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

Explanation for the correct option:

Compute the required term:

Given:

$$\begin{gathered} x+y=0...\left(i\right) \\ 3x+y-4=0...\left(ii\right) \\ x+3y-4=0...\left(iii\right) \end{gathered}$$

From $\left(i\right)$ we get $x=-y$

Put $x=-y$ in $\left(ii\right)$

$$\begin{gathered} -3y+y-4=0 \\ -2y=4 \end{gathered}$$

So $y=-2$

Then $x=2$

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

Similarly solving $\left(ii\right)$ and $\left(iii\right),$ we get $x=1,y=1.$

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

Similarly solving $\left(i\right)$ and $\left(iii\right),$ we get $x=-2,y=2$

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

$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ (distance formula)

$$\begin{gathered} =\sqrt{{\left(1-2\right)}^2+{\left(1+2\right)}^2} \\ =\sqrt{1+9} \\ =\sqrt{10} \end{gathered}$$

Similarly

$$\begin{gathered} BC=\sqrt{{\left(-2-1\right)}^2+{\left(2-1\right)}^2} \\ =\sqrt{10} \end{gathered}$$

$$\begin{gathered} AC=\sqrt{{\left(-2-2\right)}^2+{\left(2+2\right)}^2} \\ =\sqrt{32} \end{gathered}$$

Here $AB=BC$

Here the two sides of the triangle are equal. So the triangle is isosceles.

Hence, option $A$ is the correct answer.