If the straight lines x+3y=4, 3x+y=4 and x+y=0 form a triangle, then the triangle is:

Question

If the straight lines $x+3y=4,\ 3x+y=4$ and $x+y=0$ form a triangle, then the triangle is:

Accepted Answer

Vertices: $x+3y=4 \cap 3x+y=4 \Rightarrow (1,1)$; $x+3y=4\cap x+y=0 \Rightarrow (-2,2)$; $3x+y=4\cap x+y=0 \Rightarrow (2,-2)$.

$$AB=\sqrt{9+1}=\sqrt{10}, \quad BC=\sqrt{1+9}=\sqrt{10}, \quad CA=\sqrt{16+16}=\sqrt{32}$$

Since $AB=BC$: the triangle is isosceles.

Solutions · 1