Mathematics — Standard (041) — CBSE Class 10 board question

Let A(–2, 3), B(8, 3), C(6, 7).

$$AB^2 = (8-(-2))^2 + (3-3)^2 = 100 + 0 = 100$$

$$BC^2 = (6-8)^2 + (7-3)^2 = 4 + 16 = 20$$

$$AC^2 = (6-(-2))^2 + (7-3)^2 = 64 + 16 = 80$$

Since $BC^2 + AC^2 = 20 + 80 = 100 = AB^2$,

by the converse of Pythagoras theorem, $\angle C = 90°$.

Therefore, the given points are the vertices of a right-angled triangle.

Source: Chapter 7, Section 7.2 – Distance Formula

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• Examiners expect you to calculate all three squared distances using the distance formula, then verify the Pythagorean condition ($a^2 + b^2 = c^2$) with the largest value on the right.
• State the converse of Pythagoras theorem explicitly — this is what earns the concluding mark.
• Writing $BC^2 + AC^2 = AB^2$ clearly is essential; don't just say "it satisfies Pythagoras."
• No need to find actual distances (square roots); working with squared values is sufficient and cleaner.