Question

The straight lines x + y = 0, 3x + y-4 = 0, and x + 3y - 4 = 0 form a triangle which is

• A. isosceles
• B. equilateral
• C. Right-angled
• D. None of these

Answer 1

The correct option is A isosceles

Solving the given equation of lines pairwise, we get the vertices of triangle as $A(-2,2),B(2,-2),C(1,1).$ Then,
$AB=\sqrt{16+16}=4\sqrt{2}BC=\sqrt{1+9}=\sqrt{10}CA=\sqrt{9+1}=\sqrt{10}$
Hence, the triangle is isosceles.