Question

A quadrilateral has vertices at A(– 4, – 2), B(5, – 1), C(6, 5) and D(– 7, 6). Show that the mid-points of its sides form a parallelogram.

Answer 1

Let A(– 4, – 2), B(5, – 1), C(6, 5) and D(– 7, 6) are the vertices of a quadrilateral.

Mid point of a line = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

Mid point of AB (E) = `((-4 + 5)/2, (-2 - 1)/2) = (1/2, (-3)/2)`

Mid point of BC (F) = `((5 + 6)/2, (-1 + 5)/2) = (11/2, 4/2)` = `(11/2, 2)`

Mid point of CD (G) = `((6 - 7)/2, (5 + 6)/2) = ((-1)/2, 11/2)`

Mid point of AD (H) = `((-4 - 7)/2, (-2 + 6)/2) = ((-11)/2, 4/2) = ((-11)/2, 2)`

The midpoint of ABCD are E `(1/2, -3/2)`, F `(11/2, 2)`, G `(-1/2, 11/2)` and H `(-11/2, 2)`

Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`

Slope of EF = `((-3/2 - 2)/(1/2 - 11/2)) = (((-3 - 4)/2)/((1 - 11)/2)) = (((-7)/2)/((-10)/2))`

= `(-7)/2 xx (-1)/5 = 7/10`

Slope of FG = `(11/2 - 2)/(-1/2 - 11/2) = ((11 - 4)/2)/((-1 - 11)/2) = (7/2)/(-6)`

= `7/2 xx (-1)/6`

= `(-7)/12`

Slope of GH = `(2 - 11/2)/((-11)/2 + 1/2)= ((4 - 11)/2)/((-11 + 1)/2)`

= `((-7)/2)/((-10)/2)`

= `(-7)/2 xx (-2)/10`

= `7/10`

Slope of EH = `(2 + 3/2)/(-11/2 - 1/2) = ((4 + 3)/2)/((-11 - 1)/2)`

= `(7/2)/(-6)`

= `7/2 xx (-1)/6`

= `-7/12`

Slope of EF = Slope of GH =  `7/10`

∴ EF || GH   ...(1)

Slope of FG = Slope of EH = `-7/12`

∴ FG || EH  ...(2)

From (1) and (2) we get EFGH is a parallelogram.

The midpoint of the sides of the Quadrilateral ABCD is a Parallelogram.