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

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Solution

The correct option is A $1$
Given pair of lines
$15 x^{2} + 56 x y + 48 y^{2} = 0$
$x = \frac{- 56 y \pm \sqrt{3136 y^{2} - 2880 y^{2}}}{30}$

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

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

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

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

Let point $Q (a, b)$
Distance of line (1) from Q is equal to distance of line (2) from Q
$∣ ∣ ∣ ∣ \frac{3 a + 4 b}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \frac{5 a + 12 b}{\sqrt{5^{2} + 12^{2}}} ∣ ∣ ∣ ∣$

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

$39 a + 52 b = 25 a + 60 b$
$14 a = 8 b - - - - (3)$

Point Q also lie on line $7 x + 4 y + 2 = 0$
$7 a + 4 b + 2 = 0 - - - (4)$
On solving eq (3) and (4) we get coordinate of only one point $Q$

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