Question

The equation x- y = 4 and $x^2+4xy+y^2=0$ represent the sides of

• A. an equilateral triangle
• B. a right angled triangle
• C. an isosceles triangle
• D. none of these

Answer 1

The correct option is A

an equilateral triangle

Acute angle between the lines $x^2+4xy+y^2=0$ is
$ta{n}^{-1}\frac{2\sqrt{4-1}}{1+1}=ta{n}^{-1}\sqrt{3}=\frac{\pi }{3}$
Angle bisectors of $x^2+4xy+y^2=0$ are given by
$\frac{x^2-y^2}{1-1}=\frac{xy}{2}\Rightarrow x^2-y^2=0\Rightarrow x=\pm y$
As x + y = 0 is perpendicular to x - y = 4, the given triangle is isosceles with vertical angle equal to $\frac{\pi }{3}$ and hence it is equilateral.