-1(x-3) \\ & y-2=-x+3 \end{aligned} $

$ \Rightarrow x+y=5 $

Solution

(b) Given point is $(1,2)$ and slope of the required line is 1 .

$\because x+y+1=0 \Rightarrow y=-x-1 \Rightarrow m_1=-11$

$\therefore$ slope of the line $=\frac{-1}{-1}=1$

$\therefore$ Equation of required line is

$ y-2 =1(x-1) $

$ \Rightarrow y-2 =x-1 $

$ \Rightarrow y-x-1 =0 $

Solution

(c) Since, intercepts on the axes are $a,-b$ then equation of the line is $\frac{x}{a}-\frac{y}{b}=1$.

$ \begin{matrix} \Rightarrow & \frac{y}{b}=\frac{x}{a}-1 \\ \Rightarrow & y=\frac{b x}{a}-b \end{matrix} $

So, the lope of this line i.e., $m_1=\frac{b}{a}$.

Also, for intercepts on the axes as $b$ and $-a$, then equation of the line is

$ \Rightarrow \quad \begin{aligned} \frac{x}{b}-\frac{y}{a} & =1 \\ \frac{y}{a} & =\frac{x}{b}-1 \Rightarrow y=\frac{a}{b} x-a \end{aligned} $

and slope of this line i.e., $m_2=\frac{a}{b}$

$ \therefore \quad \tan \theta=\frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{a}{b} \cdot \frac{b}{a}}=\frac{\frac{b^{2}-a^{2}}{a b}}{2}=\frac{b^{2}-a^{2}}{2 a b} $

Solution

(d) Given, line is $\frac{x}{a}+\frac{y}{b}=1$

Since, the points $(2,-3)$ and $(4,-5)$ lies on this line.

$ \begin{matrix} \therefore & \frac{2}{a}-\frac{3}{b}=1 \\ \\ \text { and } & \frac{4}{a}-\frac{5}{b}=1 \end{matrix} $

On multiplying by 2 in Eq. (ii) and then subtracting Eq. (iii) from Eq. (ii), we get

$ -\frac{6}{b}+\frac{5}{b} =1 $

$ \Rightarrow \frac{-1}{b} =1 $

$ \therefore b =-1 $

On putting $b=-1$ in Eq. (ii), we get

$ \begin{aligned} \Rightarrow & \frac{2}{a}+3 =1 \\ \therefore & \frac{2}{a} =-2 \Rightarrow a=-1 \\ & (a, b) =(-1,-1) \end{aligned} $

Solution

(a) Given equation of lines

and $\quad \begin{aligned} 2 x-3 y+5 & =0 \\ 3 x+4 y & =0\end{aligned}$

From Eq. (ii), put the value of $x=\frac{-4 y}{3}$ in Eq. (i), we get

$ 2 \Big(\frac{-4 y}{3}\Big)-3 y+5 =0 $

$ \Rightarrow -8 y-9 y+15 =0 $

$ \Rightarrow y =\frac{15}{17} $

From Eq. (ii),

$ \begin{aligned} & 3 x+4 \cdot \frac{15}{17}=0 \\ & \Rightarrow x=\frac{-60}{17 \cdot 3}=\frac{-20}{17} \end{aligned} $

So, the point of intersection is $\frac{-20}{17}, \frac{15}{17}$

$\therefore$ Required distance from the line $5 x-2 y=0$ is,

$ d=\frac{|-5 \times \frac{20}{17}-2 \Big(\frac{15}{17}\Big)|}{\sqrt{25+4}}=\frac{|\frac{-100}{17}-\frac{30}{17}|}{\sqrt{29}}=\frac{130}{17 \sqrt{29}} $

$ \because \text { distance of a point } p(x_1, y_1) \text { from the line } a x+b y+c=0 \text { is } d=\frac{|a x_1+b y_1+c|}{\sqrt{a^{2}+b^{2}}} $

Solution

(a) So, the given point $A$ is $(3,-2)$.

So, the equation of line $\sqrt{3} x+y=1$.

$ \begin{aligned} \Rightarrow & y =-\sqrt{3} x+1 \\ \therefore & \text { Slope, } m_1 =-\sqrt{3} \end{aligned} $

Let slope of the required line be $m_2$.

