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.

Answer 1

Dear Student,

ABCD is parallelogram. So, AB must be parallel to CD.

So, slope of AB = slope of CD

$$\begin{gathered} \Rightarrow \frac{3-(-2)}{2-1}=\frac{-3-2}{-4-k} \\ \Rightarrow -4-k=-1 \\ \Rightarrow k=-3 \end{gathered}$$

Slope of AB = 5

Slope of AD= $\frac{-3-(-2)}{-4-1}=\frac{1}{5}$

$$\begin{gathered} \tan (\angle DAB)=\frac{5-{\displaystyle \frac{1}{5}}}{1+5.{\displaystyle \frac{1}{5}}}(U\sin gtheformulaforanglebetweenlinesofslope{m}_{1}and{m}_2,\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}) \\ \tan (\angle DAB)=\frac{24}{10}=\frac{12}{5} \\ \Rightarrow \sin (\angle DAB)=\frac{12}{\sqrt{{12}^2+5^2}}=\frac{12}{13} \\ \sin (\angle DAB)=\frac{AltitudetobaseAB}{SideAD}=\frac{h}{\sqrt{5^2+1^2}}=\frac{h}{\sqrt{26}} \\ \Rightarrow \frac{12}{13}=\frac{h}{\sqrt{26}} \\ \Rightarrow h=12\sqrt{\frac{2}{13}} \end{gathered}$$
Regards,