(a)OR(b)

Vertices of a right triangle ABC with $\displaystyle \angle \mathrm{B}=90^{\circ}$ are $\displaystyle \mathrm{A}(3,4), \mathrm{B}(1,1)$ and C$\displaystyle (-8, 7)$. Find the value of $\displaystyle \tan \mathrm{A}$.

$\displaystyle \mathrm{BC}=\sqrt{(-8-1)^{2}+(7-1)^{2}}=\sqrt{117}=3 \sqrt{13}$ $\displaystyle \mathrm{AB}=\sqrt{(3-1)^{2}+(4-1)^{2}}=\sqrt{13}$ $\displaystyle \tan \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 \sqrt{13}}{\sqrt{13}}=3$ Using distance formula, prove that the points A$\displaystyle (2, 3)$, B$\displaystyle (-7, 0)$ and $\displaystyle \mathrm{C}(-1,2)$ are collinear. \[\begin{array}{l} A B=\sqrt{(-7-2)^{2}+(0-3)^{2}}=\sqrt{90}=3 \sqrt{10} \\ B C=\sqrt{(-1+7)^{2}+(2-0)^{2}}=\sqrt{40}=2 \sqrt{10} \\ A C=\sqrt{(-1-2)^{2}+(2-3)^{2}}=\sqrt{10}=\sqrt{10} \\ A C+B C=A B \\ \therefore A, B, C \text { are collinear. } \end{array} \]