Question

By using the formula for area of a triangle in each of the following find the value of 'k' for which the points are collinear.
(i) (5, 5), (1, k), (-5, -10)

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{A}\equiv \left(5,5\right)\equiv \left(x_1,y_1\right) \\ \therefore x_1=5,y_1=5 \\ \mathrm{B}\equiv \left(1,k\right)\equiv \left(x_2,y_2\right) \\ \therefore x_2=1,y_2=k \\ \mathrm{and}\mathrm{C}\equiv \left(-5,-10\right)\equiv \left(x_3,y_3\right) \\ \therefore x_3=-5,y_3=-10 \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ \mathrm{When}\mathrm{three}\mathrm{points}\mathrm{are}\mathrm{collinear},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{these}\mathrm{three}\mathrm{points}\mathrm{is}0. \\ \mathrm{Given}:\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{are}\mathrm{collinear}. \\ \therefore \mathrm{Area}\mathrm{of}△\mathrm{ABC}=0 \\ \Rightarrow \frac{1}{2}\left[5\left\{k-\left(-10\right)\right\}+1\left(-10-5\right)-5\left(5-k\right)\right]=0 \\ \Rightarrow \frac{1}{2}\left[5\left(k+10\right)+1\left(-15\right)-5\left(5-k\right)\right]=0 \\ \Rightarrow \frac{1}{2}\left(5k+50-15-25+5k\right)=0 \\ \Rightarrow \frac{1}{2}\left(10k+10\right)=0 \\ \Rightarrow 10k+10=0 \\ \Rightarrow 10k=-10 \\ \therefore k=-1 \end{gathered}$$