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

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

Thinking Process

First of all find the slope, using the formula $=\frac{y_2-y_1}{x_2-x_1}$. Then, slope of perpendicular line is $-\frac{1}{m}$.

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 $C,m_{BC}=\frac{-1-3}{3-2}=-4$

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 $y-2=\frac{1}{4}(x-5)$.

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

Thinking Process

If the angle between the lines having the slope $m_1$ and $m_2$ is $\theta$, then $\tan \theta =|\frac{m_1-m_2}{1+m_1{m}_2}|$.

Use this formula to solve the above problem.

Solution

$\begin{array}{rl}\text{Given lines,}y& =(2-\sqrt{3})(x+5)\\ \text{Slope of this line,}{m}_1& =(2-\sqrt{3})\\ \text{and}y& =(2+\sqrt{3})(x-7)\\ \text{Slope of this line,}{m}_2& =(2+\sqrt{3})\end{array}$

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

$\tan \theta =|\frac{m_1-m_2}{1+m_1{m}_2}|$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore$ 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 $\frac{x}{6}+\frac{y}{8}=1$

Thinking Process

The perpendicular distance of a point $(x_1,y_1)$ from the line $Ax+By+C=0$, is $d^{\prime }$, where $d=|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}|$.

Solution

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

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

Taking positive sign,

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

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

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

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

Taking negative sign,

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

On putting $k=11in\mathrm{Eq}.$ (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

$\begin{array}{rl}& \tan \theta =|\frac{m_1-m_2}{1+m_1{m}_2}|\Rightarrow \tan \theta =|\frac{-\frac{b}{a}-\frac{b}{a}}{1+(\frac{-b}{a})(-\frac{b}{a})}|\\ & \Rightarrow \tan \theta =|\frac{\frac{-2b}{a}}{\frac{a^2-b^2}{a^2}}|\Rightarrow \tan \theta =\frac{2ab}{a^2-b^2}\end{array}$

Hence proved.

Thinking Process

Equation of a line passing through the point $(x_1,y_1)$ and having slope $m$ is $y-y_1=m(x-x_1)$.

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

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

Thinking Process

First of all solve the given equation of lines to get the point of intersection. Then, if a line having slope $m_1$ is parallel to another line having slope, $m_2$, then $m_1=m_2$. Now, use the formula i.e., equation of a line passing through the point $(x_1,y_1)$ with slope $m$ is $y-y_1=m(x-x_1)$

Solution

Given equation of lines

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

From Eq. (i),

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

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

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

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

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 $3x+4y=7$. 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

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

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

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

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

Solution

Given equation of line

$\begin{array}{c}ax+by+8=0\\ \\ \Rightarrow \frac{x}{\frac{-8}{a}}+\frac{y}{\frac{-8}{b}}=1\end{array}$

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

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

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

So, the intercepts are -3 and 2 .

According to the question,

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

Thinking Process

The coordinates of a point which divides the join of $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m_1:m_2$ internally is $(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$.

Solution

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

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

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

Hence, the equation of a line $AB$ is

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

Thinking Process

The equation of the line having normal distance Pfrom the origin and angle $\alpha$ which the normal makes with the positive direction of X-axis is $x\cos \alpha +y\sin \alpha =p$. Use this formula to solve the above problem.

Solution

Given that, $OC=P=4$ units

$\begin{array}{rl}& \mathrm{\angle }BAX={120}^{\circ }\\ & \text{ Let }\mathrm{\angle }COA=\alpha ,\mathrm{\angle }OCA={90}^{\circ }\\ & \because \mathrm{\angle }BAX=\mathrm{\angle }COA+\mathrm{\angle }OCA\text{ [exterior angle property] }\\ & \Rightarrow {120}^{\circ }=\alpha +{90}^{\circ }\\ & \therefore \alpha ={30}^{\circ }\\ & \text{ Now, the equation of required line is}\\ & x\cos {30}^{\circ }+y\sin {30}^{\circ }=4\\ & \Rightarrow x\cdot \frac{\sqrt{3}}{2}+y\cdot \frac{1}{2}=4\\ & \Rightarrow \sqrt{3}x+y=8\end{array}$

Solution

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

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

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

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

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

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

Taking negative sign,

$\begin{array}{rl}& 1=-(\frac{m+\frac{3}{4}}{1-\frac{3m}{4}})\Rightarrow 1-\frac{3m}{4}=-m-\frac{3}{4}\\ & \Rightarrow m-\frac{3m}{4}=-1-\frac{3}{4}\\ & \Rightarrow \frac{m}{4}=\frac{-7}{4}\Rightarrow m=-7\\ & \begin{array}{c}\therefore \text{ Equation of side }AC\text{ having slope }(\frac{1}{7})\text{ is }\end{array}\\ & y-2=\frac{1}{7}(x-2)\\ & \Rightarrow 7y-14=x-2\\ & \text{ and equation of side }AB\text{ having slope }(-7)\text{ is }\\ & \Rightarrow y-2=-7(x-2)\\ & \Rightarrow y-2=-7x+14\\ & \Rightarrow 7x+y-16=0\end{array}$

Thinking Process

Find the length of perpendicular ( $p$ ) from $(2,-1)$ to the line and use $p=l\sin {60}^{\circ }$, where $l$ is the length of the side of the triangle.

