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

Solution:

(d) right-angled

Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,

$P Q=\sqrt{(-5-0)^{2}+(3-6)^{2}}$

$=\sqrt{(-5)^{2}+(-3)^{2}}$

$=\sqrt{25+9}$

$=\sqrt{34}$ units

$Q R=\sqrt{(3+5)^{2}+(1-3)^{2}}$

$=\sqrt{(8)^{2}+(-2)^{2}}$

$=\sqrt{64+4}$

$=\sqrt{68}$

$=2 \sqrt{17}$ units

$P R=\sqrt{(3-0)^{2}+(1-6)^{2}}$

$=\sqrt{(3)^{2}+(-5)^{2}}$

$=\sqrt{9+25}$

$=\sqrt{34}$ units

$P Q^{2}+P R^{2} \Rightarrow\left\{(\sqrt{34})^{2}+(\sqrt{34})^{2}\right\}=68$

$Q R^{2} \Rightarrow(2 \sqrt{17})^{2}=68$

Thus, $P Q^{2}+P R^{2}=Q R^{2}$

Therefore, ∆PQR is right-angled.