MAH-CET 2026 — Mathematics PYQ

To determine the type of triangle formed by the given straight lines, we can find either the lengths of its sides or the relationships between the slopes of the lines. Let's name the given equations:

• Line 1 ($L_1$): $x+y=0$

• Line 2 ($L_2$): $3x+y-4=0$

• Line 3 ($L_3$): $x+3y-4=0$

Step 1: Find the vertices of the triangle

Vertices are found by calculating the points of intersection for each pair of straight lines.

• Intersection of $L_1$ and $L_2$ (Vertex A):

  From $L_1$, we have $y=-x$. Substitute this into $L_2$:

  $$3x+(-x)-4=0⟹2x=4⟹x=2$$

  Since $y=-x$, then $y=-2$.

  $$\text{Vertex A}=(2,-2)$$

• Intersection of $L_1$ and $L_3$ (Vertex B):

  From $L_1$, we have $y=-x$. Substitute this into $L_3$:

  $$x+3(-x)-4=0⟹-2x=4⟹x=-2$$

  Since $y=-x$, then $y=2$.

  $$\text{Vertex B}=(-2,2)$$

• Intersection of $L_2$ and $L_3$ (Vertex C):

  Subtracting $L_3$ from $L_2$:

  $$(3x+y-4)-(x+3y-4)=0$$

  $$2x-2y=0⟹x=y$$

  Substitute $x=y$ into $L_2$:

  $$3x+x-4=0⟹4x=4⟹x=1⟹y=1$$

  $$\text{Vertex C}=(1,1)$$

Step 2: Calculate the lengths of the sides

We use the distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$ to find the lengths of sides $AB$, $BC$, and $CA$.

• Length of side $AB$:

  $$AB=\sqrt{(-2-2{)}^2+(2-(-2){)}^2}=\sqrt{(-4{)}^2+(4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$$

• Length of side $BC$:

  $$BC=\sqrt{(1-(-2){)}^2+(1-2{)}^2}=\sqrt{(3{)}^2+(-1{)}^2}=\sqrt{9+1}=\sqrt{10}$$

• Length of side $CA$:

  $$CA=\sqrt{(2-1{)}^2+(-2-1{)}^2}=\sqrt{(1{)}^2+(-3{)}^2}=\sqrt{1+9}=\sqrt{10}$$

Step 3: Analyze side lengths

Comparing the side lengths:

• $AB=\sqrt{32}$

• $BC=\sqrt{10}$

• $CA=\sqrt{10}$

Since two sides are equal in length ($BC=CA\ne AB$), the triangle is isosceles.

Correct Answer

(b) isosceles