Question

The triangle formed the lines $x+y-4=0,3x+y=4,x+3y=4$ is

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

Answer 1

The correct option is A isosceles

$x+y=4$ ........$\left(1\right)$

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

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

Solving $\left(1\right)$ and $\left(2\right)$

$x+y-3x-y=4-4$

$\Rightarrow -2x=0$ or $x=0$

Put $x=0$ in $\left(1\right)$ we get

$y=4-x=4-0=4$

$\therefore$ the point of intersection of the lines $\left(1\right)$ and $\left(2\right)$ is at $A(0,4)$

Eqn$\left(2\right)-3\times \left(3\right)$

$3x+y-3x-9y=4-12$

$-8y=-8$

$\Rightarrow y=\frac{-8}{-8}=1$

Put $y=1$ in $\left(3\right)$ we get

$y=4-3y=4-3=1$

$\therefore$ the point of intersection of the lines $\left(2\right)$ and $\left(3\right)$ is at $B(1,1)$

Solving $\left(3\right)$ and $\left(1\right)$

$x+3y-x-y=4-4$

$\Rightarrow 2y=0$

$\Rightarrow y=0$

Put $y=0$ in $\left(1\right)$ we get

$x=4-y=4-0=4$

$\therefore$ the point of intersection of the lines $\left(3\right)$ and $\left(1\right)$ is at $C(4,0)$

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

$AB=\sqrt{{(1-0)}^2+{(1-4)}^2}=\sqrt{1+9}=\sqrt{10}$

$BC=\sqrt{{(0-1)}^2+{(4-1)}^2}=\sqrt{1+9}=\sqrt{10}$

$CA=\sqrt{{(0-0)}^2+{(4-4)}^2}=0$

Since $CA=0$ there exists no line which joins the points $C$ and $A$.

Hence $ABC$ is not a triangle.