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 $\frac{7}{5}$.

• A. $4x+3y+1=0$
• B. $3x+4y-6=0$
• C. $4x-3y+1=0$
• D. $3x+4y+6=0$

Answer 1

Answer: (C)

Given equation of lines are
$x-y+1=0\dots \left(i\right)$ and
$2x-3y+5=0\dots \left(ii\right)$
On solving $\left(i\right)$ and $\left(ii\right)$, we get
$x=2$,
$y=3$
$\therefore$ The point of intersection is $\left(2,3\right)$.
Let slope of the required line be $m$.
Then equation of line passing through $\left(2,3\right)$ is
$y-3=m\left(x-2\right)$
$\Rightarrow mx-y+3-2m=0\dots \left(iii\right)$
Now, perpendicular distance of line $\left(iii\right)$ from point $\left(3,2\right)$ is $\frac{7}{5}$.
$\Rightarrow \frac{7}{5}=\left|\frac{3m-2+3-2m}{\sqrt{1+m^2}}\right|$
$\Rightarrow \frac{49}{25}=\frac{{\left(m+1\right)}^2}{1+m^2}$
$\Rightarrow 49+49m^2=25\left(m^2+2m+1\right)$
$\Rightarrow 12m^2-25m+12=0$
$\Rightarrow m=\frac{25\pm \sqrt{49}}{24}=\frac{25\pm 7}{24}$
$\Rightarrow m=\frac{4}{3}$ or $m=\frac{3}{4}$
$\therefore$ Equation of line when $m=\frac{4}{3}$ is
$y-3=\frac{4}{3}\left(x-2\right)$
$\Rightarrow 3y-9=4x-8$
$\Rightarrow 4x-3y+1=0$
and equation of line when $m=\frac{3}{4}$ is
$y-3=\frac{3}{4}\left(x-2\right)$
$\Rightarrow 4y-12=3x-6$
$\Rightarrow 3x-4y+6=0$