Question

The triangle formed by 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

Explanation for correct option

Given,

$x+y–4=0$ ..(i)

$3x+y=4$…(ii)

$x+3y=4$ …(iii)

We get the point of intersection of lines by solving the equations.

Solving (i) and (ii), we get$x=0,y=4$

Let $A$be$(0,4)$

Solving (ii) and (iii), we get $x=1,y=1$

Let $B$ be $(1,1)$

Solving (i) and (iii), we get $x=4,y=0$

Let $C$ be $(4,0)$

Using the distance formula, we find the length of the sides.

$$\begin{gathered} AB=\sqrt{({(1–0)}^2+{(1–4)}^2)}=\sqrt{10} \\ BC=\sqrt{({(1–4)}^2+{(1–0)}^2)}=\sqrt{10} \\ AC=\sqrt{({(4–0)}^2+{(0–4)}^2)}=\sqrt{32} \end{gathered}$$

Here $AB=BC.$

So the triangle is isosceles.

Hence $\mathrm{Option}\left(\mathrm{A}\right)$ is correct.