First slide

Pair of straight lines

Question

If $x^{2} + 2 xy + {ky}^{2} + 4 x - 2 y + l = 0$ represents pair of perpendicular lines then

Moderate

A

$k = 1, l = - \frac{1}{2}$
B

$k = - 1, l = \frac{1}{2}$
C

$k = - 1, l = - \frac{1}{2}$
D

$k = 1, l = \frac{1}{2}$

Solution

The equation ${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ represents pair of lines if $|\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix}| = 0$ Given that the equation $x^{2} + 2 xy + {ky}^{2} + 4 x - 2 y + l = 0$ represents a pair of perpendicular lines
If they are perpendicular lines $a + b = 0$
Hence,
$k + 1 = 0 \Rightarrow k = - 1$
Consider the condition for representing a pair of lines, substitute $k = - 1$
$\begin{matrix} |\begin{matrix} 1 & 1 & 2 \\ 1 & - 1 & - 1 \\ 2 & - 1 & l \end{matrix}| = 1 (- l - 1) - 1 (l + 2) + 2 (- 1 + 2) \\ = - 1 - l - l - 2 + 2 \\ = - 1 - 2 l \end{matrix}$
Hence,
$\begin{matrix} 1 + 2 l = 0 \\ l = - \frac{1}{2} \end{matrix}$
Therefore, $l = - \frac{1}{2} and k = - 1$

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