Question Number 35208 by Rio Mike last updated on 16/May/18
$${The}\:{points}\:{A},{B},{C}\:{have}\:{coordinates} \\ $$$$\left(\mathrm{3},\mathrm{1}\right),\left(\mathrm{1},\mathrm{5}\right),\left(\mathrm{5},\mathrm{7}\right) \\ $$$$\left.{a}\right)\:{find}\:{the}\:{line}\:{l}_{\mathrm{1}} \:{joining}\:{the}\:{points} \\ $$$${A}\:{and}\:{B} \\ $$$$\left.{b}\right){Find}\:{the}\:{line}\:{l}_{\mathrm{2}} \:{perpendicular} \\ $$$${to}\:{l}_{\mathrm{1}} \:{and}\:{passes}\:{through}\:{the}\:{point} \\ $$$${B} \\ $$$$\left.{c}\right)\:{show}\:{that}\:{the}\:{point}\:{C}\:{lie}\:{on}\:{the} \\ $$$${line}\:{l}_{\mathrm{2}\:\:} {hence}\:{or}\:{otherwise}\:{find}\: \\ $$$${the}\:{area}\:{of}\:{the}\:{triangle}\:{ABC}. \\ $$
Answered by ajfour last updated on 16/May/18
$${y}−\mathrm{1}=\left(\frac{\mathrm{1}−\mathrm{5}}{\mathrm{3}−\mathrm{1}}\right)\left({x}−\mathrm{3}\right) \\ $$$$\Rightarrow\:\:{y}=\mathrm{7}−\mathrm{2}{x} \\ $$$$ \\ $$$${y}−\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\mathrm{2}{y}={x}+\mathrm{9}\: \\ $$$$\mathrm{2}{y}=\mathrm{5}+\mathrm{9}\:=\mathrm{2}×\mathrm{7} \\ $$$$\: \\ $$$$\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\mid\begin{vmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{7}}&{\mathrm{1}}\end{vmatrix}\mid \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mid−\mathrm{6}+\mathrm{4}−\mathrm{18}\mid\:=\:\mathrm{10}\:{sq}.\:{units}\:. \\ $$