$ \begin{aligned} & \therefore \quad \tan \theta=\Big|\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\Big| \quad \Big[\because \tan \theta=|\frac{m_1-m_2}{1+m_1 m_2}|\Big] \\ & \Rightarrow \quad \tan 60^{\circ}= \pm \Big(\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\Big) \\ & \begin{matrix} \Rightarrow & \sqrt{3}=\Big(\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\Big) \end{matrix} \quad \text { [taking positive sign] } \\ & \Rightarrow \quad \sqrt{3}-3 m_2=-\sqrt{3}-m_2 \\ & \Rightarrow \quad 2 \sqrt{3}=2 m_2 \\ & \Rightarrow \quad m_2=\sqrt{3} \end{aligned} $

$\therefore \quad$ Equation of line passing through $(3,-2)$ is

$ y+2=\sqrt{3}(x-3) $

$ \Rightarrow \sqrt{3} x-y-2-3 \sqrt{3} =0 $

$ \Rightarrow \sqrt{3}-3 m_2 =\sqrt{3}+m_2 $

$ \Rightarrow m_2 =0 $

$\therefore$ The equation of line is $y+2=0(x-3)$

$\Rightarrow \quad y+2=0$

So, the required equation of lines are $\sqrt{3} x-y-2-3 \sqrt{3}=0$ and $y+2=0$.

Solution

(a) Let slope of the line be $m$.

$\because \quad$ Equation of line passing through $(1,0)$ is

$ \begin{aligned} &\Rightarrow y-0=m(x-1) \\ & y-m x+m=0 \end{aligned} $

Since, the distance from origin is $\frac{\sqrt{3}}{2}$.

Then,

$ \begin{aligned} \Rightarrow & \frac{\sqrt{3}}{2} =\frac{m}{\sqrt{1+m^{2}}} \\ \Rightarrow & \frac{3}{4} =\frac{m^{2}}{1+m^{2}} \\ \Rightarrow & 3+3 m^{2} =4 m^{2} \\ \Rightarrow & m^{2} =3 \\ \Rightarrow & m = \pm \sqrt{3} \end{aligned} $

So, the first equation of line is

$ \begin{aligned} & & y=\sqrt{3}(x-1) \\ \Rightarrow & & \sqrt{3} x-y-\sqrt{3}=0 \end{aligned} $

and the second equation of line is

$ \begin{aligned} & & y=-\sqrt{3}(x-1) \\ \Rightarrow & & \sqrt{3} x+y-\sqrt{3}=0 \end{aligned} $

Solution

(b) Given, equation of the lines are and $\quad \begin{aligned} y & =m x+c_1 \\ y & =m x+c_2\end{aligned}$

$\therefore$ Distance between them is given by

$ d=\frac{|c_1-c_2|}{\sqrt{1+m^{2}}} $

Solution

(b) Given, equation of the line is

$ y=3 x+4 $

$\therefore$ Slope of this line, $m_1=3$

So, the slope of line $O P$ is $-\frac{1}{3}$.

$\therefore$ Equation of line $O P$ is

$ \begin{matrix} & y-3 =-\frac{1}{3}(x-2) \\ \Rightarrow & 3 y-9 =-x+2 \\ \Rightarrow & x+3 y-11 =0 \end{matrix} $

Using the value of $y$ from Eq. (i) in Eq. (ii), we get

$ x+3(3 x+4)-11=0 $

$ \begin{matrix} \Rightarrow & x+9 x+12-11=0 \\ \Rightarrow & 10 x+1=0 \Rightarrow x=-\frac{1}{10} \end{matrix} $

Put $x=\frac{-1}{10}$ in Eq. (i), we get

$ y=\frac{-3}{10}+4=\frac{-3+40}{10}=\frac{37}{10} $

So, the foot of perpendicular is $\Big(-\frac{1}{10}, \frac{37}{10}\Big)$.

Solution

(a) Since, the coordinates of the middle point are $P(3,2)$.

$ \begin{aligned} & \therefore \quad 3=\frac{1 \cdot 0+1 \cdot a}{1+1} \\ & \Rightarrow \quad 3=\frac{a}{2} \Rightarrow a=6 \\ & \text { Similarly, } \quad b=4 \\ & \therefore \quad \text { Equation of the line is } \frac{x}{6}+\frac{y}{4}=1 \\ & \Rightarrow \quad 2 x+3 y=12 \end{aligned} $

Solution

(c) Since, the line passes through $(1,2)$ and parallel to the line $y=3 x-1$. So, slope of the required line $m=3$. Hence, the equation of line is

