Question

# The points A ( 1 , -2 ) , B ( 2 , 3 ), C ( k , 2 )and D ( - 4 , - 3 ) are the vertices of a parallelogram. Find the value of k and the altitude of the parallelogram corresponding to the base AB.

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Solution

## Dear Student, ABCD is parallelogram. So, AB must be parallel to CD. So, slope of AB = slope of CD $⇒\frac{3-\left(-2\right)}{2-1}=\frac{-3-2}{-4-k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒-4-k=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒k=-3$ Slope of AB = 5 Slope of AD= $\frac{-3-\left(-2\right)}{-4-1}=\frac{1}{5}$ $\mathrm{tan}\left(\angle DAB\right)=\frac{5-\frac{1}{5}}{1+5.\frac{1}{5}}\left(U\mathrm{sin}gtheformulaforanglebetweenlinesofslope{m}_{1}and{m}_{2},\mathrm{tan}\theta =\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\angle DAB\right)=\frac{24}{10}=\frac{12}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\left(\angle DAB\right)=\frac{12}{\sqrt{{12}^{2}+{5}^{2}}}=\frac{12}{13}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\angle DAB\right)=\frac{AltitudetobaseAB}{SideAD}=\frac{h}{\sqrt{{5}^{2}+{1}^{2}}}=\frac{h}{\sqrt{26}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{12}{13}=\frac{h}{\sqrt{26}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒h=12\sqrt{\frac{2}{13}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ Regards,

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