Question

Show that points P(2,-2). (7,3), R(11.-1) and S (6.-6) are vertices of a
parallelogram​

Answer 1

The given points are P(2, -2), Q(7, 3), R(11, -1) and S (6, -6).

PQ=

(3−(−2))

2

+(7−2)

2

=

25+25

=5

2

QR=

(7−11)

2

+(3−(−1))

2

=

16+16

=4

2

RS=

(11−6)

2

+(−1−(−6))

2

=

25+25

=5

2

PS=

(2−6)

2

+(−2−(−6))

2

=

16+16

=4

2

So, PQ = RS and QR = PS

Thus, opposite sides are equal.

Hence, the given points form a parallelogram