Question

The equations $x-y=4$ 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. a right angled isosceles triangle

Answer 1

The correct option is A an equilateral triangle

$x^2+4xy+y^2=0\Rightarrow {\left(\frac{y}{x}\right)}^2+4\frac{y}{x}+1=0\Rightarrow \frac{y}{x}=-2\pm \sqrt{3}$
Acute angle between the lines is ${\tan }^{-1}\frac{\left(2\sqrt{4-1}\right)}{1+1}=\frac{\pi }{3}$
Now angle made by the line $x-y=4$ and $\frac{y}{x}=-2\pm \sqrt{3}$
$={\tan }^{-1}\left|\frac{-2\pm \sqrt{3}-1}{1+(-2\pm \sqrt{3})}\right|={\tan }^{-1}\left|\frac{\sqrt{3}(-\sqrt{3}\pm 1)}{-1\pm \sqrt{3}}\right|=\frac{\pi }{3}$
Hence the lines will form a equilateral triangle

Alternate Solution:
Acute angle between pair of lines $x^2+4xy+y^2=0$
$={\tan }^{-1}\frac{2\sqrt{2^2-1}}{2}=\frac{\pi }{3}$
the line $y=x+c$ makes equal angle with $x^2+4xy+y^2=0$
$[\because y=x\text{doesn't change the pair of lines}{x}^2+4xy+y^2=0]$
$\therefore$ all angles of the triangle are equal to $60°$
Hence triangle is equilateral triangle