The triangle formed by the pair of lines $x^{2} - 4 y^{2} = 0$ and the line $x - a = 0$ is always

equilateral

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isosceles

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right angled

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scalene

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Solution

The correct option is B isosceles
$x^{2} - 4 y^{2} = 0$
$(x - 2 y) (x + 2 y) = 0$
$y = \frac{x}{2}$ and $y = \frac{- x}{2}$
So, triangle formed by $y = \frac{x}{2}, y = \frac{- x}{2}$ and $x = a$ is
$Δ O A B$ is isosceles triangle.

Suggest Corrections

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