First slide

Pair of straight lines

Question

The equation to the pair of lines perpendicular to $x^{2} + xy + y^{2} = 0$ and passing through (2,1),is $x^{2} - xy + y^{2} + lx + my + n = 0$ then $l + m + n =$

Easy

A

-1
B

0
C

-4
D

4

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 $x^{2} + xy + y^{2} = 0$ and passing through the point (2,1), is ${(x - 2)}^{2} - (x - 2) (y - 1) + {(y - 1)}^{2} = 0$
Simplifying the equation
$\begin{matrix} x^{2} - 4 x + 4 - xy + x + 2 y - 2 + y^{2} - 2 y + 1 = 0 \\ x^{2} - xy + y^{2} - 3 x + 3 = 0 \end{matrix}$
Comparing the above equation with $x^{2} - xy + y^{2} + lx + my + n = 0$
We get $l = - 3, m = 0, n = 3$
Therefore, $l + m + n = 0$

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