$ y-2=3(x-1) $

Solution

(a) Equation of $O B$ is

$ y-0 =\frac{1-0}{1-0}(x-0) $

$ \Rightarrow y =x $

$ \text { and equation of } A C \text { is } $

$ \Rightarrow y-0 =\frac{1-0}{0-1}(x-1) $

$ \Rightarrow y+y-1 =0 $

Solution

(b) Equation of straight lines are

$ \begin{aligned} \qquad y & =m x+c, \text { parameter }=2 \\ \frac{x}{a}+\frac{y}{b} & =1, \text { parameter }=2 \\ y-y_1 & =m(x-x_1), \text { parameter }=2 \\ \text { and } \quad x \cos w & +y \sin w=p, \text { parameter }=2 \end{aligned} $

It is clear that from Eqs. (i), (ii), (iii) and (iv), for specifying a straight line clearly two parameters should be known.

Solution

(b) Let the reflection of $A(4,1)$ in $y=x$ is $B(h, k)$.

Now, mid-point of $A B$ is $\Big(\frac{4+h}{2}, \frac{1+k}{2}\Big)$ which lies on $y=x$.

i.e., $\quad \frac{4+h}{2}=\frac{1+k}{2} \Rightarrow h-k=-3$

So, the slope of line $y=x$ is 1 .

$ \text { Slope of } A B =\frac{h-4}{k-1} $

$ \Rightarrow 1 \cdot \frac{h-4}{k-1} =-1$

$ \Rightarrow h-4 =1-k $

$ \Rightarrow h+k =5 $

$ \text { and } 2 h =2 \Rightarrow h=1 $

On putting $h=1$ in Eq. (ii), we get

$ k=4 $

So, the point is $(1,4)$.

Hence, after translation the point is $(1+2,4)$ or $(3,4)$.

Solution

(c) The given equation of lines are

$ 4 x+3 y+10 =0 $

$ \Rightarrow 5 x-12 y+26 =0 $

$ \Rightarrow 7 x+24 y-50 =0 $

Let the point $(h, k)$ which is equidistant from these lines.

Distance from line (i) $=\frac{|4 h+3 k+10|}{\sqrt{16+9}}$

Distance from line (ii) $=\frac{|5 h-12 k+26|}{\sqrt{25+144}}$

Distance from the line (iii) $=\frac{|7 h+24 k-50|}{\sqrt{7^{2}+24^{2}}}$

So, the point $(h, k)$ is equidistant from lines (i), (ii) and (iii).

$ \begin{matrix} \therefore & \frac{4 h+3 k+10}{\sqrt{16+9}}=\frac{5 h-12 k+26}{\sqrt{25+144}}=\frac{7 h+24 k-50}{\sqrt{49+576}} \\ \\ \Rightarrow & \frac{|4 h+3 k+10|}{5}=\frac{|5 h-12 k+26|}{13}=\frac{|7 h+24 k-50|}{25} \end{matrix} $

Clearly, if $h=0, k=0$, then $\frac{10}{5}=\frac{26}{13}=\frac{50}{25}=2$

Hence, the required point is $(0,0)$.

Solution

(d) Given line is $y=3-3 x$.

Then, slope of the required line $=\frac{1}{3}$

$\because \quad$ Equation of the required line is

$ y-2 =\frac{1}{3}(x-2) $

$ \Rightarrow 3 y-6 =x-2 $

$ \Rightarrow x-3 y+4 =0 $

$ \text { For y-intercept, } \text { put } x =0, $

$ \Rightarrow 0-3 y+4 =0$

$ y =\frac{4}{3} $

Solution

(b) Let point $A(x_1, y_1)$ lies on the line $3 x+4 y+5=0$, then $3 x_1+4 y_1+5=0$

Now, perpendicular distance from $A$ to the line

$ \begin{aligned} 3 x+4 y+2 & =0 \\ \Rightarrow \quad & \frac{|3 x_1+4 y_1+2|}{\sqrt{9+16}}=\frac{|-5-2|}{\sqrt{9+16}}=\frac{-7}{5} \end{aligned} $

Let point $B(x_2, y_2)$ lies on the line $3 x+4 y-5=0$ i.e., $3 x_2+4 y_2-5=0$. Now, perpendicular distance from $B$ to the line $3 x+4 y+2=0$,

$ \frac{|3 x_2+4 y_2+2|}{\sqrt{9+16}}=\frac{|+5-2|}{\sqrt{9+16}}=\frac{3}{5} $

Hence, the required ratio is $\frac{3}{5}: \frac{7}{5}$ i.e., $3: 7$.

