Question

Show that the points A(1,-2),B(3,6) C(5,10) and D(3,2) are the vertices of a parallelogram. the first to answer will get brainliest so hurry up

Answer 1

Answer:

Step-by-step explanation:

Slopes of parallel lines are equal.  

$m_{AB}$ = (6 + 2) / (3 - 1) = 4

$m_{CD}$ = (10 - 2) / (5 - 3) = 4

AB║CD .

AB = $\sqrt{(3-1)^2+(6+2)^2}$ = √68 = 2√17

CD = $\sqrt{(5-3)^2+(10-2)^2}$ = √68 = 2√17

AB = CD

$m_{AD}$ = (2 + 2) / (3 - 1) = 2

$m_{BC}$ = (10 - 6) / (5 - 3) =2

AD║BC

AD = $\sqrt{(1-3)^2+(-2-2)^2}$ = 2√5

BC = $\sqrt{(5-3)^2+(10-6)^2}$ = 2√5

AD = BC

Thus, ABCD is parallelogram.