Question

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

• A. isosceles
• B. right -angled
• C. obtuse-angled
• D. equilateral

Answer 1

Answer: (A), (C)

isosceles
obtuse-angled

Let the given lines be represented by $AB,BC$ and $CA$, respectively, Then $A(-2,2),B(2,-2)$ and $C(1,1)$ are the vertices of a triangle $ABC$.
Also, note that $AB=4\sqrt{2},AC=\sqrt{10}andBC=\sqrt{10}.$
This shows that $△ABC$ is isosceles; it is clearly not right-angled or equilateral. Since
$\cos C=\frac{B{C}^2+A{C}^2-A{B}^2}{2\left(BC\right)\left(AC\right)}=\frac{10+10-32}{2\sqrt{1}0\sqrt{1}0}<0$
$\Rightarrow △ABC$ is obtuse-angled.