Question

In the following examples, can the segment joining the given points form a triangle ? If triangle is formed, state the type of the triangle considering sides of the triangle.
(1) L(6,4) , M(–5,–3) , N(–6,8)

(2) P(–2,–6) , Q(–4,–2), R(–5,0)

(3) $\mathrm{A}\left(\sqrt{2},\sqrt{2}\right),\mathrm{B}\left(–\sqrt{2},–\sqrt{2}\right),\mathrm{C}\left(–\sqrt{6},\sqrt{6}\right)$

Answer 1

(1) L(6,4) , M(–5,–3) , N(–6,8)
$$\begin{gathered} \mathrm{LM}=\sqrt{{\left(6+5\right)}^2+{\left(4+3\right)}^2}=\sqrt{121+49}=\sqrt{170}=2\sqrt{35} \\ \mathrm{MN}=\sqrt{{\left(-5+6\right)}^2+{\left(-3-8\right)}^2}=\sqrt{1+121}=\sqrt{122}=2\sqrt{61} \\ \mathrm{LN}=\sqrt{{\left(6+6\right)}^2+{\left(4-8\right)}^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10} \end{gathered}$$
LM, MN and LN form a scalene triangle.

(2) P(–2,–6) , Q(–4,–2), R(–5,0)
$$\begin{gathered} \mathrm{PQ}=\sqrt{{\left(-2+4\right)}^2+{\left(-6+2\right)}^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5} \\ \mathrm{QR}=\sqrt{{\left(-4+5\right)}^2+{\left(-2+0\right)}^2}=\sqrt{1+4}=\sqrt{5} \\ \mathrm{PR}=\sqrt{{\left(-2+5\right)}^2+{\left(-6+0\right)}^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5} \\ \mathrm{PQ}+\mathrm{QR}=\mathrm{PR} \end{gathered}$$
Since the sum of the two sides is not greater than the third side so, the given vertices do not form a triangle.

(3) $\mathrm{A}\left(\sqrt{2},\sqrt{2}\right),\mathrm{B}\left(–\sqrt{2},–\sqrt{2}\right),\mathrm{C}\left(–\sqrt{6},\sqrt{6}\right)$
$$\begin{gathered} \mathrm{AB}=\sqrt{{\left(\sqrt{2}+\sqrt{2}\right)}^2+{\left(\sqrt{2}+\sqrt{2}\right)}^2}=\sqrt{8+8}=\sqrt{16}=4 \\ \mathrm{BC}=\sqrt{{\left(-\sqrt{2}+\sqrt{6}\right)}^2+{\left(-\sqrt{2}-\sqrt{6}\right)}^2}=\sqrt{16}=4 \\ \mathrm{AC}=\sqrt{{\left(\sqrt{2}+\sqrt{6}\right)}^2+{\left(\sqrt{2}-\sqrt{6}\right)}^2}=\sqrt{16}=4 \\ \mathrm{AB}=\mathrm{BC}=\mathrm{AC} \end{gathered}$$
So, these vertices form an equilateral triangle.