Question

Show that points A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4) are the vertices of a parallelogram ABCD.

Answer 1

We know that, slope of line = `(y_2 - y_1)/(x_2 - x_1)`

Slope of side AB = `(2 - (-7))/(-1 - (-4))`

= `(2 + 7)/(-1 + 4)`

= `9/3`

= 3   ...(i)

Slope of side BC = `(5 - 2)/(8 - (-1))`

= `3/(8 + 1)`

= `3/9`

= `1/3`   ...(ii)

Slope of side CD = `(-4 - 5)/(5 - 8)`

= `(-9)/(-3)`

= 3   ...(iii)

Slope of side AD = `(-4 - (-7))/(5 - (-4))`

= `(-4 + 7)/(5 + 4)`

=`3/9`

= `1/3`   ...(iv)

∴ Slope of side AB = Slope of side CD   ...[From (i) and (iii)]

∴ side AB || side CD

∴ Slope of side BC = Slope of side AD   ...[From (ii) and (iv)]

∴ side BC || side AD

Both the pairs of opposite sides of ABCD are parallel.

∴ ▢ABCD is a parallelogram.

∴ Points A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4) are the vertices of a parallelogram.