Solution

(c) Let $A B C$ be the equilateral triangle with vertex $A(h, k)$. Let the coordinates of $D$ are $(\alpha, \beta)$.

We know that, $2: 1$ from the vertex $A$.

$ \begin{matrix} \therefore & 0 =\frac{2 \alpha+h}{3} \text { and } 0=\frac{2 \beta+k}{3} \\ \\ \Rightarrow & 2 \alpha =-h \\ \\ \text { and } & 2 \beta =-k \\ \\ \text { Also, } D(\alpha, \beta) \text { lies on the line } & x+y-2=0 . \\ \\ \therefore & \alpha+\beta-2=0 \\ \\ & A D \perp B C \end{matrix} $

Since, the slope of line $B C$ i.e., $m_{B C}=-1$

and slope of the line $A G$ i.e., $m_{A G}=\frac{k-0}{h-0}=\frac{k}{h}$

$ \begin{aligned} \Rightarrow & (-1) \cdot \Big(\frac{k}{h}\Big) & =-1 \\ \Rightarrow & h =k \end{aligned} $

From Eqs. (i) and (iii),

$ \begin{aligned} & \qquad \begin{aligned} 2 \alpha & =-h \text { and } 2 \beta=-h \\ \therefore & \alpha=\beta \end{aligned} \\ & \text { From Eq. (ii), } \\ & \text { If } \alpha=1 \text {, then } \beta=1 \\ & \text { From Eq. (i), } h=-2, k=-2 \\ & \text { So, the vertex } A \text { is }(-2,-2) . \end{aligned} $

Solution

Given line is

$ a x+b y+c=0 $

Since, $a, b$ and $c$ are in AP, then

$ b=\frac{a+c}{2} $

$ \Rightarrow \quad a-2 b+c=0 $

On comparing Eqs. (i) and (ii), we get

$ x=1, y=2 $

So, $(1,-2)$ lies on the line.

Solution

Let equation of line is

$ \frac{x}{a}+\frac{y}{a}=1 $

Since, this line passes through $(1,-2)$.

$ \Rightarrow \quad \begin{aligned} \frac{1}{a}-\frac{2}{a} & =1 \\ 1-2 & =a \Rightarrow a=-1 \end{aligned} $

$\therefore$ Required equation of the line is

$ -x-y=1 $

$ \Rightarrow \quad x+y+1=0 $

Solution

Since, the given point $P(3,2)$ and line is $x-2 y=3$.

Slope of this line is $m_1=\frac{1}{2}$

Let the slope of the required line is $m$.

$ \begin{matrix} \text { Then, } & \tan \theta=\Big|\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}\Big| \\ \\ \Rightarrow & 1= \pm \Big(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\Big) \end{matrix} $

Taking positive sign,

$ 1+\frac{m}{2}=m-\frac{1}{2} $

$ \begin{aligned} \Rightarrow m-\frac{m}{2} & =1+\frac{1}{2} \\ \Rightarrow\frac{m}{2} & =\frac{3}{2} \Rightarrow m=3 \end{aligned} $

Taking negative sign,

$ \begin{aligned} & & 1=-\Big(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\Big) \\ \Rightarrow & 1+\frac{m}{2} =-m+\frac{1}{2} \\ \Rightarrow & m+\frac{m}{2} =\frac{1}{2}-1 \\ \Rightarrow & \frac{3 m}{2} =\frac{-1}{2} \Rightarrow m=\frac{-1}{3} \end{aligned} $

$\therefore \quad$ First equation of the line is

$ y-2 =3(x-3) $

$ 3 x-y-7 =0 $

and second equation of the line is

$ \begin{matrix} y-2 & =-\frac{1}{3}(x-3) \\ \Rightarrow & 3 y-6 & =-x+3 \\ \Rightarrow & x+3 y-9 & =0 \end{matrix} $

Solution

$ \begin{gathered} \text{Given line is}\quad 3 x-4 y-8=0 \\ \text{For point (3,4),\quad}9-4 \cdot 4-8 \\ \Rightarrow 9-16-8 \\ \Rightarrow 9-24 \\ \Rightarrow -15<0 \end{gathered} $

For point $(2,-6)$,

$ \begin{aligned} & 6+24-8 \\ & 22>0 \end{aligned} $

Since, the value are of opposite sign.

Hence, the points $(3,4)$ and $(2,-6)$ lies on opposite side to the line.

