**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.

**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) = (y_{2} - y_{1})/(x_{2} - x_{1})

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

Slope of AB m = (-4-3.5)/(10-2.5) = -7.5/7.5 m = -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 BC m = (-2.5+4)/(2.5-10) = 1.5/(-7.5) = -15/75 m = -1/5 |
Slope of DA m = (5-3.5)/(-5-2.5) = 1.5/(-7.5) = -15/75 m = -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 AB m = (-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 BC m = (-3+3)/(1+2) m = 0/3 m = 0----(3) |
Slope of DA m = (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)

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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