Question

Equation of the line equidistant from the lines $2x+y+4=0,3x+6y-5=0$ is

• A. $3x-3y+17=0$
• B. $5x+7y-5=0$
• C. $3x-3y+19=0$
• D. $9x-9y+17=0$

Answer 1

Answer: (A)

$3x-3y+17=0$

equation of the line equidistant from the lines $2x+y+4=0$
$3x+6y-5=0$ is equation of angle bisector

equation of line equidistant from $A{x}_1+B{y}_1+C_1=0$ and $A{x}_2+B{y}_2+C_2=0$ is ,

$\left|\frac{A{x}_1+B{y}_1+C_1}{\sqrt{A_1^2+B_1^2}}\right|=\pm \left|\frac{A{x}_2+B{y}_2+C_2}{\sqrt{A_2^2+B_2^2}}\right|$

So, $\left|\frac{3x+6y-5}{\sqrt{45}}\right|=\pm \left|\frac{2x+y+4}{\sqrt{5}}\right|$
$|3x+6y-5|=\pm 3|2x+y+4|$
$3x-3y+17=0$ or $9x+9y+7=0$
$3x-3y+17=0$ and $9x+9y+7=0$ are the angle bisectors of given line
i.e. this lines are equidistant from given lines.