First slide

Pair of straight lines

Question

The area of the triangle formed by pair of lines ${(ax + by)}^{2} - 3 {(bx - ay)}^{2} = 0$ and the line $ax + by + c = 0$ is

Moderate

A

$\frac{c^{2}}{a^{2} + b^{2}}$
B

$\frac{c^{2}}{\sqrt{2} (a^{2} + b^{2})}$
C

$\frac{c^{2}}{\sqrt{3} (a^{2} + b^{2})}$
D

$\frac{c^{2}}{2 (a^{2} + b^{2})}$

Solution

The area of the equilateral triangle formed by the lines $ax + by + c = 0$ and the pair of lines passing through the origin is $\frac{c^{2}}{\sqrt{3} (a^{2} + b^{2})}$
given lines
$ax + by + c = 0$ ……….(1)
By splitting the combined equation ${(ax + by)}^{2} - 3 {(bx - ay)}^{2} = 0$
$ax + by + \sqrt{3} (bx - ay) = 0$
$(a + \sqrt{3} b) x + (b - \sqrt{3} a) y = 0$ …………..(2)
$ax + by - \sqrt{3} (bx - ay) = 0$
$(a - \sqrt{3} b) x + (b + \sqrt{3} a) y = 0$ ………….(3)
Solving (1) and (2) vertex is $A = (\frac{c (b - \sqrt{3} a)}{\sqrt{3} (a^{2} + b^{2})}, - \frac{c (a + \sqrt{3} b)}{\sqrt{3} (a^{2} + b^{2})})$
Solving (2) and (3) , vertex is $0 = (0, 0)$
Solving (3) and (1) Vertex is $B = (- \frac{c (b + \sqrt{3} a)}{\sqrt{3} (a^{2} + b^{2})}, \frac{c (a + \sqrt{3} b)}{\sqrt{3} (a^{2} + b^{2})})$
$Area of the triangle OAB = \frac{1}{2} |x, y_{2} - x_{2} y_{1}| = \frac{c^{2}}{\sqrt{3} (a^{2} + b^{2})} square units$

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