Question

Let the image of point (1, 2) with respect to x-axis, y-axis and line y = -x are A, B and C respectively. Then area of $\triangle$ABC (in square units) is -

• A. 5
• B. 0
• C. 15
• D. 17

Answer 1

Answer: (A)

5

Answer 2

Let the given point be $P(1,2)$.

1. Reflection about x-axis: $A = (1, -2)$.

2. Reflection about y-axis: $B = (-1, 2)$.

3. Reflection about the line $y = -x$: The reflection of $(x, y)$ about $y = -x$ is $(-y, -x)$. Hence, $C = (-2, -1)$.

4. Area of $\triangle ABC$:

   Using the determinant formula:

   $$\text{Area} = \frac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$$

   Substituting:

   $$\text{Area} = \frac{1}{2} \left| 1\,(2 - (-1)) + (-1)\,((-1) - (-2)) + (-2)\,((-2) - 2) \right|$$ $$= \frac{1}{2} \left| 1\,(3) + (-1)\,(1) + (-2)\,(-4) \right|$$ $$= \frac{1}{2} \left| 3 - 1 + 8 \right| = \frac{1}{2} \times 10 = 5$$

Reflect $(1,2)$ to get $A(1,-2)$, $B(-1,2)$, and $C(-2,-1)$; then apply the determinant formula to find the area = 5 sq. units.