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

Answer: (A)

an equilateral triangle

$(y+m_{1}x)(y+m_{2}x)=0$

$y^2+(m_1+m_2)xy+m_2{m}_1{x}^2=0$

Comparing coefficients we get

$m_1+m_2=4$

$m_1{m}_2=1$

This implies

$m_1^2+1=4m_1$

$m_1^2-4m_1+1=0$

Therefore $m_1=2-\sqrt{3}$ and $m_1=2+\sqrt{3}$

$\tan A=2-\sqrt{3}$ and $\tan A=2+\sqrt{3}$

Hence $A={15}^0$ and $A={75}^0$

Corresponding values of $m_2={75}^{\circ }$ and ${15}^{\circ }$

Therefore the angle between the lines is ${75}^0-{15}^0$

$={60}^0$

The equation of the angle bisectors of the lines is $x=\pm y$

The line $x=-y$ is perpendicular to $x-y=4$

Hence the above is an isosceles triangle with the vertical angle being ${60}^0$

Hence the triangle is an equilateral triangle.