Solution

Let the coordinaters of the point are $(h, k)$,

$\therefore \quad$ Distance between $(3,-2)$ and $(h, k)$,

$ d_1^{2}=(3-h)^{2}+(-2-k)^{2} $

Now, distance of the point $(h, k)$ from the line $5 x-12 y=3$ is,

$ d_2=\Big|\frac{5 h-12 k-3}{\sqrt{25+144}}\Big|=\Big|\frac{5 h-12 k-3}{13}\Big| $

Given that, $ d_1^{2}=d_2 $

$ \begin{aligned} & \Rightarrow \quad(3-h)^{2}+(2+k)^{2}=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad 9-6 h+h^{2}+4+4 k+k^{2}=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad h^{2}+k^{2}-6 h+4 k+13=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad 13 h^{2}+13 k^{2}-78 h+52 k+169=5 h-12 k-3 \\ & \Rightarrow \quad 13 h^{2}+13 k^{2}-83 h+64 k+172=0 \end{aligned} $

$\therefore$ Locus of this point is

$ \quad 13 x^{2}+13 y^{2}-83 x+64 y+172=0 $

Solution

Given equation of the line is

$ x \sin \theta+y \cos \theta=p $

Let the mid-point of $A B$ is $p(h, k)$.

So, the mid-point of $A B$ are

$ \quad\Big(\frac{a}{2}, \frac{b}{2}\Big) $

Since, the point $(a, 0)$ lies on the line (i), then

$ a \sin \theta+0=p $

$ \Rightarrow \quad a \sin \theta=p \Rightarrow a=\frac{p}{\sin \theta} $

and the point $(0, b)$ also lies on the line, then

$0+b\cos \theta=p$

$ \Rightarrow \quad b \cos \theta=p \Rightarrow b=\frac{p}{\cos \theta} $

Now $ \text { mid-point of } A B=\Big(\frac{a}{2}, \frac{b}{2}\Big) \text { or } \Big(\frac{p}{2 \sin \theta}, \frac{p}{2 \cos \theta}\Big) $

$\because$ $ \frac{p}{2 \sin \theta}=h \Rightarrow \sin \theta=\frac{p}{2 h} $

and

$ \frac{p}{2 \cos \theta}=k \Rightarrow \cos \theta=\frac{p}{2 k} $

$\therefore \quad \sin ^{2} \theta+\cos ^{2} \theta=\frac{p^{2}}{4 h^{2}}+\frac{p^{2}}{4 k^{2}}$

$ \Rightarrow \quad 1=\frac{p^{2}}{4}\Big( \frac{1}{h^{2}}+\frac{1}{k^{2}}\Big) $

Locus of the mid-point is

$ \begin{aligned} 4 & =p^{2} \Big(\frac{1}{x^{2}}+\frac{1}{y^{2}}\Big) \\ \Rightarrow \quad 4 x^{2} y^{2} & =p^{2}\Big(x^{2}+y^{2}\Big) \end{aligned} $

Solution

True

We know that, if the vertices of a triangle have integral coordinates, then the triangle cannot be equilateral. Hence, the given statement is true.

Since, in equilatteral triangle, we get $\tan 60^{\circ}=\sqrt{3}=$ Slope of the line, so with integral coordinates as vertices, the triangle cannot be equilateral.

Solution

False

Given points are $A(-2,1), B(0,5)$ and $C(1,2)$.

Now, $ \text { slope of } A B=\frac{5-1}{0+2}=2 $

Slope of

$ \begin{aligned} & B C=\frac{2-5}{-1-0}=3 \\ & A C=\frac{2-1}{-1+2}=1 \end{aligned} $

Slope of

Since, the slopes are different.

Hence, $A, B$ and $C$ are not collinear. So, statement is false.

Solution

False

Given point $p(a \cos ^{3} \theta, a \sin ^{3} \theta)$ and the line is $x \sec \theta+y cosec \theta=a$

$\because \quad$ Slope of this line $=\frac{-\sec \theta}{cosec \theta}=-\tan \theta$

and

$ \text { slope of required line }=\frac{1}{\tan \theta}=\cot \theta $

$\therefore \quad$ Equation of the required line is

$ \begin{matrix} \Rightarrow & y-a \sin ^{3} \theta=\cot \theta(x-a \cos ^{3} \theta) \\ \Rightarrow & y \sin \theta-a \sin ^{4} \theta=x \cos \theta-acos^{4} \theta \\ \Rightarrow & x \cos \theta-y \sin \theta=a \cos ^{4} \theta-asin^{4} \theta \\ \Rightarrow & x \cos \theta-y \sin \theta=a[(\cos ^{2} \theta+\sin ^{2} \theta)(\cos ^{2} \theta-\sin ^{2} \theta)] \\ \Rightarrow & x \cos \theta-y \sin \theta=a \cos ^{2} \theta \end{matrix} $

Hence, the given statement is false.

