Question

Find the circumcentre of triangle form by the points $(3,2,-5)$,$(-3,8,-5)$,$(-3,2,-5)$

• A. $(0,5,5)$
• B. $(0,5,-5)$
• C. $(0,-5,0)$
• D. None of these

Answer 1

The correct option is B $(0,5,-5)$

$A(3,2,-5),B(-3,8,-5),C(-3,2,-5)$

$AB=\sqrt{6^2+6^2+0}=6\sqrt{2}$

$BC=\sqrt{6^2+0+0}=6$

$CA=\sqrt{6^2+0+0}=6$

It is a right angled $\Delta$ at $C$ because

$(AB{)}^2=(BC{)}^2+(CA{)}^2$

Circumference of right angled $\Delta$ lies on midpoint of hypotenuse AB

$C(\frac{3+-3}{2},\frac{2+8}{2},\frac{-5+-5}{2})$

$C(0,5,-5)$ is circumcentre of the triangle