The vertices of a quadrilateral are at A(3,8),B(-5,-2),C(2,-7) and D(9,7).?

Coordinates of the mid-point of AD = ((3+9)/2, (8+7)/2) = (6, 7.5)
Coordinates of the mid-point of DC= ((2+9)/2, (-7+7),2) = (5.5, 0)
Slope of the line joining the mid-points of AD and DC = (7.5 - 0)/(5.5 - 6) = -15

Coordinates of the mid-point of AB = ((3-5)/2, (8-2)/2) = (-1, 3)
Coordinates of the mid-point of BC= ((-5+2)/2, (-2-7)/2) = (-1.5, -4.5)
Slope of the line joining the mid-points of AD and DC = (-1 + 1.5)/(3 + 4.5) = 1/15

(Slope of the line joining the mid-points of AD and DC) × (Slope of the line joining the mid-points of AD and DC)
= (-15) × (1/15)
= -1

Thus, the line joining the mid-points of AD and DC // the line joining the midpoints of AB and BC.

ABCD is a quadrilateral. The diagonal AC divides it into two triangles ABC and ADC both with base AC.
We know that a line joining the mid point of two two sides is parallel to the third side.
Hence the line joining the mid point of AB and BC is parallel to AC
Also the line joining the mid point of AD and DC is parallel to AC
i.e line joining the mid point of AB and BC is parallel to the line joining the mid point of AD and DC both being parallel to the same line AC

A (3 ; 8) B (- 5 ; - 2) C (2 ; - 7) D (9 ; 7)


Let's calculate the coordinates of the midpoint of [AD] → point M

xM = (xA + xD)/2 = (3 + 9)/2 = 6

yM = (yA + yD)/2 = (8 + 7)/2 = 15/2

→ M (6 ; 15/2)


Let's calculate the coordinates of the midpoint of [DC] → point N

xN = (xD + xC)/2 = (9 + 2)/2 = 11/2

yN = (yD + yC)/2 = (7 - 7)/2 = 0

→ N (11/2 ; 0)


How to get the equation of the line (MN) that passes through M (6 ; 15/2) N (11/2 ; 0) ?

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept

To calculate m

m = (yN - yM) / (xN - xM)

m = [0 - (15/2)] / [(11/2) - 6]

m = - (15/2) / [(11/2) - (12/2)]

m = - (15/2) / - (1/2)

m = (15/2) / (1/2)

m = 15 ← the slope of the line (MN) is 15



Let's calculate the coordinates of the midpoint of [AB] → point R

xR = (xA + xB)/2 = (3 - 5)/2 = - 1

yR = (yA + yB)/2 = (8 - 2)/2 = 3

→ R (- 1 ; 3)


Let's calculate the coordinates of the midpoint of [BC] → point S

xS = (xB + xC)/2 = (- 5 + 2)/2 = - 3/2

yS = (yB + yC)/2 = (- 2 - 7)/2 = - 9/2

→ S (- 3/2 ; - 9/2)


How to get the equation of the line (RS) that passes through R (- 1 ; 3) S (- 3/2 ; - 9/2) ?

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept

To calculate m

m = (yS - yR) / (xS - xR)

m = [- (9/2) - 3] / [- (3/2) + 1]

m = [- (9/2) - (6/2)] / [- (3/2) + (2/2)]

m = - (15/2) / - (1/2)

m = (15/2) / (1/2)

m = 15 ← the slope of the line (RS) is 15




Two lines are parallel if they have the same slope.

As the slope of the line ((MN) is similar to the slope of the line (RS), you can conclude that these two lines are parallel.