Solution

True

$\begin{matrix} \text { Given that, } & \\ \begin{aligned} & x+2 y-10=0 \\ & \text { and }\end{aligned} \\ 2 x+y+5 =0\end{matrix} $

From Eq. (i), put the value of $x=10-2 y$ in Eq. (ii), we get

$ \begin{aligned} \Rightarrow & 20-4 y+y+5 =0 \\ \Rightarrow & 20-3 y+5 =0 \\ \therefore & x+\frac{50}{3}-10 =0 \\ \Rightarrow & x+\frac{20}{3}=0 \Rightarrow x =\frac{-20}{3} \end{aligned} $

So, the point of intersection is $\Big(-\frac{20}{3}, \frac{25}{3}\Big)$.

If the line $5 x+4 y=0$ passes through the point $\Big(-\frac{20}{3}, \frac{25}{3}\Big)$, then this point should lie on this line.

$ \because \quad 5 \frac{-20}{3}+\frac{4(25)}{3}=\frac{-100}{3}+\frac{100}{3}=0 $

So, this point lies on the given line.

Hence, the statement is true.

Solution

True

Let $A B C$ be an equilateral triangle with vertex $A(2,3)$, and equation of $B C$ is $x+y=2$. i.e., slope $=-1$.

Let slope of line $A B$ is $m$.

Since, the angle between line $A B$ and $B C$ is $60^{\circ}$.

$ \begin{aligned} & \therefore \quad \tan 60^{\circ}=\Big|\frac{m+1}{1-m} \Big|\\ & \Rightarrow \quad \sqrt{3}= \pm \Big(\frac{m+1}{1-m}\Big) \quad \text { [taking positive sign] } \\ & \Rightarrow \quad \sqrt{3}-\sqrt{3} m=m+1 \\ & \Rightarrow \quad \sqrt{3}-1=m+\sqrt{3} m \\ & \Rightarrow \quad \sqrt{3}-1=m(1+\sqrt{3}) \\ & \therefore \quad m=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \\ & =\frac{3+1-2 \sqrt{3}}{3-1}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3} \end{aligned} $

Similarly, slope of $A B=2+\sqrt{3}$

$\therefore$ Equation of other two side is

[taking negative sign]

$ y-3=(2 \pm \sqrt{3})(x-2) $

Hence, the statement is true.

Solution

True

Given equation of lines are $4x+y-1=0$

$ \text { and } \quad 7 x-3 y-35=0 $

From Eq. (i), on putting $y=1-4 x$ in Eq. (ii), we get

$ 7 x-3+12 x-35=0 $

$ \Rightarrow \quad 19 x-38=0 \Rightarrow x=2 $

On putting $x=2$ in Eq. (i), we get

$ 8+y-1=0 \Rightarrow y=-7 $

Now, the equation of a line passing through $(3,5)$ and $(2,-7)$ is

$ \begin{aligned} y-5 =\frac{-7-5}{2-3}(x-3) \\ \Rightarrow y-5 =12(x-3) \\ \Rightarrow 12 x-y-31 =0 \end{aligned} $

Distance from $(0,0)$ to the line (iii),

$ d_1=\frac{|-31|}{\sqrt{144+1}}=\frac{31}{\sqrt{145}} $

$\therefore$ Distance from $(8,34)$ to the line (iii),

$ \begin{aligned} d_2 & =\frac{|96-34-31|}{\sqrt{145}}=\frac{31}{\sqrt{145}} \\ \because \quad d_1 & =d_2 \end{aligned} $

Hence, the statement is true.

Solution

True

Given that, equation of line is

$ \frac{x}{a}+\frac{y}{b}=1 $

Equation of line passing through origin and perpendicular to line (i) is

$ \frac{x}{b}-\frac{y}{a}=0 $

Now, foot of perpendicular is the point of intersection of lines (i) and (ii). To find its locus we have to eliminate the variable $a$ and $b$.

