Question

The perpendicular distance between the lines $9x^2- 24xy + 16y^2 + 21x- 28y+ 10 = 0$ is

• A. 44747
• B. 44625
• C. 44656
• D. 44566

Answer 1

Answer: (B)

44625

Answer 2

Since, $9x^2 - 24 xy + 16y^2 + 21x - 28y + 10 = 0$
$\Rightarrow \:\: (3x)^2 + (4y)^2 - 24xy + 21x - 28y + 10 = 0$
$\Rightarrow \:\: (3x - 4y)^2 + 7(3x - 4y) + 10 = 0$
Let $3x-4y = t$
$\therefore \:\: t^2 + 7t + 10 = 0 \Rightarrow \: (t + 5)(t + 2) = 0$
$\Rightarrow \:\: t = - 5 , -2$
$\therefore \:\:\: 3x - 4y = - 5 , - 2$
$\therefore$ Lines are $3x - 4y + 5 = 0$ and $3x - 4y + 2 = 0$
$\therefore$ Distance between these lines
$= \frac{\left|5-2\right|}{\sqrt{3^{2} +4^{2}}} =\frac{3}{\sqrt{9+16}} = \frac{3}{5}$