Question

prove that the points (-7,-3), (5, 10), (5, 8) and (3, -5) taken in order are the corners of a parallelogram.

Answer 1

As we get
`$\mathrm{AB}=\mathrm{CD}=√(313)$` and
`$\mathrm{BC}=\mathrm{DA}=√(104)$` it is proven that a given quadrilateral is a parallelogram.

Let A, B, C and D is a quadrilateral.

It have a coordinates as follows:

Point A = (-7,-3)

Point B = (5,10)

Point C = (15,8)

Point D = (3,-5)

Here we have to use the distance formula

`$\mathrm{d}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}$`

By substituting the values here,

`& \mathrm{AB}^2=(5+7)^2+(10+3)^2=12^2+13^2=144+169=313`

`& \mathrm{BC}^2=(15-5)^2+(8-10)^2=10^2+(-2)^2=100+4=104 \\`

`& \mathrm{CD}^2=(3-15)^2+(-5,-8)^2=(-12)^2+(-13)^2=144+169=313 \\`

`& \mathrm{DA}^2=(3+7)^2+(-5+3)^2=10^2+(-2)^2=100+4=104`

So,
`$\mathrm{AB}=\mathrm{CD}=√(313)$` and
`$\mathrm{BC}=\mathrm{DA}=√(104)$`

i.e., The opposite sides are equal. Hence ABCD is a parallelogram.