Question

ABC is a triangle formed by the vertices (0, 5), (6, 13), and (6, 3). If AD⊥BC, then what is the length of AD?

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{area}\mathrm{of}∆\mathrm{whose}\mathrm{vertices}\mathrm{are}\left(x_1,y_1\right);\left(x_2,y_2\right);\left(x_3,y_3\right)\mathrm{is} \\ ∆=\frac{1}{2}\left|\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\right| \\ \mathrm{So},\mathrm{area}\mathrm{of}∆\mathrm{whose}\mathrm{vertices}\mathrm{are}\mathrm{A}\left(0,5\right);\mathrm{B}\left(6,13\right);\mathrm{C}\left(6,3\right)\mathrm{is} \\ \mathrm{area}\mathrm{of}∆\mathrm{ABC}=\frac{1}{2}\left|\left[0\left(13-3\right)+6\left(3-5\right)+6\left(5-13\right)\right]\right| \\ \mathrm{area}\mathrm{of}∆\mathrm{ABC}=\frac{1}{2}\left|\left(0+\left(-12\right)+\left(-48\right)\right)\right| \\ \mathrm{area}\mathrm{of}∆\mathrm{ABC}=\frac{1}{2}\left|-60\right|=\frac{1}{2}\times 60=30\mathrm{square}\mathrm{units} \\ \mathrm{Now},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{distance}\mathrm{between}\mathrm{points}\left(x_1,y_1\right)\mathrm{and}\left(x_2,y_2\right)\mathrm{is} \\ \mathrm{distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ \mathrm{So},\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{joining}\mathrm{the}\mathrm{points}\left(6,3\right)\mathrm{and}\left(6,13\right)\mathrm{is} \\ \mathrm{Now},\mathrm{BC}=\sqrt{{\left(6-6\right)}^2+{\left(13-3\right)}^2}=\sqrt{{10}^2}=10\mathrm{units} \\ \mathrm{N}\mathrm{o}\mathrm{w},\mathrm{A}\mathrm{D}\perp \mathrm{B}\mathrm{C}. \\ \mathrm{Now},\mathrm{area}\mathrm{of}∆\mathrm{ABC}=30\mathrm{square}\mathrm{units} \\ \Rightarrow \frac{1}{2}\times \mathrm{AD}\times \mathrm{BC}=30 \\ \Rightarrow \frac{1}{2}\times \mathrm{AD}\times 10=30 \\ \Rightarrow \mathrm{AD}=\frac{60}{10}=6\mathrm{units} \end{gathered}$$