Solution

Given that, equilateral $\mathrm{△}ABC$ having equation of base is $x+y=2$.

$\ln \mathrm{△}ABD$,

$\begin{array}{rl}\sin {60}^{\circ }& =\frac{AD}{AB}\\ AD& =AB\sin {60}^{\circ }=AB\frac{\sqrt{3}}{2}\\ \Rightarrow AD& =AB\frac{\sqrt{3}}{2}\end{array}$

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

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

From Eq. (i),

$\begin{array}{rl}& \frac{1}{\sqrt{2}}=AB\frac{\sqrt{3}}{2}\\ & AB=\sqrt{\frac{2}{3}}\end{array}$

Thinking Process

Let the slope of the line be $m$. Then, the equation of the line passing through the fixed point $P(x_1,y_1)$ is $y-y_1=m(x-x_1)$. Taking the algebraic sum of perpendicular distances equal to zero, we get $y_1-1=m(x_1-1)$. Thus, $(x_1,y_1)$ is $(1,1)$.

Solution

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

$\therefore$ 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

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

$\text{ Now, }\frac{-y_1-2m+m{x}_1+2-y_1+m{x}_1+1-y_1-m+m{x}_1}{\sqrt{1+m^2}}=0$

$\Rightarrow -3y_1-3m+3m{x}_1+3=0$

$\Rightarrow -y_1-m+m{x}_1+1=0$

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 $A(1,2)$.

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

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

and

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

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

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

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

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

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

Thinking Process

If a line is $\frac{x}{a}+\frac{y}{b}=1$, where $\frac{1}{a}+\frac{1}{b}=$ constant $=\frac{1}{k}$ (say). This implies that $\frac{k}{a}+\frac{k}{b}=1\Rightarrow$ line passes through the fixed point $(k,k)$.

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.

Thinking Process

If the point $(h,k)$ divides the join of $A(x_1,y_2)$ and $B(x_2,y_2)$ internally, in the ratio $m_1:m_2$. Then, first of all find the coordinates of $A$ and $B$ using section formula for internal division i.e., $h=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},k=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}$. Then, find the equation of required line.

Solution

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

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

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

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

Solution

Given equation of lines

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

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

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

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

Let slope of the required line be $m$.

$\therefore$ Equation of line is

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

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

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

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

Thinking Process

Given that $|x|+|y|=1$, which gives four sides of a square.

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 .

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

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

Thinking Process

Lines are $y=\sqrt{3}x+2$ and $y=-\sqrt{3}x+2$ according as $x\ge 0$ or $x<0$. Y-axis is the bisector of the angles between the lines. $P_1,P_2$ are the points on these lines at a distance of 5 units from the point of intersection of these lines which have a point on $Y$-axis as common foot of perpendiculars drawn from these points. The $y$-coordinate of the foot of the perpendiculars is given by $2+5\cos {30}^{\circ }$.

Solution

Given equation of lines are

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

$[\because x\ge 0]$

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

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

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

Here,

$\begin{array}{rl}OC& =2\\ \text{In }\mathrm{△}DEC\frac{CD}{CE}& =\cos {30}^{\circ }\\ CD=5\cos 30\\ & =5\cdot \frac{\sqrt{3}}{2}\end{array}$

$\begin{array}{rl}\therefore & CD=5\cos {30}^{\circ }\\ & =5\cdot \frac{\sqrt{3}}{2}\\ \Rightarrow & OD=OC+CD=2+5\frac{\sqrt{3}}{2}\end{array}$

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

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 $p$

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

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

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

Solution

(a) Given that,

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

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

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

$\Rightarrow 5y=3x-15$

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

• Option (b) $3y-5x+15=0$ is incorrect because the slope of the line is given as $\frac{3}{5}$, which means the coefficient of $x$ should be 3 and the coefficient of $y$ should be 5. This option reverses these coefficients.
• Option (c) $5y-3x-15=0$ is incorrect because the constant term should be $+15$ to match the intercept of $-3$ on the $Y$-axis when rearranged in the form $y=mx+c$. This option has $-15$ as the constant term, which does not satisfy the given intercept condition.

