Question

5) Show that the OPQRS formed by P(2.1).Q(-1,3). R(-5-3) and S(-2,-5) is a
rectangle​

Answer 1

Answer:

Given:

• Points = P(2.1), Q(-1,3), R(-5-3), S(-2,-5)

Find:

• Prove that it is a rectangle

Solution:

By using distance formula,

$\: \: \: \: \: \: { \boxed{ \sf{Distance = \sqrt{ {( x_{2} - x_{1} ) }^{2} + {( y_{2} - y_{1}) }^{2} } }}}$

${ \sf{PQ = \sqrt{ {( - 1 - 2)}^{2} + {(3 - 1)}^{2} } }} \\ \\ { \sf{PQ = \sqrt{ {( - 3)}^{2} + {(2)}^{2} } }} \\ \\ \sf{PQ = \sqrt{9 + 4} } \\ \\ { \sf{PQ = \sqrt{12} }} \\ \\ { \sf{ PQ = 2 \sqrt{3} }}$

${ \sf{QR = \sqrt{ {( - 5 - ( - 1))}^{2} + {( - 3 - 3)}^{2} } }} \\ \\ { \sf{QR = \sqrt{ {( - 4)}^{2} + {( - 6)}^{2} } }} \\ \\ { \sf{QR = \sqrt{16 + 36} }} \\ \\ { \sf{QR = \sqrt{52} }} \\ \\ { \sf{QR = 2 \sqrt{13} }}$

${ \sf{RS = \sqrt{ {( - 2 - ( - 5))}^{2} + { {( - 5 - ( - 3)}^{2} } } }} \\ \\ { \sf{RS = \sqrt{ {(3)}^{2} + {(2)}^{2} } }} \\ \\ { \sf{RS = \sqrt{9 + 4} }} \\ \\ { \sf{RS = \sqrt{12} }} \\ \\ { \sf{RS = 2\sqrt{3} }}$

${ \sf{PS = \sqrt{ {( - 2 - 2)}^{2} + {( - 5 - 1)}^{2} } }} \\ \\ { \sf{PS = \sqrt{ {( - 4)}^{2} + {( - 6)}^{2} } }} \\ \\ { \sf{PS = \sqrt{16 + 36} }} \\ \\ { \sf{PS = \sqrt{52} }} \\ \\ { \sf{PS = 2 \sqrt{13} }}$

Here opposite sides are equal

PQ = RS and QR = PS

${ \sf{PR = \sqrt{ {( - 5 - 2)}^{2} + {( - 3 - 1)}^{2} } }} \\ \\ { \sf{PR = \sqrt{ {( - 7)}^{2} + {( - 4)}^{2} } }} \\ \\ { \sf{PR = \sqrt{49 + 16} }} \\ \\ { \sf{PR = \sqrt{65} }}$

${ \sf{QS = \sqrt{ {( - 5 - 3)}^{2} + {( - 2 + 1)}^{2} } }} \\ \\ { \sf{QS = \sqrt{ {( - 8)}^{2} + {( - 1)}^{2} } }} \\ \\ { \sf{QS = \sqrt{64 + 1} }} \\ \\ { \sf{QS = \sqrt{65} }}$

Here diagonals are equal

PR = QR

• According to rectangle properties, opposite sides are equal and diagonals are equal

Therefore,

• Given points forms a rectangle