Question

If (b₂ – b₁) (b₃ – b₁) + (a₂ – a₁) (a₃ – a₁) = 0, then circumcentre of the triangle having vertices (a₁, b₁), (a₂, b₂) and (a₃, b₃) is

• A. $\left( \frac{a_{1} + a_{2}}{2},\frac{b_{1} + b_{2}}{2} \right)$
• B. $\left( \frac{a_{1} + a_{2} + a_{3}}{3},\frac{b_{1} + b_{2} + b_{3}}{3} \right)$
• C. $\left( \frac{a_{2} + a_{3}}{2},\frac{b_{2} + b_{3}}{2} \right)$
• D. $\left( \frac{a_{1} + a_{3}}{2},\frac{b_{1} + b_{3}}{2} \right)$

Answer 1

Answer: (C)

$\left( \frac{a_{2} + a_{3}}{2},\frac{b_{2} + b_{3}}{2} \right)$

Answer 2

(b₂ – b₁) (b₃ – b₁) = – (a₂ – a₁) (a₃ – a₁)

$\left( \frac{b_{2} - b_{1}}{a_{2} - a_{1}} \right)\left( \frac{b_{3} - b_{1}}{a_{3} - a_{1}} \right)$ = – 1 Q m₁m₂ = –1

$\left( \frac{a_{2} + a_{3}}{2},\frac{b_{2} + b_{3}}{3} \right)$