Question

The ratio in which the line $3x+4y+2=0$ divides the distance between $3x+4y+5=0$ and $3x+4y-5=0$.

• A. $7:3$
• B. $3:7$
• C. $2:3$
• D. None of these

Answer 1

The correct option is B $3:7$

Lines $3x+4y+2=0$ & $3x+4y+5=0$ are on the same side of the origin. The distance between these Iines is

$d_1=\left|\frac{2-5}{\sqrt{3^2+4^2}}\right|=\frac{3}{5}$

Lines 3x + 4y + 2 = 0 and 3x + 4y - 5 = 0 are on the opposite sides of the origin. The distance between these lines is

$d_2=\left|\frac{2+5}{\sqrt{3^2+4^2}}\right|=\frac{7}{5}$

Thus, $3x+4y+2=0$ divides the distance between $3x+4y+5=0$ and $3x+4y-5=0$ in the ratio $d_1:d_2$, i.e. $3:7$.