Question

Find the equation of straight lines passing through $(1,1)$ and which are at a distance of $3$ units from $(-2,3)$.

Answer 1

Let the equation of line be $x+by+c=0$

The line passes through $(1,1)$

$1+b+c=0$

$c=-(b+1)$

$x+by-(b+1)=0$

This line is at a distance of $3$ units from $(-2,3)$

$\left|\frac{-2+3b-(b+1)}{\sqrt{1+b^2}}\right|=3$

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

$\Rightarrow \frac{4b^2+9-6b}{1+b^2}=9$ $\Rightarrow 4b^2+9-6b=9+9b^2$

$\Rightarrow 5b^2+12b=0$ $\Rightarrow b(5b+12)=0$

$b=0$ (or) $b=\frac{-12}{5}$

So lines would be

$x-1=0$ & $x-\left(\frac{12}{5}\right)y+\frac{7}{5}=0$

$=5x-12y+7=0$.