First slide

Pair of straight lines

Question

The area of the triangle formed by the lines $3 x^{2} - 4 xy + y^{2} = 0, 2 x - y + 6 = 0$ is

Moderate

A

9
B

36
C

45
D

54

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 $3 x^{2} - 4 xy + y^{2} = 0, 2 x - y + 6 = 0$
The values are $a = 3, 2 h = - 4, b = 1, l = 2, m = - 1, n = 6$
Therefore, the area of the triangle is A
$\begin{array}{l} A = |\frac{{(6)}^{2} \sqrt{{(2)}^{2} - (3) (1)}}{3 {(- 1)}^{2} + 4 (2) (- 1) + 1 {(2)}^{2}}| \\ = |\frac{36}{1}| \\ = 36 \end{array}$
Therefore, the area of the triangle is $36 square units$

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