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 the triangle DEF.show that the quadrilateral ABCD is a Parallelogram

Answer 1

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.

Answer 2

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

and points A(-3,6),B(0,7)and C(1,9) are the mid points of the sides DE, EF and FD.

So,from mid point formula

$\frac{x_{1}+ x_{2}}{2} = - 3 \\ \\ x_{1} + x_{2} = - 6 \: \: \: eq1 \\ \\ \frac{y_{1} + y_{2}}{2} = 6 \\ \\ y_{1} + y_{2} = 12 \: \: \: eq2$
by the same way

$x_{2} + x_{3} = 0 \: \: \: eq3\\ \\ y_{2} + y_{3} = 14...eq4 \\ \\ x_{1} + x_{3} = 2 \: \: \: eq5 \\ \\ y_{1} + y_{3} = 18 \: \: \: eq6$
from eq3

$x_{3} = - x_{2}$
put this value in eq 5,and solve eq1 and eq3

$x_{1} + x_{2} = - 6 \\ x_{1} - x_{2} = 2 \\ \\ 2x_{1} = - 4 \\ \\ x_{1} = - 2$
from eq4

$y_{3} = 14 - y_{2}$
put this value in eq6 and solve eq2 and eq6 for y1

$y_{1}+ y_{2} = 12 \\ \\ y_{1} - y_{2} = 4 \\ \\ 2y_{1} = 16 \\ \\ y_{1} = 8$
So,points D(-2,8)

If ABCD is a parallelogram then AB || DC, and opposite sides are equal

So,
$AB = \sqrt{( { - 3)}^{2} + ( {6 - 7)}^{2} } \\ \\ AB= \sqrt{10} \\ \\ DC = \sqrt{ {( - 2 - 1)}^{2} + {(9 -8 )}^{2} } \\ \\ = \sqrt{10}$

AB= DC

by the same way

$AD = \sqrt{ {( - 2 + 3)}^{2} + ( {8 - 6)}^{2} } \\ \\ = \sqrt{5} \\ \\ BC = \sqrt{5}$
AD=BC

hence ABCD is a parallelogram.

Hope it helps you.