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

Given lines

$x+y=0$-----$\left(1\right)$

$3x+y-4=0$-----$\left(2\right)$

$x+3y-4=0$-----$\left(3\right)$

On solving equation $\left(1\right)$ and $\left(2\right)$ we get

$-x=-3x+4\Rightarrow x=2\Rightarrow y=-2$

$\therefore$ intersection point $A(2,-2)$

On solving equation $\left(2\right)$ and $\left(3\right)$ we get

$-3x+4=\frac{4-x}{3}\Rightarrow -9x+12=4-x\Rightarrow x=1\Rightarrow y=1$

$\therefore$ intersection point $B(1,1)$

On solving equation $\left(3\right)$ and $\left(1\right)$ we get

$-x=\frac{4-x}{3}\Rightarrow -3x=4-x\Rightarrow x=-2\Rightarrow y=2$

$\therefore$ intersection point $C(-2,2)$

So $AB,BC,AC$ form a triangle $ABC$

$AB=\sqrt{(1-2{)}^2+(1+2{)}^2}=\sqrt{10}$

$BC=\sqrt{(-1-2{)}^2+(2-1{)}^2}=\sqrt{10}$

$AC=\sqrt{(-2-2{)}^2+(2+2{)}^2}=\sqrt{32}$

$\therefore AB=BC$

Since two sides of triangle are equal then $ABC$ is an isosceles triangle