Question

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

• A. isosceles
• B. equilateral
• C. right angled
• D. none of these

Answer 1

The correct option is A isosceles

$x+y=0$ ..... $\left(i\right)$

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

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

Solving lines $\left(i\right)$ and $\left(ii\right)$, we get

$-x=-3x+4⟹x=2⟹y=-2$

$\therefore \left(i\right)$ and $\left(ii\right)$ intersect at $A=(2,-2)$

Solving lines $\left(ii\right)$ and $\left(iii\right)$, we get

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

$\therefore \left(ii\right)$ and $\left(iii\right)$ intersect at $B=(1,1)$

Solving lines $\left(i\right)$ and $\left(iii\right)$, we get

$-x=\frac{4-x}{3}⟹-3x=4-x⟹x=-2⟹y=2$

$\therefore \left(i\right)$ and $\left(iii\right)$ intersect at $C=(-2,2)$

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

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

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

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

$\therefore AB=BC$

Since, two sides of a triangle are equal then the triangle formed by $A,B,C$ is isosceles triangle.

Hence, option A is correct.