Question

The straight lines $x+y=0,5x+y=4$ and $x+5y=4$ form

• A. an isosceles triangle
• B. an equilateral triangle
• C. a scalene triangle
• D. a right angles triangle

Answer 1

The correct option is A

an isosceles triangle

Explanation for the correct option:

Compute the required term:

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

$5x+y=4..\left(ii\right)$

$x+5y=4\dots \left(iii\right)$

Solving $\left(i\right)$ and $\left(ii\right),$ we get $x=1,y=-1$

So $A=(1,-1)$

Solving $\left(ii\right)$ and $\left(iii\right),$ we get $x=\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,y=\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

So $B=(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.)$

Solving $\left(i\right)$ and $\left(iii\right),$ we get $x=-1,y=1$

So $C=(-1,1)$

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

$=\raisebox{1ex}{$\sqrt{26}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Similarly $BC=\raisebox{1ex}{$\sqrt{26}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

$AC=2\sqrt{2}$

Here $AB=BC$

Here the two sides of the triangle are equal, so the triangle is isosceles.

Hence option $A$ is the correct answer.