Question

(−2, 4), (4, 8), (10, 7) and (11, –5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Answer 1

Let the given points be A(−2, 4), B(4, 8), C(10, 7) and D(11, −5).

Let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

Co-ordinates of P are ${\displaystyle (\frac{-2+4}{2},\frac{4+8}{2})=(1,6)}$

Co-ordinates of Q are ${\displaystyle (\frac{4+10}{2},\frac{8+7}{2})=(7,\frac{15}{2})}$

Co-ordinates of R are ${\displaystyle (\frac{10+11}{2},\frac{7-5}{2})=(\frac{21}{2},1)}$

Co-ordinates of S are ${\displaystyle (\frac{-2+11}{2},\frac{4-5}{2})=(\frac{9}{2},\frac{-1}{2})}$

Slope of PQ =${\displaystyle \frac{\frac{15}{2}-6}{7-1}=\frac{\frac{15-12}{2}}{6}=\frac{3}{12}=\frac{1}{4}}$

Slope of RS =${\displaystyle \frac{\frac{-1}{2}-1}{\frac{9}{2}-\frac{21}{2}}=\frac{\frac{-1-2}{2}}{\frac{9-21}{2}}=\frac{-3}{-12}=\frac{1}{4}}$

Since, slope of PQ = Slope of RS, PQ || RS.

Slope of QR =${\displaystyle \frac{1-\frac{15}{2}}{\frac{21}{2}-7}=\frac{\frac{2-15}{2}}{\frac{21-14}{2}}=-\frac{13}{7}}$

Slope of SP =${\displaystyle \frac{6+\frac{1}{2}}{1-\frac{9}{2}}=\frac{\frac{12+1}{2}}{\frac{2-9}{2}}=-\frac{13}{7}}$

Since, slope of QR = Slope of SP, QR || SP.

Hence, PQRS is a parallelogram.