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 angled triangle

Answer 1

The correct option is A an isosceles triangle

Given lines are
$x+y=0$ .....(i)
$5x+y=4$ .....(ii)
and $x+5y=4$ .....(iii)

On solving above equations pairwise, we get
$A(1,-1),B(\frac{2}{3},\frac{2}{3}),C(-1,1)$

Distance between two point $(x_1,y_1)$ and $(x_2,y_2)$ is given by,

D$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Now, $AB=\sqrt{{(\frac{2}{3}-1)}^2+{(\frac{2}{3}+1)}^2}$

$=\sqrt{\frac{1}{9}+\frac{25}{9}}=\sqrt{\frac{26}{9}}=\frac{\sqrt{26}}{3}$

$BC=\sqrt{{(-1-\frac{2}{3})}^2+{(1-\frac{2}{3})}^2}$

$=\sqrt{\frac{25}{9}+\frac{1}{9}}=\sqrt{\frac{26}{9}}=\frac{\sqrt{26}}{3}$

and $CA=\sqrt{{(1+1)}^2+{(-1-1)}^2}$

$=\sqrt{4+4}=2\sqrt{2}$

Here, we see that $AB=BC$
Hence, given lines form an isosceles triangle.