Question

Find the equation of lines through the point of intersection of the lines $x-y+1=0$ and $2x-3y+5=0$ whose distance from the point $(3,2)is\frac{7}{5}$.

Answer 1

The equation of the lines through the point of intersection of thelines $x-y+1=0$ and $2x-3y+5=0$ is given by

$x-y+1+a(2x-3y+5)=0)$ $\Rightarrow (1+2a)x+y(-3a-1)+5a+1=0$

...(i)

The distance of the above line from the point is given by

$\frac{3(2a+1)+2(-3a-1)+5a+1}{\sqrt{(2a+1{)}^2+(-3a-1{)}^2}}$

$\therefore \frac{|3(2a+1)+2(-3a-1)+5a+1|}{\sqrt{(2a+1{)}^2+(-3a-1{)}^2}}=\frac{7}{5}$

$\Rightarrow \frac{|5a+2|}{\sqrt{13a^2+10a+2}}=\frac{7}{5}$

$\Rightarrow 25(5a+2{)}^2=49(13a^2+10a+2)$

$\Rightarrow 6a^2-5a-1=0$

$\Rightarrow a=1,-\frac{1}{6}$

Substituting the value of a in (i), we get $3x-4y+6=0$ and $4x-3y+1=0$