Question

(-2,4),(4,8),(10,7),(11,-5)are the vertices of a quadrilateral.
show that the quadrilateral obtained on joining the midpoint of its sides is a parallelogram

Answer 1

Here is the required answer

Answer 2

Proved below.

Step-by-step explanation:

Given:

Let A = (−2,4), B = (4,8), C = (10,7) and D = (11,−5) are the vertices of a quadrilateral.

So using midpoint formula $(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2})$

Let midpoint of AB be E,

$E=(\frac{4-2}{2},\frac{8+4}{2})$

E=(1,6)

Let midpoint of BC be F,

$F=(\frac{10+4}{2},\frac{7+8}{2})$

$F=(7,\frac{15}{2} )$

Let midpoint of CD be G,

$G=(\frac{10+11}{2}, \frac{7-5}{2})$

$G=(\frac{21}{2},1)$

Let midpoint of AD be H,

$H=\frac{11-2}{2},\frac{-5+4}{2}$

$G=(\frac{9}{2},\frac{-1}{2} )$

Now  for proving parallelogram showing opposite side are at same slope will be sufficient

So,

slope of EF=$\frac{y_2-y_1}{x_2-x_1}$

                  =$\frac{\frac{15}{2}-6 }{7-1}$

                  =$\frac{\frac{15-12}{2}}{6}$

                  =$\frac{15-12}{2\times6}$

                  =$\frac{3}{12}$

Slope of EF=$\frac{1}{4}$

Slope of GH=$\frac{\frac{-1}{2}-1 }{\frac{9}{2}-\frac{21}{2}  }$

                   =$\frac{\frac{-1-2}{2} }{\frac{9-21}{2} }$

                   =$\frac{-3}{-12}$

Slope of GH= $\frac{1}{4}$            

So slope of these opposite side are equal.

Similarly, shows that slope of other two opposite side are equal and hence they are parallelogram.

Hence proved.