Solution

(a) Let equation of line be

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

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

• Option (b) 0: The slope of 0 would imply a horizontal line, which does not cut off intercepts of equal lengths on the axes. A horizontal line would only intersect the y-axis and not the x-axis.

• Option (c) 2: A slope of 2 would imply a steep line that does not cut off intercepts of equal lengths on the axes. The intercepts would not be symmetric and equal in length.

• Option (d) $\sqrt{3}$: A slope of $\sqrt{3}$ would also imply a steep line that does not cut off intercepts of equal lengths on the axes. The intercepts would again be unequal.

Solution

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

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

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

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

$\Rightarrow x+y=5$

• Option (a) $x-y=5$: This equation represents a line with a slope of 1, which is not perpendicular to the line $y=x$ (which also has a slope of 1). Therefore, it cannot be the correct equation.

• Option (c) $x+y=1$: Although this equation represents a line with a slope of -1 (which is perpendicular to the line $y=x$), it does not pass through the point (3,2). Substituting (3,2) into the equation $x+y=1$ gives $3+2=5$, not 1.

• Option (d) $x-y=1$: This equation represents a line with a slope of 1, which is not perpendicular to the line $y=x$ (which also has a slope of 1). Therefore, it cannot be the correct equation.

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

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

$\Rightarrow y-2=x-1$

$\Rightarrow y-x-1=0$

• Option (a) $y-x+1=0$ is incorrect because when you substitute the point $(1,2)$ into the equation, it does not satisfy the equation. Specifically, $2-1+1\ne 0$.

• Option (c) $y-x+2=0$ is incorrect because when you substitute the point $(1,2)$ into the equation, it does not satisfy the equation. Specifically, $2-1+2\ne 0$.

• Option (d) $y-x-2=0$ is incorrect because when you substitute the point $(1,2)$ into the equation, it does not satisfy the equation. Specifically, $2-1-2\ne 0$.

Solution

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

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

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

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

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

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

• Option (a): The expression (\frac{a^{2}-b^{2}}{a b}) does not account for the factor of 2 in the denominator that arises from the formula for the tangent of the angle between two lines. The correct formula involves dividing by 2, which is missing in this option.

• Option (b): The expression (\frac{b^{2}-a^{2}}{2}) is incorrect because it does not include the product of (a) and (b) in the denominator. The correct formula for the tangent of the angle between the lines should have (2ab) in the denominator, not just 2.

• Option (d): This option is incorrect because the correct answer is provided in option (c). Therefore, “None of these” is not a valid choice.

Solution

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

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

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

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

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

• Option (a) $(1,1)$: If (a = 1) and (b = 1), the line equation becomes (\frac{x}{1} + \frac{y}{1} = 1) or (x + y = 1). Substituting the points ((2, -3)) and ((4, -5)) into this equation:

  • For ((2, -3)): (2 + (-3) = -1 \neq 1)
  • For ((4, -5)): (4 + (-5) = -1 \neq 1) Therefore, the points do not lie on the line (x + y = 1).

• Option (b) $(-1,1)$: If (a = -1) and (b = 1), the line equation becomes (\frac{x}{-1} + \frac{y}{1} = 1) or (-x + y = 1). Substituting the points ((2, -3)) and ((4, -5)) into this equation:

  • For ((2, -3)): (-2 + (-3) = -5 \neq 1)
  • For ((4, -5)): (-4 + (-5) = -9 \neq 1) Therefore, the points do not lie on the line (-x + y = 1).

• Option (c) $(1,-1)$: If (a = 1) and (b = -1), the line equation becomes (\frac{x}{1} + \frac{y}{-1} = 1) or (x - y = 1). Substituting the points ((2, -3)) and ((4, -5)) into this equation:

  • For ((2, -3)): (2 - (-3) = 5 \neq 1)
  • For ((4, -5)): (4 - (-5) = 9 \neq 1) Therefore, the points do not lie on the line (x - y = 1).

Thinking Process

First of all find the point of intersection of the given first two lines, then get the perpendicular distance from this point to the third line. Using formula i.e., distance of a point $(x_1,y_1)$ from the line $ax+by+c=0$ is $d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$.

Solution

(a) Given equation of lines

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

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

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

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

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

From Eq. (ii),

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

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

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

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

$\because \text{ distance of a point }p(x_1,y_1)\text{ from the line }ax+by+c=0\text{ is }d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

• Option (b): The numerator in the distance formula is incorrectly calculated. The correct numerator should be 130, not 13. This option likely results from a miscalculation or simplification error.

• Option (c): The denominator in the distance formula is incorrectly simplified. The correct denominator should include the term (\sqrt{29}), not just 7. This option likely results from an incorrect application of the distance formula.

• Option (d): This option is a catch-all for any answer that does not match the correct calculation. Since option (a) is correct, option (d) is not applicable.