First slide

Pair of straight lines

Question

If $a + b = 2 h$ , then the area of the triangle formed by the lines ${ax}^{2} + 2 hxy + {by}^{2} = 0$ and the line $x - y + 2 = 0$ in square units is

Moderate

A

$|\frac{a - b}{a + b}|$
B

$|\frac{a + b}{a - b}|$
C

$|\frac{a^{2} + b^{2}}{a - b}|$
D

$|\frac{a^{2} + b^{2}}{a + b}|$

Solution

The area of the triangle formed by the lines ${ax}^{2} + 2 hxy + {by}^{2} = 0$ and $lx + my + n = 0$ is
$|\frac{n^{2} \sqrt{h^{2} - ab}}{{am}^{2} - 2 hlm + {bl}^{2}}|$
Here $a = a, 2 h = 2 h, b = b, l = 1, m = - 1, n = 2$
Substitute the appropriate values in the above formula and then use the condition $a + b = 2 h$
Hence, the required area is $\frac{4 \sqrt{h^{2} - ab}}{|a + 2 h + b|} = \frac{4 \sqrt{\frac{{(a + b)}^{2}}{4} - ab}}{|a + b + a + b|} = |\frac{a - b}{a + b}|$

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