Question

If the points A (x,y), B($-$5, 7) and C($-$4, 5) are collinear then show that 2x + y + 3 =0

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{A}\equiv \left(x,y\right)\therefore x_1=x,y_1=y \\ \mathrm{B}\equiv \left(-5,7\right)\therefore x_2=-5,y_2=7 \\ \mathrm{and}\mathrm{C}\equiv \left(-4,5\right)\therefore x_3=-4,y_3=5 \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ \mathrm{When}\mathrm{three}\mathrm{points}\mathrm{are}\mathrm{collinear},\mathrm{then} \\ \mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{these}\mathrm{three}\mathrm{points}=0 \\ \mathrm{Given}:\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{are}\mathrm{collinear}. \\ \therefore \mathrm{Area}\mathrm{of}△\mathrm{ABC}=0 \\ \Rightarrow \frac{1}{2}\left[x\left(7-5\right)-5\left(5-y\right)-4\left(y-7\right)\right]=0 \\ \Rightarrow \left[x\left(2\right)-5\left(5-y\right)-4\left(y-7\right)\right]=0 \\ \Rightarrow 2x-25+5y-4y+28=0 \\ \therefore 2x+y+3=0 \end{gathered}$$