Two bags A (-7, -3)B(5, 10)C(15, 8)D(3, -5) are vertices of parallelogram
Answers
How can you prove that the points A (-7,-3) B (5,10) C (15,8) and D (3 ,-5) taken in the order on the vertices of a parallelogram and draw a graph?
To prove that the points A(-7, - 3); B(5, 10); C(15,8) and D(3, - 5) are the four vertices of a parallelogram, we need to prove that the pair of opposite sides (AB and DC) and (AD and BC) are parallel ie have the same slope and the two diagonals bisect each other.
Slope of straight line AB = [10- (-3)]/[5 - (-7)] = 13/12
Slope of straight line CD = [8 - (-5)]/[15 - 3] = 13/12
Therefore, the straight lines AB and CD are parallel to each other.
Slope of straight line AD = [ - 5 - (-3)]/[3 - (-7)] = - 2/10 = - 1/5
Slope of straight line BC = [8 - 10]/[15 - 5]= - 2/10 = - 1/5
Therefore the straight lines AD and BC are parallel to each other.
So opposite sides of the quadrilateral are parallel to each other.
Coordinates of the midpoint of diagonal AC = [{(-7 +15)/2}, {(-3 +8)/2}] = (+4, +5/2)
Coordinates of the midpoint of diagonal BD = [ {(3+5)/2}, {(-5 +10)/2}]= ( +4, +5/2)
This means that the diagonals of the quadrilateral divide indeed divide each other because midpoint of both diagonals is same.
Therefore ABCD is a parallelogram.