First slide

Pair of straight lines

Question

The area of the triangle formed by the lines $x^{2} + 3 xy + y^{2} = 0, x + y + 1 = 0$ is

Moderate

A

$- \frac{2}{\sqrt{5}}$
B

$\frac{\sqrt{5}}{2}$
C

$- \frac{\sqrt{5}}{2}$
D

$\frac{2}{\sqrt{5}}$

Solution

The area of the triangle formed by the lines ${ax}^{2} + 2 hxy + {by}^{2} = 0$ and $lx + my + n = 0$ is $|\frac{n^{2} \sqrt{h^{2} - ab}}{{am}^{2} - 2 hlm + {bl}^{2}}|$
For the equations $x^{2} + 3 xy + y^{2} = 0, x + y + 1 = 0$
The values are $a = 1, 2 h = 3, b = 1, l = 1, m = 1, n = 1$
Therefore, the area of the triangle is A
$\begin{array}{l} A = |\frac{{(1)}^{2} \sqrt{{(\frac{3}{2})}^{2} - (1) (1)}}{1 {(1)}^{2} - 3 (1) (1) + 1 {(1)}^{2}}| \\ = |\frac{\sqrt{\frac{5}{4}}}{1}| \\ = \frac{\sqrt{5}}{2} \end{array}$
Therefore, the area of the triangle is $\frac{\sqrt{5}}{2}$

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