First slide

Pair of straight lines

Question

The product of perpendicular from (1,1) to the pair of lines $2 x^{2} + 3 xy + y^{2} = 0$ is equal to the product of perpendiculars from (0,0) to the pair of lines $x^{2} + 2 xy + 3 y^{2} + x + 2 y + k = 0$ then k =

Moderate

A

$\frac{5}{\sqrt{12}}$
B

$- \frac{12}{\sqrt{5}}$
C

$\frac{12}{\sqrt{5}}$
D

$- \frac{5}{\sqrt{12}}$

Solution

The product of the perpendiculars drawn from the point $P (x_{1}, y_{1})$ to the pair of lines represented by the equation ${ax}^{2} + 2 hxy + {by}^{2} = 0$ is $|\frac{{ax}_{1}^{2} + 2 {hx}_{1} y_{1} + {by}_{1}^{2}}{\sqrt{{(a - b)}^{2} + 4 h^{2}}}|$
The product of the perpendiculars drawn from the point $P (0, 0)$ to the pair of lines represented by the equation ${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ is $|\frac{c}{\sqrt{{(a - b)}^{2} + 4 h^{2}}}|$
Given that product of perpendicular from (1,1) to the pair of lines $2 x^{2} + 3 xy + y^{2} = 0$ is equal to the product of perpendiculars from (0,0) to the pair of lines
$x^{2} + 2 xy + 3 y^{2} + x + 2 y + k = 0$
Hence,
$\begin{matrix} |\frac{2 {(1)}^{2} + 3 (1) (1) + {(1)}^{2}}{\sqrt{{(2 - 1)}^{2} + {(3)}^{2}}}| = |\frac{k}{\sqrt{{(1 - 3)}^{2} + 2 h^{2}}}| \\ |\frac{6}{\sqrt{10}}| = |\frac{k}{\sqrt{8}}| \end{matrix}$
Cross multiply and then simplify $k = \frac{12}{\sqrt{5}}$
Therefore, $k = \frac{12}{\sqrt{5}}$

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