Question

Equation of the line through the point of intersection of the lines $3x+2y+4=0$ and $2x+5y-1=0$ whose distance from $(2,-1)$ is $2$, is

• A. $2x-y+5=0$
• B. $4x+3y+5=0$
• C. $x+2=0$
• D. $3x+y+5=0$

Answer 1

Answer: (B)

$4x+3y+5=0$

We have,

$3x+2y+4=0\to \left(1\right)$

$2x+5y-1=0\to \left(2\right)$

From $\left(1\right)$ & $\left(2\right)$,

$11y-11=0$

$y=1$ and $x=-2$

So, equation of line passing through $(-2,1)$

$y-1=m(x+2)$

$y-mx=2m+1\to \left(3\right)$

Line $3,$ distance from $(2,-1)$ is $2$

$\therefore 2=\left|\frac{-1-2m-2m-1}{\sqrt{1+m^2}}\right|$

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

$1+m^2=4m^2+4m+1$

$3m^2+4m=0$

$m=0$ or $m=\frac{-4}{3}$

So, equation of line is

$y+\frac{4}{3}x=\frac{-5}{3}$

$\Rightarrow 3y+4x+5=0$