Question

(3) Show that points P (2,-2), Q (7.3), R(11.-1) and S (6, -6) and vertices of a parallelograrn.​

Answer 1

Answer:

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Answer 2

Solution :-

Here, Quadrilateral PQRS is given

Coordinatos of given points = P( 2,-2 ) , Q( 7, 3) , R( 11 , -1 ) and S ( 6 , -6 )

By using distance formula ,

√( x2 - x1 )^2 + ( y2 - y1 )^2

Therefore,

PQ = √( 7 - 2 )^2 + ( 3 - ( -2 ))^2

PQ = √ ( 5 )^2 + ( 3 + 2 )^2

PQ = √ (5 )^2 + ( 5)^2

PQ = √ 25 + 25

PQ = √ 50

PQ = 5√2

QR = √( 11 - 7 )^2 + ( -1 - 3 ))^2

QR = √ ( 4 )^2 + ( -4 )^2

QR = √ 16 + 16

QR = √ 32

QR = 4√2

RS = √( 6 - 11 ))^2 + ( (-6) - ( -1 ))^2

RS = √ ( 6 - 11 )^2 + ( ( -6 ) + 1 )^2

RS = √ ( -5 )^2 + ( -5 )^2

RS = √25 + 25

RS = √50

RS = 5√2

SP = √( 2 - 6 )^2 + ( -2 - ( -6 )^2

SP = √ ( 4 )^2 + ( 4 )^2

SP = √16 + 16

SP = √32

SP = 4√2

Here, In quadrailateral PQRS

Opposite sides are equal that is

PQ = RS and QR = SP .

Hence, The given point forms parallelogram $.$