Question

Show that (9, 0), (9, 6), (-9, 6) and (-9, 0) are the vertices of a rectangle

Answer 1

Solution

Given

the points are (9,0) , (9,6) , (-9,6) and (-9,0)

Let us consider these points as A , B ,C and D respectively .

So that ,

• A(9,0)

• B(9,6)

• C(-9,6)

• D(-9,0)

Formula to be used

$Distance \: between \: two\: points \: ( x_{1},y_{1}) \: and \: ( x_{2},y_{2})is \\ = \sqrt{ (x_{2} - x_{1}) ^{2} + (y_{2} - y_{1})^{2} }$

So Applying the distance formula in the given Data we have

$AB = \sqrt{(9 - 9) {}^{2} + (6 - 0) {}^{2} } \\ \implies AB = \sqrt{36} \\ \implies AB = 6 \: units$

$BC = \sqrt{( - 9 - 9) ^{2} + (6 - 6) {}^{2} } \\ \implies BC = \sqrt{( - 18) ^{2} } \\ \implies BC = \sqrt{( - 1) ^{2} \times ( {18})^{2} } \\ \implies BC = 18 \: units$

$CD = \sqrt{( - 9 + 9) {}^{2} + (0 - 6) ^{2} } \\ \implies CD = \sqrt{( - 6) ^{2} } \\ \implies CD = \sqrt{ {( - 1)}^{2} \times (6) {}^{2} } \\ \implies CD = 6 \: units$

$AD = \sqrt{( - 9 - 9)^{2} + (0 - 0) ^{2} } \\ \implies AD = \sqrt{( - 18) {}^{2} } \\ \implies AD = \sqrt{( { - 1)}^{2} \times (18) {}^{2} } \\ \implies AD = 18 \: units$

Thus the opposite sides

AB = CD = 6 units

BC = AD = 18 units

Now for diagonals

$AC = \sqrt{( - 9 - 9) ^{2} + (6 - 0) ^{2} } \\ \implies AC = \sqrt{(18 )^{2} + (6 )^{2} } \\ \implies AC = \sqrt{324 + 36} \\ \implies AC = \sqrt{360} \\ \implies AC = \sqrt{36 \times 10} \\ \implies AC = 6 \sqrt{10} \: units$

$BD = \sqrt{(9 + 9) ^{2} + (0 - 6) ^{2} } \\ \implies BD = \sqrt{(18 {)}^{2} + ( {6})^{2} } \\ \implies BD = \sqrt{324 + 36} \\ \implies BD = \sqrt{360} \\ \implies BD = 6 \sqrt{10} \: units$

Thus the diagonals AC = BD = 6√10 units

Now from the properties of a rectangle we have

• The opposite sides of a rectangle are equal

• The diagonals of a rectangle are equal

Therefore , ABCD is a rectangle

$Shown$