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

Lines 3x+4y+2=0 and 3x+4y+5=0 are on the same side of the origin. The distance between these lines is
$d_1=\left|\begin{array}{c}\frac{2-5}{\sqrt{3^2+4^2}}\end{array}\right|=\frac{3}{5}$
Lines 3x+4y+2=0 and 3x+4y-5=0 are on the opposite sides of the origin. The distance between these lines is
$d_2=\left|\begin{array}{c}\frac{2+5}{\sqrt{3^2+4^2}}\end{array}\right|=\frac{7}{5}$
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 .