Question

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

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

Answer 1

The correct option is B

$3:7$

Find the ratio in which distance is divided:

Given, $3x+4y+2=0....\left(1\right)$

$3x+4y+5=0.....\left(2\right)$

$3x+4y-5=0.....\left(3\right)$

From the given equations we can see that all three equations have an equal slope.

So, the distance between two parallel lines, $d=\left|\frac{c_1-c_2}{\sqrt{{a_1}^2+{b_1}^2}}\right|$.

The distance between $\left(1\right)$ and $\left(2\right)$

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

The distance between $\left(1\right)$ and $\left(3\right)$

$=d_2=\left|\frac{2+5}{5}\right|=\frac{7}{5}$

Therefore, ratio in which the line $3x+4y+2=0$ divides the distance between $3x+4y+5=0$and $3x+4y-5=0$ is$=\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{7}{5}}}=\frac{3}{7}$

Hence, the correct option is (B).