First slide

Pair of straight lines

Question

The product of the perpendicular from (1,1) to the pair of lines $x^{2} + 2 xy + y^{2} + kx + 2 y + 5 = 0$ is $\sqrt{2}$ then $k = (k < 0)$

Moderate

A

$2 \sqrt{2} - 11$
B

$2 \sqrt{2} + 11$
C

$11 - 2 \sqrt{2}$
D

$- 11 - 2 \sqrt{2}$

Solution

The product of the perpendiculars drawn from the point $P (x_{1}, y_{1})$ to the pair of lines represented by the equation ${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ is
$|\frac{{ax}_{1}^{2} + 2 {hx}_{1} y_{1} + {by}_{1}^{2} + 2 {gx}_{1} + 2 {fy}_{1} + c}{\sqrt{{(a - b)}^{2} + 4 h^{2}}}|$
For the equation $x^{2} + 2 xy + y^{2} + kx + 2 y + 5 = 0$ the product of the perpendiculars from the point (1,1) is $\sqrt{2}$ .
$\begin{array}{l} \sqrt{2} = |\frac{{(1)}^{2} + 2 (1) (1) + {(1)}^{2} + k (1) + 2 (1) + 5}{\sqrt{{(1 - 1)}^{2} + {(2)}^{2}}}| \\ = |\frac{11 + k}{\sqrt{4}}| \\ = |\frac{11 + k}{2}| \end{array}$
Squaring on both sides and then simplify, take only positive value for k
$\begin{matrix} 11 + k = 2 \sqrt{2} \\ k = 2 \sqrt{2} - 11 \end{matrix}$
Therefore, the value of k is $2 \sqrt{2} - 11$

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