Chapter: Co-ordinate Geometry· Mathematics Part 2 (Geometry)
Answer: \(AC^2 = AB^2 + BC^2\), so ΔABC is right-angled at B.
\(AB^2 = (1 + 3)^2 + (-2 - 2)^2 = 16 + 16 = 32\)
\(BC^2 = (9 - 1)^2 + (6 + 2)^2 = 64 + 64 = 128\)
\(AC^2 = (9 + 3)^2 + (6 - 2)^2 = 144 + 16 = 160\)
\(AB^2 + BC^2 = 32 + 128 = 160 = AC^2\)
By the converse of Pythagoras theorem, ΔABC is a right-angled triangle and the right angle is opposite the longest side AC, i.e. \(\angle ABC = 90^\circ\).