First slide

Pair of straight lines

Question

The equation to the pair of lines passing through (1,2)and perpendicular to $x^{2} + 2 xy + y^{2} = 0$ is

Easy

A

$x^{2} - 2 xy + y^{2} - 2 x - 2 y + 1 = 0$
B

$x^{2} - 2 xy + y^{2} + 2 x - 2 y + 1 = 0$
C

$x^{2} - 2 xy + y^{2} + 2 x + 2 y + 1 = 0$
D

$x^{2} + 2 xy + y^{2} + 2 x - 2 y + 1 = 0$

Solution

The combined equation of pair of lines which are perpendicular to the pair of lines ${ax}^{2} + 2 hxy + {by}^{2} = 0$ and passing through the point $(x_{1}, y_{1})$ is
$b {(x - x_{1})}^{2} - 2 h (x - x_{1}) (y - y_{1}) + a {(y - y_{1})}^{2} = 0$
Hence the equation of pair of lines perpendicular to the pair of lines $x^{2} + 2 xy + y^{2} = 0$ and passing through the point d is ${(x - 1)}^{2} - 2 (x - 1) (y - 2) + {(y - 2)}^{2} = 0$
Simplifying the equation
$\begin{matrix} {(x - 1)}^{2} - 2 (x - 1) (y - 2) + {(y - 2)}^{2} = 0 \\ x^{2} - 2 x + 1 - 2 (xy - 2 x - y + 2) + (y^{2} - 4 y + 4) = 0 \\ x^{2} - 2 x + 1 - 2 xy + 4 x + 2 y - 4 + y^{2} - 4 y + 4 = 0 \\ x^{2} - 2 xy + y^{2} + 2 x - 2 y + 1 = 0 \end{matrix}$
Therefore, the required equation is $x^{2} - 2 xy + y^{2} + 2 x - 2 y + 1 = 0$

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