Question

In the following example, can the segment joining the given point form a triangle? If triangle is formed, state the type of the triangle considering side of the triangle.

 P(–2, –6) , Q(–4, –2), R(–5, 0)

Answer 1

 P(–2, –6) , Q(–4, –2), R(–5, 0)

By distance formula,

\[\mathrm{d}(\mathrm{P},\mathrm{Q}) = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\]

= \[\sqrt{[-4-(-2)]^{2}+[-2-(-6)]^{2}}\]

= \[\sqrt{\left(-4+2\right)^{2}+\left(-2+6\right)^{2}}\]

= \[\sqrt{\left(-2\right)^{2}+4^{2}}\]

= \[\sqrt{4+16}\]

= \[\sqrt{20}\]

∴ d(P, Q) = \[2\sqrt{5}\]          ...(i)

\[\mathrm{d}(\mathrm{Q},\mathrm{R})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\]

= \[\sqrt{\left[-5-(-4)\right]^{2}+\left[0-(-2)\right]^{2}}\]

= \[\sqrt{\left(-5+4\right)^{2}+\left(0+2\right)^{2}}\]

= \[\sqrt{(-1)^{2}+2^{2}}\]

= \[\sqrt{1 + 4}\]

∴ d(Q, R) = \[\sqrt{5}\]           ...(ii)

\[\mathrm{d}(\mathrm{P},\mathrm{R})= \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\]

= \[\sqrt{\left[-5-(-2)\right]^{2}+\left[0-(-6)\right]^{2}}\]

= \[\sqrt{\left(-5+2\right)^{2}+\left(0+6\right)^{2}}\]

= \[\sqrt{\left(-3\right)^{2}+6^{2}}\]

= \[\sqrt{9 + 36}\]

= \[\sqrt{45}\]

= \[3\sqrt{5}\]

∴ d (P, R) = \[3\sqrt{5}\]           ...(iii)

On adding (i) and (ii),

∴ d(P, Q) + d(Q, R) = \[2\sqrt{5}\] + \[\sqrt{5}\] 

= \[3\sqrt{5}\]

∴ d(P, Q) + d(Q, R) = d (P, R)          …[From (iii)]

∴ Points P, Q, and R are collinear points.

We cannot construct a triangle through 3 collinear points.

∴ The segments joining the points P, Q and R will not form a triangle.