Question

show that quadrilateral PQRS formed by vertices P(-2,5), Q(7,10) , R(12,11) , and S(3,_4) is not a parallelogram.​

Answer 1

Answer:

Step-by-step explanation:

Slopes of parallel lines are equal. Let's find slope of each side of quadrilateral PQRS:

Slope of PS is

$m_{PS}$ = (-4 - 5) / (3 - (-2)) = -9/5

Slope of PQ is

$m_{PQ}$ =  (10 - 5) / (7 - (-2)) = 5/9

5/9 is opposite reciprocal to (- 9/5)

====> PS ⊥ PQ

$m_{QR}$ = (11 - 10) / (12 - 7) = 1/5

$m_{RS}$ = (-4 - 11) / (3 - 12) = 5/3

Quadrilateral PQRS is not a parallelogram.