CUET PG 2022 — Mathematics PYQ

Step 1: Find the Vertices

1. Intersection of $L_1:x+y=0$ and $L_2:3x+y-4=0$

Subtracting $L_1$ from $L_2$:

Substitute $x=2$ in $L_1$: $2+y=0⟹y=-2$.

Vertex A = $(2,-2)$

2. Intersection of $L_1:x+y=0$ and $L_3:x+3y-4=0$

Subtracting $L_1$ from $L_3$:

Substitute $y=2$ in $L_1$: $x+2=0⟹x=-2$.

Vertex B = $(-2,2)$

3. Intersection of $L_2:3x+y-4=0$ and $L_3:x+3y-4=0$

Multiply $L_3$ by $3$: $3x+9y-12=0$.

Subtract $L_2$ from this:

Substitute $y=1$ in $L_3$: $x+3(1)-4=0⟹x=1$.

Vertex C = $(1,1)$

Step 2: Calculate the Length of the Sides

Using the distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$:

• Length AB:

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

• Length BC:

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

• Length AC:

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

Step 3: Analyze the Triangle Type

1. Isosceles: Since $BC=AC=\sqrt{10}$, the triangle has two equal sides.

2. Right-angled check: Let's check the slopes $m_1$ and $m_2$ of the lines.

3. Slope of $L_2$ ($3x+y-4=0$): $m_2=-3$

4. Slope of $L_3$ ($x+3y-4=0$): $m_3=-1/3$

5. Since $m_2\times m_3\ne -1$, it is not a right-angled triangle at Vertex C.

Final Answer:

The triangle formed is an Isosceles triangle.