Question

The points P(0, 6), Q(−5, 3) and R(3, 1) are the vertices of a triangle, which is

(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

Answer 1

(d) right-angled

Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,
$$\begin{gathered} PQ=\sqrt{{\left(-5-0\right)}^2+{\left(3-6\right)}^2} \\ =\sqrt{{\left(-5\right)}^2+{\left(-3\right)}^2} \\ =\sqrt{25+9} \\ =\sqrt{34}\text{units} \\ QR=\sqrt{{\left(3+5\right)}^2+{\left(1-3\right)}^2} \\ =\sqrt{{\left(8\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{64+4} \\ =\sqrt{68} \\ =2\sqrt{17}\text{units} \\ PR=\sqrt{{\left(3-0\right)}^2+{\left(1-6\right)}^2} \\ =\sqrt{{\left(3\right)}^2+{\left(-5\right)}^2} \\ =\sqrt{9+25} \\ =\sqrt{34}\text{units} \\ P{Q}^2+P{R}^2\Rightarrow \left\{{\left(\sqrt{34}\right)}^2+{\left(\sqrt{34}\right)}^2\right\}=68 \\ Q{R}^2\Rightarrow {\left(2\sqrt{17}\right)}^2=68 \end{gathered}$$
$\text{Thus,}P{Q}^2+P{R}^2=Q{R}^2$
Therefore, ∆PQR is right-angled.