# Find the equations of the lines through

Question:

Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5.

Solution:

Given two lines are $x-y+1=0 \ldots \ldots$ (i)

And $2 x-3 y+5=0 \ldots$ (ii)

Now, point of intersection of these lines can be find out as: Multiplying equation (i) by 2 , we get

Multiplying equation (i) by 2, we get

$2 x-2 y+2=0 \ldots \ldots$ (iii)

On subtracting equation (iii) from (ii), we get

$2 x-2 y+2-2 x+3 y-5=0$

$\Rightarrow y-3=0$

$\Rightarrow y=3$

On putting value of $y$ in equation (ii), we get

$2 x-3(3)+5=0$

$\Rightarrow 2 x-9+5=0$

$\Rightarrow 2 x-4=0$

$\Rightarrow 2 x=4$

$\Rightarrow x=2$

So, the point of intersection of given two lines is $(x, y)=(2,3)$ Let $m$ be the slope of the required line

$\therefore$ Equation of the line is

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

$\Rightarrow \mathrm{y}-3=\mathrm{mx}-2 \mathrm{~m}$

$\Rightarrow \mathrm{mx}-\mathrm{y}-2 \mathrm{~m}+3=0 \ldots$ (iv)

Since, the perpendicular distance from the point $(3,2)$ to the line is $7 / 5$ then

$\mathrm{d}=\left|\frac{\mathrm{m}(3)-2+3-2 \mathrm{~m}}{\sqrt{(\mathrm{m})^{2}+(1)^{2}}}\right|$

$\Rightarrow \frac{7}{5}=\left|\frac{3 m+1-2 m}{\sqrt{m^{2}+1}}\right|$

$\Rightarrow \frac{7}{5}=\frac{m+1}{\sqrt{m^{2}+1}}$

Squaring both the sides, we get

$\Rightarrow \frac{49}{25}=\frac{(m+1)^{2}}{m^{2}+1}$

On cross multiplication we get

$\Rightarrow 49\left(m^{2}+1\right)=25(m+1)^{2}$

Computing and simplifying we get

$\Rightarrow 49 m^{2}+49=25\left(m^{2}+1+2 m\right)$

$\Rightarrow 49 m^{2}+49=25 m^{2}+25+50 m$

$\Rightarrow 25 m^{2}+25+50 m-49 m^{2}-49=0$

$\Rightarrow-24 m^{2}+50 m-24=0$

$\Rightarrow-12 m^{2}+25 m-12=0$

$\Rightarrow 12 m^{2}-25 m+12=0$

$\Rightarrow 12 m^{2}-16 m-9 m+12=0$

Taking $m$ common we get

$\Rightarrow 4 m(3 m-4)-3(3 m-4)=0$

$\Rightarrow(3 m-4)(4 m-3)=0$

$\Rightarrow 3 m-4=0$ or $4 m-3=0$

$\Rightarrow 3 m=4$ or $4 m=3$

$\Rightarrow \mathrm{m}=\frac{4}{3} \mathrm{Or} \mathrm{m}=\frac{3}{4}$

$\therefore \mathrm{m}=\frac{4}{3}, \frac{3}{4}$

Putting the value of $m=4 / 3$ in equation (iv), we get

$\frac{4}{3} x-y-2\left(\frac{4}{3}\right)+3=0$

Simplifying and computing we get

$\Rightarrow \frac{4}{3} x-y-\frac{8}{3}+3=0$

$\Rightarrow \frac{4}{3} \mathrm{x}-\mathrm{y}=\frac{8-9}{3}$

$\Rightarrow \frac{4}{3} x-y=-\frac{1}{3}$

$\Rightarrow \frac{4}{3} x-y+\frac{1}{3}=0$

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

Putting the value of $m=3 / 4$ in equation (iv), we get

$\frac{3}{4} x-y-2\left(\frac{3}{4}\right)+3=0$

Simplifying and computing we get

$\frac{3}{4} x-y-2\left(\frac{3}{4}\right)+3=0$

Simplifying and computing we get

$\Rightarrow \frac{3}{4} x-y-\frac{3}{2}+3=0$

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

$\Rightarrow \frac{3}{4} x-y=\frac{3-6}{2}$

$\Rightarrow \frac{3}{4} x-y+\frac{3}{2}=0$

$\Rightarrow 3 x-4 y+6=0$

Hence, the required equation are $4 x-3 y+1=0$ and $3 x-4 y+6=0$