First slide

Pair of straight lines

Question

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

Easy

A

$- \frac{2}{9}$
B

$\frac{2}{9}$
C

$\frac{9}{2}$
D

$- \frac{9}{2}$

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 $x^{2} + 2 xy + y^{2} + 2 x + 2 y + 1 = 0$ the product of the perpendiculars from the point (1,1) is d
$\begin{array}{l} d = |\frac{{(1)}^{2} + 2 (1) (1) + {(1)}^{2} + 2 (1) + 2 (1) + 1}{\sqrt{{(1 - 1)}^{2} + {(2)}^{2}}}| \\ = |\frac{9}{\sqrt{4}}| \\ = |\frac{9}{2}| \end{array}$
Therefore, the product of the perpendiculars drawn from the point (1,1) to the pair of lines
$x^{2} + 2 xy + y^{2} + 2 x + 2 y + 1 = 0 is \frac{9}{2} units$

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