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

Using the distance formula:

$$AB = \sqrt{(-5-1)^2 + (0-0)^2} = \sqrt{36} = 6$$

$$BC = \sqrt{(-2+5)^2 + (5-0)^2} = \sqrt{9+25} = \sqrt{34}$$

$$CA = \sqrt{(1+2)^2 + (0-5)^2} = \sqrt{9+25} = \sqrt{34}$$

Since $BC = CA = \sqrt{34}$ but $AB = 6 \neq \sqrt{34}$, triangle ABC is an isosceles triangle.

Source: Chapter 7, Section 7.2 (Distance Formula)

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• The examiner wants you to calculate all three sides using the distance formula and then compare them.
• If two sides are equal → isosceles; all three equal → equilateral; none equal → scalene; check Pythagoras for right-angled.
• Always state the conclusion clearly. One mark is typically for correct calculation; the second for the correct identification of triangle type.