# PROVING THE GIVEN VERTICES OF A PARALLELOGRAM USING SLOPE

Proving the Given Vertices of a Parallelogram Using Slope :

Here we are going to see, how to prove the given vertices of parallelogram using slope.

## Proving the Given Vertices of a Parallelogram Using Slope - Questions

Question 1 :

Show that the given points form a parallelogram :

A(2.5, 3.5) , B(10,-4), C(2.5,-2.5) and D(-5,5)

Solution :

In a parallelogram, opposite sides will be parallel, by proving that slope of opposite sides are equal, we may say that opposite sides are parallel.

Slope  (m)  =  (y2 - y1)/(x2 - x1)

A(2.5, 3.5) , B(10,-4),C(2.5,-2.5) and D(-5,5)

 Slope of ABm  =  (-4-3.5)/(10-2.5)  =  -7.5/7.5m  =  -1 Slope of CD :m  =  (5-(-2.5))/(-5-2.5)  =  (5+2.5)/(-7.5)m  =  7.5/(-7.5)m  =  -1 Slope of BCm  =  (-2.5+4)/(2.5-10)  =  1.5/(-7.5)=  -15/75m  =  -1/5 Slope of DAm  =  (5-3.5)/(-5-2.5)  =  1.5/(-7.5)=  -15/75m  =  -1/5

Slope of AB  =  Slope of CD

Slope of BC  =  Slope of DA

Hence the given points form a parallelogram.

Question 2 :

If the points A(2, 2), B(–2, –3), C(1, –3) and D(x, y) form a parallelogram then find the value of x and y.

Solution :

Since the given points form a parallelogram,

Slope of AB  =  Slope of CD

Slope of BC  =  Slope of DA

A(2, 2), B(–2, –3), C(1, –3) and D(x, y)

 Slope of ABm  =  (-3-2)/(-2-2)  =  -5/(-4)m  =  5/4 ---(1) Slope of CD :m  =  (y-(-3))/(x - 1)m  =  (y + 3)/(x - 1) ---(2) Slope of BCm  =  (-3+3)/(1+2)m  =  0/3m = 0----(3) Slope of DAm  =  (y - 2)/(x - 2) ----(4)

(1)  =  (2)

5/4  =  (y + 3)/(x - 1)

5(x - 1)  =  4(y + 3)

5x - 5  =  4y + 12

5x - 4y  =  12 + 5

5x - 4y  =  17   -----(5)

(3)  =  (4)

0  =  (y - 2)/(x - 2)

y - 2  =  0

y  =  2

By applying the value of y in (5), we get

5x - 4(2)  =  17

5x - 8  =  17

5x  =  17 + 8

5x  =  25

x  =  5

Hence the values of x and y are 5 and 2 respectively.

Question 3 :

Let A(3, -4), B(9, -4) , C(5, -7) and D(7, -7) . Show that ABCD is a trapezium.

Solution :

A trapezium will always contain two parallel sides and two non parallel sides.

Slope of AB :

m = (-4 +4)/(9-3)

m = 0/6  =  0

Slope of BC :

m = (-7 +4)/(5-9)

m = -3/(-4)  =  3/4

Slope of CD :

m = (-7 + 7)/(7 - 5)

m = 0/2  =  0

Slope of DA :

m = (-7 + 4)/(7 - 3)

m = -3/4

In the given trapezium, sides AB and CD are parallel.

Question 4 :

A quadrilateral has vertices A(-4,-2) , B(5,-1) , C(6,5) and D(-7,6) . Show that the mid-points of its sides form a parallelogram

Solution :

Midpoint of the side AB  =  P

=  (-4 + 5)/2, (-2 - 1)/2

=  P (1/2, -3/2)

Midpoint of the side BC  =  Q

=  (5 + 6)/2, (-1 + 5)/2

=  Q (11/2, 4/2)

=  Q (11/2, 2)

Midpoint of the side CD  =  R

=  (-7 + 6)/2, (6 + 5)/2

=  R (-1/2, 11/2)

Midpoint of the side DA  =  S

=  (-7 - 4)/2, (-2 + 6)/2

=  S (-11/2, 4/2)

=  S (-11/2, 2)

## Slopes of Opposite Sides :

Slope of PQ  =  (7/2) / 5  ==>  7/10

Slope of RS  =  (-7/2)/(-5)  =  7/10

Slope of PQ  =  (7/2) / (-5)  ==>  -7/10

Slope of RS  =  (7/2)/(-5)  =  -7/10

Question 5 :

PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1,1) and (2,-1) respectively, find the coordinates of P.

Solution :

After having gone through the stuff given above, we hope that the students would have understood, "Proving the Given Vertices of a Parallelogram Using Slope".

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