First slide

Pair of straight lines

Question

If $O, A, B$ are vertices of a triangle whose sides are given by ${(5 x + 12 y)}^{2} - 3 {(12 x - 5 y)}^{2} = 0$ and $5 x + 12 y - 78 = 0$ then AB =

Moderate

A

$2 \sqrt{3}$
B

$3 \sqrt{3}$
C

$4 \sqrt{3}$
D

$5 \sqrt{3}$

Solution

The triangle formed by the lines ${(5 x + 12 y)}^{2} - 3 {(12 x - 5 y)}^{2} = 0$ and $5 x + 12 y - 78 = 0$ is equilateral triangle.
The area of The area of the equilateral triangle formed by the lines $ax + by + c = 0$ and the pair of lines ${(ax + by)}^{2} - 3 {(bx - ay)}^{2} = 0$ is $\frac{c^{2}}{\sqrt{3} (a^{2} + b^{2})}$
Hence, the area of the triangle is $\frac{{(78)}^{2}}{\sqrt{3} (5^{2} + 12^{2})}$
If the length of the side of the equilateral triangle is a then its area is $\frac{\sqrt{3} a^{2}}{4}$
it implies that
$\begin{matrix} \frac{\sqrt{3} a^{2}}{4} = \frac{{(78)}^{2}}{\sqrt{3} (5^{2} + 12^{2})} \\ a^{2} = \frac{4 \cdot 78 \cdot 78}{3 (169)} \\ = \frac{4 \cdot 6 \cdot 6}{3} \\ = 48 \end{matrix}$
Therefore, $a = 4 \sqrt{3}$

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