Question

Using distance formula show that the points L(4, 1), M(1,-3) and N(-2, -5) are collinear.

Answer 1

Given: L(4, –1), M(1, –3) and N(–2, –5).
From these points, x₁ = 4, y₁ = –1, x₂ = 1, y₂ = –3, x₃ = –2, y₃ = –5.
By distance formula,

$$\begin{gathered} \mathrm{LM}=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2} \\ =\sqrt{(1-4{)}^2+[-3-(-1){]}^2} \\ =\sqrt{(-3{)}^2+(-2{)}^2} \\ =\sqrt{9+4} \\ =\sqrt{13} \end{gathered}$$

$$\begin{gathered} \mathrm{MN}=\sqrt{(x_3-x_2{)}^2+(y_3-y_2{)}^2} \\ =\sqrt{(-2-1{)}^2+[-5-(-3){]}^2} \\ =\sqrt{(-3{)}^2+(-2{)}^2} \\ =\sqrt{9+4} \\ =\sqrt{13} \end{gathered}$$

$$\begin{gathered} \mathrm{LN}=\sqrt{(x_3-x_1{)}^2+(y_3-y_1{)}^2} \\ =\sqrt{(-2-4{)}^2+[-5-(-1){]}^2} \\ =\sqrt{(-6{)}^2+(-4{)}^2} \\ =\sqrt{36+16} \\ =\sqrt{52} \\ =2\sqrt{13} \end{gathered}$$

$$\begin{gathered} \mathrm{Here},\mathrm{LM}+\mathrm{MN}=\sqrt{13}+\sqrt{13} \\ =2\sqrt{13} \\ \mathrm{And}\mathrm{LN}=2\sqrt{13} \\ \Rightarrow \mathrm{LM}+\mathrm{MN}=\mathrm{LN} \\ \therefore \mathrm{Points}\mathrm{L},\mathrm{M}\mathrm{and}\text{N}\mathrm{are}\mathrm{collinear}. \end{gathered}$$