Question

Determine whether the points are collinear.

P(–2, 3), Q(1, 2), R(4, 1)

Answer 1

By distance formula,

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

= \[\sqrt{\left[1- (-2)\right]^{2}+\left(2 - 3\right)^{2}}\]

= \[\sqrt{(1+ 2)^{2}+(2 - 3)^2}\]

= \[\sqrt{(3)^{2}+(- 1)^2}\]

= \[\sqrt{9 + 1}\]

∴ \[\mathrm{d}(\mathrm{P},\mathrm{Q}) = \sqrt{10}\]          ...(i)

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

= \[\sqrt{(4 - 1)^{2} + (1 - 2)^{2}}\]

= \[\sqrt{3^{2} + (-1)^2}\]

= \[\sqrt{9 + 1}\]

∴ \[\mathrm{d}(\mathrm{Q},\mathrm{R}) = \sqrt{10}\]          ...(ii)

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

= \[\sqrt{[4 - (-2)]^{2} + (1 - 3)^{2}}\]

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

= \[\sqrt{36 + 4}\]

= \[\sqrt{40}\]

= \[2\sqrt{10}\]

∴ \[\mathrm{d}(\mathrm{P},\mathrm{R}) = 2\sqrt{10}\]          ...(iii)

On adding (i) and (ii),

\[\mathrm{d}(\mathrm{P},\mathrm{Q}) + \mathrm{d}(\mathrm{Q},\mathrm{R}) = \sqrt{10} + \sqrt{10} = 2\sqrt{10}\]

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

∴ Points P, Q and R are collinear.

Answer 2

Proof:

Let P(−2, 3) ≡ (x₁, y₁), Q(1, 2) ≡ (x₂, y₂) and R(4, 1) ≡ (x₃, y₃)

Slope of line PQ = `(y_2 - y_1)/(x_2 - x_1)`

= `(2 - 3)/(1 - (-2)) = (-1)/(1 + 2) = -1/3`        ...(1)

Slope of line QR = `(y_3 - y_2)/(x_3 - x_2)`

= `(1 - 2)/(4 - 1) = -1/3`        ...(2)

From (1) and (2),

the slope of PQ = the slope of line QR and point Q lies on both the lines.

∴ points P, Q and R are collinear.