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Question Number 110797 by ZiYangLee last updated on 30/Aug/20

$$\mathrm{If}\:\overset{\rightarrow} {{p}}=\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}\:\mathrm{and}\:\overset{\rightarrow} {{q}}=\begin{pmatrix}{{c}}\\{{d}}\end{pmatrix}, \\$$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\: \\$$$$\overset{\rightarrow} {{p}},\overset{\rightarrow} {{q}}\:\mathrm{and}\:\overset{\rightarrow} {{p}}−\overset{\rightarrow\:} {{q}}\mathrm{is}\:\frac{\left({ad}−{bc}\right)}{\mathrm{2}}. \\$$$$\\$$$$\mathrm{Hints}:\:\mathrm{Use}\:\mathrm{cosine}\:\mathrm{rule}\:\mathrm{and}\:\mathrm{sine}\:\mathrm{rule} \\$$

Commented by kaivan.ahmadi last updated on 30/Aug/20

$${let}\:\theta\:{be}\:{the}\:{angel}\:{between}\:\overset{\rightarrow} {{p}}\:{and}\:\overset{\rightarrow} {{q}}\: \\$$$$\\$$$$\overset{\rightarrow} {{p}}−\overset{\rightarrow} {{q}}=\begin{pmatrix}{{a}−{c}}\\{{b}−{d}}\end{pmatrix} \\$$$$\mid\overset{\rightarrow} {{p}}−\overset{\rightarrow} {{q}}\mid^{\mathrm{2}} =\mid\overset{\rightarrow} {{p}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {{q}}\mid^{\mathrm{2}} −\mathrm{2}\mid\overset{\rightarrow} {{p}}\mid.\mid\overset{\rightarrow} {{q}}\mid.{cos}\theta\Rightarrow \\$$$$\left({a}−{c}\right)^{\mathrm{2}} +\left({b}−{d}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }.\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }.{cos}\theta \\$$$$\Rightarrow−{ac}−{bd}=\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}.{cos}\theta\Rightarrow \\$$$${cos}\theta=\frac{−{ac}−{bd}}{\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}}\Rightarrow{sin}\theta=\sqrt{\mathrm{1}−\frac{\left({ac}−{bd}\right)^{\mathrm{2}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}}= \\$$$$\sqrt{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)−\left({ac}−{bd}\right)^{\mathrm{2}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}}=\sqrt{\frac{{a}^{\mathrm{2}} {d}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}{abcd}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}}= \\$$$$\sqrt{\frac{\left({ad}−{bc}\right)^{\mathrm{2}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}}=\frac{\left({ad}−{bc}\right)}{\mid\overset{\rightarrow} {{p}}\mid.\mid\overset{\rightarrow} {{q}}\mid}\:\:;\left[{ad}−{bc}\geqslant\mathrm{0}\right] \\$$$${now}\:{we}\:{have} \\$$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\mid\overset{\rightarrow} {{p}}\mid.\mid\overset{\rightarrow} {{q}}\mid.{sin}\theta=\frac{\mathrm{1}}{\mathrm{2}}\mid\overset{\rightarrow} {{p}}\mid.\mid\overset{\rightarrow} {{q}}\mid.\frac{\left({ad}−{bc}\right)}{\mid\overset{\rightarrow} {{p}}\mid.\mid\overset{\rightarrow} {{q}}\mid}=\frac{\mathrm{1}}{\mathrm{2}}\left({ad}−{bc}\right) \\$$

Commented by ZiYangLee last updated on 31/Aug/20

$$\mathrm{NICE} \\$$