Question

The circumcentre of the triangle with vertices \( (-2,3), (2,-1), (4,0) \) is:

• A. \( \left( \dfrac{3}{2}, \dfrac{5}{2} \right) \)
• B. \( \left( \dfrac{3}{2}, \dfrac{-5}{2} \right) \)
• C. \( \left( \dfrac{-3}{2}, \dfrac{5}{2} \right) \)
• D. \( \left( \dfrac{-3}{2}, \dfrac{-5}{2} \right) \)

Hint:

To find the circumcentre, use the perpendicular bisectors of any two sides and find their intersection point.

Answer 1

The Correct Option is A

To find the circumcentre of a triangle with vertices \( A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) \), we solve the perpendicular bisectors of any two sides. Let \( A = (-2, 3), B = (2, -1), C = (4, 0) \). 1. Midpoint of \( AB \): \[ M_{AB} = \left( \dfrac{-2 + 2}{2}, \dfrac{3 + (-1)}{2} \right) = (0, 1) \] Slope of \( AB = \dfrac{-1 - 3}{2 - (-2)} = \dfrac{-4}{4} = -1 \Rightarrow \) Perpendicular slope = 1 Equation of perpendicular bisector of \( AB \): \[ y - 1 = 1(x - 0) \Rightarrow y = x + 1 \tag{1} \] 2. Midpoint of \( BC \): \[ M_{BC} = \left( \dfrac{2 + 4}{2}, \dfrac{-1 + 0}{2} \right) = (3, -0.5) \] Slope of \( BC = \dfrac{0 - (-1)}{4 - 2} = \dfrac{1}{2} \Rightarrow \) Perpendicular slope = -2 Equation of perpendicular bisector of \( BC \): \[ y + 0.5 = -2(x - 3) \Rightarrow y = -2x + 6 - 0.5 = -2x + 5.5 \tag{2} \] Solving equations (1) and (2): \[ x + 1 = -2x + 5.5 \Rightarrow 3x = 4.5 \Rightarrow x = \dfrac{3}{2},\quad y = x + 1 = \dfrac{5}{2} \] So, the circumcentre is: \[ \boxed{\left( \dfrac{3}{2}, \dfrac{5}{2} \right)} \]