Question

Prove that the points (7, 10), (−2, 5) and (3, −4) are the vertices of an isosceles right triangle. [CBSE 2013]

Answer 1

Let the given points be A(7, 10), B(−2, 5) and C(3, −4).

Using distance formula, we have

$\mathrm{AB}=\sqrt{{\left(-2-7\right)}^2+{\left(5-10\right)}^2}=\sqrt{{\left(-9\right)}^2+{\left(-5\right)}^2}=\sqrt{81+25}=\sqrt{106}$ units

$\mathrm{BC}=\sqrt{{\left[3-\left(-2\right)\right]}^2+{\left(-4-5\right)}^2}=\sqrt{5^2+{\left(-9\right)}^2}=\sqrt{25+81}=\sqrt{106}$ units

$\mathrm{CA}=\sqrt{{\left(3-7\right)}^2+{\left(-4-10\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-14\right)}^2}=\sqrt{16+196}=\sqrt{212}$ units

Thus, AB = BC = $\sqrt{106}$ units

∴ ∆ABC is an isosceles triangle.

Also,

AB² + BC² = 106 + 106 = 212

and CA² = 212

∴ AB² + BC² = CA²

So, ∆ABC is right angled at B. (Converse of Pythagoras theorem)

Hence, the given points are the vertices of an isosceles right triangle.