On squaring and adding Eqs. (i) and (ii), we get

$ \begin{matrix} \Rightarrow & \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{2 x y}{a b}+\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}-\frac{2 x y}{a b} =1 \\ \\ \Rightarrow & x^{2} \Big(\frac{1}{a^{2}}+\frac{1}{b^{2}}\Big)+y^{2} \Big(\frac{1}{a^{2}}+\frac{1}{b^{2}}\Big) =1 \\ \\ \Rightarrow \frac{x^{2}}{c^{2}}+\frac{y^{2}}{c^{2}} =1 \\ \\ x^{2}+y^{2} =c^{2} & [\because \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}] \end{matrix} $

Hence, the statement is true.

Solution

False

Given lines are

and $\quad \begin{aligned} & a x+2 y+1=0 \\ & b x+3 y+1=0\end{aligned}$

From Eq. (i), on putting $y=\frac{-a x-1}{2}$ in Eq. (ii), we get

$ b x-\frac{3}{2}(a x+1)+1=0 $

$ \Rightarrow 2 b x-3 a x-3+2=0 $

$ \Rightarrow x(2 b-3 a)=1 \Rightarrow x=\frac{1}{2 b-3 a} $

Now, using $x=\frac{1}{2 b-3 a}$ in Eq. (i), we get

$ \frac{a}{2 b-3 a}+2 y+1 =0 $

$ \Rightarrow 2 y =-\Big[\frac{a+2 b-3 a}{2 b-3 a}\Big] $

$ \Rightarrow 2 y =\frac{-(2 b-2 a)}{2 b-3 a} $

$ \Rightarrow y =\frac{(a-b)}{2 b-3 a} $

So, the point of intersection is $\Big(\frac{1}{2 b-3 a}, \frac{a-b}{2 b-3 a}\Big)$.

Since, this point lies on $c x+4 y+1=0$, then

$ \frac{c}{2 b-3 a}+\frac{4(a-b)}{2 b-3 a}+1=0 $

$ \Rightarrow c+4 a-4 b+2 b-3 a=0 $

$ \Rightarrow -2 b+a+c=0 \Rightarrow 2 b=a+c $

Hence, the given statement is false.

Solution

False

Given points are $A(3,-4), B(-2,6), P(-3,6)$ and $Q(9,-18)$.

Now $ \begin{aligned} & \text { slope of } A B=\frac{6+4}{-2-3}=-2 \\ & \text { slope of } P Q=\frac{-18-6}{9+3}=-2 \end{aligned} $

and

So, line $A B$ is parallel to line $P Q$.

Solution

(i) Let the coordinate of point $P(x_1, y_1)$ on the line $x+5 y=13 i$.e.,

$ P(13-5 y_1, y_1) $

$\therefore$ Distance of $P$ from the line $12 x-5 y+26=0$,

$ \begin{aligned} & \qquad 2=\Big|\frac{12(13-5 y_1)-5 y_1+26}{\sqrt{144+25}}\Big| \\ & \Rightarrow \quad 2= \pm \frac{156-60 y_1-5 y_1+26}{13} \\ & \Rightarrow \quad-65 y_1=-156 \quad \text { [taking positive sign] } \\ & \begin{aligned} y_1 & =\frac{156}{65}=\frac{12}{5} \\ x_1 & =13-5 y_1 \\ & =13-12=1 \end{aligned} \\ & \text { So, the coordinate of is } P \begin{aligned} & \Big(1, \frac{12}{5}\Big) . \end{aligned} \\ & \text { Similarly, the coordinates of } Q \text { are }\Big(-3, \frac{16}{5}\Big) . \end{aligned} $

(ii) Let coordinates of the point on the line $x+y=4$ be $(4-y_1, y_1)$. Distance from the line $4 x+3 y-10=0$.

$ 1=\Big|\frac{4(4-y_1)+3 y_1-10}{\sqrt{16+9}}\Big| $

$ \Rightarrow 1= \pm \frac{16-4 y_1+3 y_1-10}{5} $

$ \Rightarrow 5=6-y_1 $

$ \Rightarrow y_1=1 $

$ \text { If } y_1=1 \text {, then } x_1=3 $

So, the point is $(3,1)$.

Similarly, taking negative sign the point is $(-7,11)$.

(iii) Given point $A(-2,5)$ and $B(3,1)$.

Now, the point $P$ divides line joining the point $A$ and $B$ in $1: 2$.

