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
$x - y+ 1=0\quad\ldots\left(i\right)$ and
$2x - 3y + 5 = 0 \quad\ldots\left(ii\right)$
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
$y - 3 = m\left(x - 2\right)$
$\Rightarrow mx-y + 3- 2m = 0 \quad\ldots\left(iii\right)$
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
$y-3=\frac{3}{4}\left(x-2\right)$
$\Rightarrow 4y-12=3x-6$
$\Rightarrow 3x - 4y + 6 = 0$