First slide

Pair of straight lines

Question

The equation to the pair of lines perpendicular to $2 x^{2} + 3 xy + y^{2} = 0$ and passing through $(2, 1)$ is $x^{2} - 3 xy + 2 y^{2} + lx + my + n = 0$ then $l + m + n =$

Moderate

A

-1
B

0
C

-4
D

1

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 $2 x^{2} + 3 xy + y^{2} = 0$ and passing through the point $(2, 1)$ is ${(x - 2)}^{2} - 3 (x - 2) (y - 1) + 2 {(y - 1)}^{2} = 0$
Simplifying the equation
$\begin{matrix} {(x - 2)}^{2} - 3 (x - 2) (y - 1) + 2 {(y - 1)}^{2} = 0 \\ x^{2} - 4 x + 4 - 3 xy + 3 x + 6 y - 6 + 2 y^{2} - 4 y + 2 = 0 \\ x^{2} - 3 xy + 2 y^{2} - x + 2 y = 0 \end{matrix}$
Comparing the above equation with $x^{2} - 3 xy + 2 y^{2} + lx + my + n = 0$
We get $l = - 1, m = 2, n = 0$
Therefore, $l + m + n = 1$

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