HOW TO PROVE THE GIVEN FOUR POINTS FORM A PARALLELOGRAM USING SLOPE

Example 1 :

Using the concept of slope, show that the vertices (1 , 2), (-2 , 2), (-4 , -3) and (-1, -3) taken in order form a parallelogram.

Solution :

Let the given points be A(1 , 2) B(-2 , 2) C (-4 , -3) and D (-1, -3)

Slope of AB  =  (y₂ - y₁)/(x₂ - x₁)

A(1 , 2) B(-2 , 2) 

  =  (2 - 2) / (-4 - 1)

  =  0/(-5)  =  0  ---(1)

Slope of BC  =  (y₂ - y₁)/(x₂ - x₁)

B(-2 , 2) and C (-4 , -3) 

  =  (-3 - 2) / (-4 + 2)

  =  -5/(-2)  =  5/2  ---(2)

Slope of CD  =  (y₂ - y₁)/(x₂ - x₁)

C (-4 , -3) and D (-1, -3)

  =  (-3 + 3) / (-1 + 4)

  =  0/5  =  0 ---(3)

D (-1, -3) and A(1, 2)

  =  (2 + 3) / (1 + 1)

  =  5/2---(4)

(1)  =  (3) and (2)  =  (4)

Hence the given points form a parallelogram.

Example 2 :

Show that the opposite sides of a quadrilateral with vertices A(-2 ,-4), B(5 , -1), C(6 , 4) and D(-1, 1) taken in order are parallel.  

Solution :

Slope of AB  =  (y₂ - y₁) / (x₂ - x₁)

A(-2 ,-4), B(5 , -1)

  =  (-1 + 4) / (5 + 2)

  =  3/7  ----(1)

Slope of BC  =  (y₂ - y₁) / (x₂ - x₁)

B(5 , -1) and C(6 , 4) 

  =  (4 + 1) / (6 - 5)

  = 5/1

=  5 ----(2)

Slope of CD  =  (y₂ - y₁) / (x₂ - x₁)

 C(6 , 4) and D(-1, 1)

  =  (1 - 4) / (-1 - 6)

  = -3/(-7)

=  3/7 ----(3)

D(-1, 1) and A(-2, -4)

  =  (-4 - 1) / (-2 + 1)

  = -5/(-1)

=  5 ----(4)

(1)  =  (3)

(2)  =  (4)

Hence the opposite sides in a quadrilateral are parallel.

Example 3 :

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 :

A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6)

Lep P, Q, R and S be the mid-points of the sides AB, BC, CD and DA respectively.

Midpoint of the side AB :

= P(⁽⁻⁴ ⁺ ⁵⁾⁄₂, ⁽⁻² ⁻ ¹⁾⁄₂)

= P (½, -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 SR :

Slope of PQ :

Slope of PQ = Slope of SR

Slope of SP :

Slope of RQ :

Example 4 :

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 :

Let (a, b) be the coordinates of P.

Slope of SP = Slope of RQ

Slope of SQ (use the coordinates of S and M) :

= ⁽⁻¹ ⁻ ¹⁾⁄₍₂ ₋ ₁₎

= ⁻²⁄₁

= -2

Slope of PR (use the coordinates of P and M) :

= ⁽⁻¹ ⁻ ᵇ⁾⁄₍₂ ₋ ₐ₎

In a rhombus, always diagonals are perpendicular.

So, the product of the slopes of the disgonals SQ and PS is equal to -1.

-2 ⋅ ⁽⁻¹ ⁻ ᵇ⁾⁄₍₂ ₋ ₐ₎ = -1

⁽² ⁺ ²ᵇ⁾⁄₍₂ ₋ ₐ₎ = -1

2 + 2b = -1(2 - a)

2 + 2b = -2 + a

2b + 4 = a

a = 2b + 4 ----(1)

Distance between S(1, 1) and M(2, -1) :

SM = √[(x₂ - x₁)² + (y₂ - y₁)²]

SM = √[(2 - 1)² + (-1 - 1)²]

SM = √[1² + (-2)²]

SM = √(1 + 4)

SM = √5

In a rhombus, always diagonals bisect each other. So, M is the midpoint of QS and PR.

QS = 2(SM)

QS = 2√5

Given : QS = 2PR.

2PR = 2√5

PR = √5

2(PM) = √5

Squaring both sides,

Substitute a = 2b + 4.

4(4b² + 8b + 4 + 1 + 2b + b²) = 5

4(5b² + 10b + 5) = 5

4 ⋅ 5(b² + 2b + 1) = 5

Divide both sides by 5.

4(b² + 2b + 1) = 1

4(b + 1)² = 1

(b + 1)² = ¼

b + 1 = ±√¼

b + 1 = ±√¼

(a, b) = (3, -½)  or  (1, -³⁄₂)

The coordinates of P are (3, -½) or (1, -³⁄₂).

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