Question

The area of the triangle if A (-1, -1), B(-1, 3) and C (2, -1) are the vertices of the triangle is ____________

(a) 8 units

(b) 4 units

(c) 6 units

(d) 5 units

This question was addressed to me during an online interview.

I'd like to ask this question from Geometry topic in division Coordinate Geometry of Mathematics – Class 10

Answer 1

Right choice is (c) 6 units

To explain: Distance between A (-1, -1) and B (-1, 3) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-1 + 1)^2 + (3 + 1)^2} \)

= \( \sqrt {4^2} \)

= √16

= 4

Distance between B (-1, 3) and C(2, -1) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2 + 1)^2 + (-1-3)^2} \)

= \( \sqrt {3^2 + (-4)^2} \)

= \( \sqrt {9 + 16} \)

= 5

Distance between A (-1, -1) and C (2, -1) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2 + 1)^2 + (-1 + 1)^2} \)

= \( \sqrt {3^2 + 0^2} \)

= 3

Now, AC^2 + AB^2 = 4^2 + 3^2 = 16 + 9 = 25

BC^2 = 5^2 = 25

Hence, it is a right-angled triangle, right-angled at A.

Area of triangle = \( \frac {1}{2}\) × base × height = \( \frac {1}{2}\) × 4 × 3 = 6 units