Question

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

Answer 1

The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4). 

Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

According to the distance formula,

`"AB" =sqrt([-1-(-4)]^2 +[2-(-7)]^2`

`∴ "AB" = sqrt(3^2+9^2)`

`∴ "AB" =sqrt(9+81) `

`∴ "AB" = sqrt90`                      ...(1)

`"BC" =sqrt([8-(-1)]^2+(5-2)^2)`

`∴ "BC" =sqrt(9^2+3^2)`

`∴ "BC" = sqrt(81+9)`

`∴ "BC" = sqrt90`                     ...(2)

`"CD" = sqrt((5-8)^2 +(-4-5)^2)`

`∴ "CD" =sqrt((-3)^2 +(-9)^2)`

 `∴ "CD" =sqrt(9+81)`

`∴ "CD" = sqrt90 `          ....... (3)

`"AD" = sqrt([5-(-4)]^2+ [-4-(-7)]^2)`

`∴ "AD" =sqrt(9^2+3^2)`

`∴ "AD" =sqrt(81+9)`

`∴ "AD" = sqrt90`                      ....... (4)

From (1), (2), (3), and (4)

AB = BC = CD = AD

Thus, all sides are equal.

Now,

AC = `sqrt([8 - (-4)]^2 + [5 - (-7)]^2)`

= `sqrt((12)^2 + (12)^2)`

= `sqrt(144 + 144)`

= `sqrt288`

= `sqrt(144 xx 2)`

= `12sqrt2`

BD = `sqrt([5 - (-1)]^2 + [(-4) - 2]^2)`

=`sqrt((6)^2 + (6)^2)`

= `sqrt(36 + 36)`

= `sqrt72`

= `sqrt(36 xx 2)`

= `6sqrt2`

∴ diagonals are not equal.

In a quadrilateral, if all the sides are equal, then it is a rhombus.

∴ `square` ABCD is a rhombus.

∴ Points A, B, C and D are the vertices of rhombus ABCD.