Question

Number of points lying on the line $7x+4y+2=0$ which is equidistant from the lines $15x^2+56xy+48y^2=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$1$

Given pair of lines

$15x^2+56xy+48y^2=0$

$x=\frac{-56y\pm \sqrt{3136y^2-2880y^2}}{30}$

$x=\frac{-56y\pm \sqrt{256}y}{30}$

$x=\frac{-56y\pm 16y}{30}$

$30x=-40y$ and $30x=-72y$

$3x+4y=0---\left(1\right)$ and $5x+12y=0-----\left(2\right)$

Let point $Q(a,b)$

Distance of line (1) from Q is equal to distance of line (2) from Q

$\left|\frac{3a+4b}{\sqrt{3^2+4^2}}\right|=\left|\frac{5a+12b}{\sqrt{5^2+{12}^2}}\right|$

$\frac{3a+4b}{5}=\frac{5a+12b}{13}$

$39a+52b=25a+60b$

$14a=8b----\left(3\right)$

Point Q also lie on line $7x+4y+2=0$

$7a+4b+2=0---\left(4\right)$

On solving eq (3) and (4) we get coordinate of only one point $Q$