Question

A quadrilateral has vertices (4, 1), (1, 7), (-6, 0) and (-1, -9). Show that the mid-points of the sides of this quadrilateral form a parallelogram.

Answer 1

$\text{Let ABCD be the given quadrilateral}\text{E is mid point of AB}\text{F is mid point of BC}\text{G is mid point of CD}\text{H is mid point of AD}\text{Using mid point formula}(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\text{Coordinates of}E=(\frac{4+1}{2},\frac{1+7}{2})=(\frac{5}{4},4)\text{Coordinates of}F=(\frac{1-6}{2},\frac{7+0}{2})=(\frac{-5}{2},\frac{7}{2})\text{Coordinates of G}=(\frac{-6-1}{2},\frac{0-9}{2})=(\frac{-7}{2},\frac{9}{2})\text{Coordinates of H}=(\frac{-1+4}{2},\frac{-9+1}{2})=(\frac{3}{2},-4)\text{In order to prove PQRS is a}||gm,\text{it is sufficient to show that:}EF||GHandEF=GH\text{Now,}\text{Slope of}EF=\frac{\frac{7}{2}-4}{\frac{-5}{2}-\frac{5}{2}}=\frac{1}{10}\text{Slope of}GH=\frac{-4+\frac{9}{2}}{\frac{3}{2}+\frac{7}{2}}=\frac{1}{10}\text{Clearly, slope of EF = slope of GH}\therefore EF||GHEF=\sqrt{{(\frac{-5}{2}-\frac{5}{2})}^2+{(\frac{7}{2}-4)}^2}=\frac{\sqrt{101}}{2}GH=\sqrt{{(\frac{3}{2}+\frac{7}{2})}^2+{(-4+\frac{9}{2})}^2}=\frac{\sqrt{101}}{2}\therefore EF=GHThus,EF||GHandEF=GH$