First slide

Pair of straight lines

Question

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

Easy

A

$- \frac{4}{3}$
B

$\frac{4}{3}$
C

$\frac{3}{4}$
D

$- \frac{3}{4}$

Solution

The product of the perpendiculars drawn from the point $P (x_{1}, y_{1})$ to the pair of lines represented by the equation ${ax}^{2} + 2 hxy + {by}^{2} = 0$ is $|\frac{{ax}_{1}^{2} + 2 {hx}_{1} y_{1} + {by}_{1}^{2}}{\sqrt{{(a - b)}^{2} + 4 h^{2}}}|$
Given that the product of the perpendiculars from the point $(- 1, 2)$ to the pair of lines $2 x^{2} + kxy + y^{2} = 0$ is 2
$\begin{matrix} 2 = |\frac{2 {(- 1)}^{2} + k (- 1) (2) + {(2)}^{2}}{\sqrt{{(2 - 1)}^{2} + {(k)}^{2}}}| \\ 2 = |\frac{6 - 2 k}{\sqrt{k^{2} + 1}}| \end{matrix}$
Squaring on both sides and then simplify
$\begin{matrix} k^{2} + 1 = {(3 - k)}^{2} \\ 9 - 6 k = 1 \\ k = \frac{8}{6} \end{matrix}$
Therefore, the value of k is $\frac{4}{3}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App