$ \begin{matrix} \because & x_1=\frac{1 \cdot 3+2(-2)}{1+2}=\frac{3-4}{3}=\frac{-1}{3} \\ \\ \text { and } & y_1=\frac{1 \cdot 1+2 \cdot 5}{1+2}=\frac{11}{3} \end{matrix} $

So, the coordinates of $P$ are $\Big(\frac{-1}{3}, \frac{11}{3}\Big)$

Thus, the point $Q$ divided the line joining $A$ to $B$ in $2: 1$.

$ \begin{matrix} \because & x_2=\frac{2 \cdot 3+1(-2)}{2+1}=\frac{4}{3} \\ \\ \text { and } & y_2=\frac{2 \cdot 1+1 \cdot 5}{2+1}=\frac{7}{3} \end{matrix} $

Hence, the coordinates of $Q$ are $\Big(\frac{4}{3}, \frac{7}{3}\Big)$.

Hence, the correct matches are (i) $\rightarrow$ (c), (ii) $\rightarrow$ (a), (iii) $\rightarrow$ (b).

Solution

(i) Given equation of the line is

$ (2 x+3 y+4)+\lambda(6 x-y+12)=0 $

If line is parallel to $Y$-axis i.e., it is perpendicular to $X$-axis

$ \therefore \quad \text { Slope }=m=\tan 90^{\circ}=\infty $

$\begin{aligned} & \text { From line (i), } x(2+6 \lambda)+y(3-\lambda)+4+12 \lambda=0 \\ & \text { and slope } \\ & =\frac{-(2+6 \lambda)}{3-\lambda} \\ & \Rightarrow \quad \frac{-2-6 \lambda}{3-\lambda}=\infty \\ & \Rightarrow \quad \frac{-2-6 \lambda}{3-\lambda}=\frac{1}{0} \Rightarrow \lambda=3 \\ & \end{aligned}$

(ii) If the line (i) is perpendicular to the line $7 x+y-4=0$ or $y=-7 x+4$

$ \begin{matrix} \because & \frac{-(2+6 \lambda)}{(3-\lambda)}(-7) & =-1 \\ \Rightarrow & 14+42 \lambda =-3+\lambda \\ \Rightarrow & 41 \lambda =-17 \\ \Rightarrow & \lambda =-\frac{17}{41} \end{matrix} $

(iii) If the line (i) passes through the point $(1,2)$.

$ \begin{matrix} \text { Then, } & (2+6+4)+\lambda(6-2+12) =0 \\ \Rightarrow & 12+16 \lambda=0 \Rightarrow \lambda =-\frac{3}{4} \end{matrix} $

(iv) If the line is parallel to $X$-axis the slope $=0$.

$ \begin{matrix} \text { Then, } & \frac{-(2+6 \lambda)}{3-\lambda}=0 \\ \\ \Rightarrow & -(2+6 \lambda)=0 \Rightarrow \lambda=-\frac{1}{3} \end{matrix} $

So, the correct matches are (i) $\rightarrow$ (d), (ii) $\rightarrow$ (c), (iii) $\rightarrow$ (a), (iv) $\rightarrow$ (b).

Solution

Given equation of the lines are

$ 2 x-3 y =0 $

and $\quad$

$ 4 x-5 y =2 $

From Eq. (i), put $x=\frac{3 y}{2}$ in Eq. (ii), we get

$\Rightarrow$ $ 4 \cdot \frac{(3 y)}{2}-5 y=2 $

$\Rightarrow$ $ 6 y-5 y=2 $

Now, put $y=2$ in Eq. (i), we get

$ y=2 $

So, the intersection points are $(3,2)$.

$ x=3 $

(i) The equation of the line passes through the point $(3,2)$ and $(2,1)$, is

$ y-2 =\frac{1-2}{2-3}(x-3) $

$ \Rightarrow y-y-1 =0 $

$ \Rightarrow x-2 =(x-3) $

(ii) If the required line is perpendicular to the line $x+2 y+1=0$

$\because$ Slope of the required line $=2$

$\therefore$ Equation of the line is

$ y-2 =2(x-3) $

$ \Rightarrow 2 x-y-4 =0 $

(iii) If the required line is parallel to the line $3 x-4 y+5=0$, then the slope of the required line $=\frac{3}{4}$

$\therefore$ Equation of the required line is

$ y-2=\frac{3}{4}(x-3) $

$ \Rightarrow 4 y-8=3 x-9 $

$ \Rightarrow 3 x-4 y-1=0 $

(iv) If the line is equally inclined to the $X$-axis, then

$ m= \pm \tan 45^{\circ}= \pm 1 $

$\therefore$ Equation of the line is