Question

The points A(-3,6), B(0,7) and C(1,9) are the mid-points of the sides DE, EF and FD
of a triangle DEF. Show that the quadrilateral ABCD is a parallellogram.​

Answer 1

Step-by-step explanation:

A (-3,6) is the midpoint of DE , B(0,7) is the midpoint of EF and C(1,9) is the midpoint of FD.

Let D = (x1, y1) , E = (x2 , y2) and F = (x3 , y3)

so, A is the midpoint of DE

so, (-3, 6) = {(x1 + x2)/2, (y1 + y2)/2}

x1 + x2 = -6

y1 + y2 = 12

Similarly, B is the midpoint of EF

so, x2 + x3 = 0

y2 + y3 = 14

and C is the midpoint of FD

so, x3 + x1 = 2

y1 + y3 = 18

now, solving equations,

x1 = (x1 + x2 + x3) - (x2 + x3)

=-2 + 0 = -2

x2 = (x1 + x2 + x3) - (x1 + x3)

= -2 -2 = -4

x3 = (x1 + x2 + x3) - (x1 + x2)

= -2 - 6 = -8

Similarly, y1 = (y1 + y2 + y3)- (y2 + y3)

= 22 - 14 = 8

y2 = (y1 + y2 + y3) - (y1 + y3)

= 22 - 18 = 4

y3 = (y1 + y2 + y3) - (y1 + y2)

= 22 - 12 = 10

hence, D(-2,8) , E(-4, 4) and F(-8, 10)

now, we have to show that ABCD is a parallelogram.

we know, diagonals of parallelogram intersect at midpoint.

e.g., midpoint of diagonal AC = midpoint of diagonal BD.

midpoint of AC = {(-3+1)/2, (6+9)/2} = (-1,15/2)

midpoint of BD = {(0-2)/2, (7+8)/2} = (-1,15/2)

hence, it is clear that, ABCD is a parallelogram.