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Coordinate Geometry

Slope Problems Related to Quadrilaterals

In this article, we shall study to prove given quadrilateral to be or parallelogram, or rhombus, or square, or rectangle using slopes.

Example – 01:

Using slopes show that the points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Solution:

Let A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2) be the given points

Slope of side AB = (0 + 1)/(4 + 2) = (1)/(6) = 1/6  …………. (1)

Slope of side BC = (3 – 0)/(3 – 4) = (3)/(-1) = – 3  …………. (2)

Slope of side CD = (2 – 3)/(-3 – 3) = (-1)/(-6) = 1/6  …………. (3)

Slope of side AD = (2 + 1)/(-3 + 2) = (3)/(-1) = – 3  …………. (4)

From equations (1) and (3)

Slope of side AB = Slope of side CD

Thus side AB ∥ side CD ………. (5)

Slope of side BC = Slope of side AD

Thus side BC ∥ side AD ………. (6)

From equations (5) and (6)

In □ ABCD

side AB ∥ side CD and side BC ∥ side AD

Thus the pairs of opposite sides are parallel.

Hence □ ABCD is a parallelogram

Example – 02:

Using slopes show that the points (-2, -1), (1, 0), (4, 3) and (1, 2) are the vertices of a parallelogram.

Solution:

Le A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) be the given points

Slope of side AB = (0 + 1)/(1 + 2) = (1)/(3) = 1/3  …………. (1)

Slope of side BC = (3 – 0)/(4 – 1) = (3)/(3) = 1  …………. (2)

Slope of side CD = (2 – 3)/(1 – 4) = (-1)/(-3) = 1/3  …………. (3)

Slope of side AD = (2 + 1)/(1 + 2) = (3)/(3) = 1  …………. (4)

From equations (1) and (3)

Slope of side AB = Slope of side CD

Thus side AB ∥ side CD ………. (5)

Slope of side BC = Slope of side AD

Thus side BC ∥ side AD ………. (6)

From equations (5) and (6)

In □ ABCD

side AB ∥ side CD and side BC ∥ side AD

Thus the pairs of opposite sides are parallel.

Hence □ ABCD is a parallelogram

Example – 03:

Points A(8, 5), B(9, -7), C(-4, 2), and D(2, 6) are the vertices of quadrilateral ABCD. If P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA respectively. Using slopes show that quadrilateral PQRS is a parallelogram.

Solution:

P is the midpoint of side AB. Points A(8, 5), B(9, -7)

By midpoint formula P((8 + 9)/2, (5 -7)/2) = P(17/2, – 1)

Q is the midpoint of side BC. Points B(9, -7), C(-4, 2)

By midpoint formula Q((9 – 4)/2, (-7 + 2)/2) = Q(5/2, – 5/2)

R is the midpoint of side CD. Points C(-4, 2), D(2, 6)

By midpoint formula R((-4 + 2)/2, (2 + 6)/2) = R(-1, 4)

S is the midpoint of side DA. Points D(2, 6), A(8, 5)

By midpoint formula S((2 + 8)/2, (6 + 5)/2) = S(5, 11/2)

Now, P(17/2, -1), Q(5/2, – 5/2), R(-1, 4) and S(5, 11/2) be the vertices of quadrilateral PQRS

Slope of side PQ = (-5/2 + 1)/(5/2 – 17/2) = (- 3/2)/(-12/2) = 1/4  …………. (1)

Slope of side QR = (4 + 5/2)/(-1 – 5/2) = (13/2)/(- 7/2) = – 13/7  …………. (2)

Slope of side RS = (11/2 – 4)/(5 + 1) = (3/2)/(6) = 1/4  …………. (3)

Slope of side SP = (11/2 + 1)/(5 – 17/2) = (13/2)/(- 7/2) = – 13/7  …………. (4)

From equations (1) and (3)

Slope of side PQ = Slope of side RS

Thus side PQ ∥ side RS ………. (5)

Slope of side QR = Slope of side SP

Thus side QR ∥ side SP ………. (6)

From equations (5) and (6)

In □ PQRS

side PQ ∥ side RS and side QR ∥ side SP

Thus the pairs of opposite sides are parallel.

Hence □ PQRS is a parallelogram

Example – 04:

Using slopes show that the points (5, 7), (4, 12), (9, 11), and (10, 6) are the vertices of a rhombus.

Solution:

Le A(5, 7), B(4, 12), C(9, 11) and D(10, 6) be the given points

Slope of side AB = (12 – 7)/(4 – 5) = 5/(-1) = – 5  …………. (1)

Slope of side BC = (11 – 12)/(9 – 4) = -1/5   …………. (2)

Slope of side CD = (6 – 11)/(10 – 9) = (-5)/(1) = – 5  …………. (3)

Slope of side AD = (6 – 7)/(10 – 5) = – 1/5  …………. (4)

Slope of diagonal AC = (11 – 7)/(9 – 5) = 4/4 = 1  …………. (5)

Slope of side BD = (6 – 12)/(10 – 4) = (- 6)/(6)  = – 1  …………. (6)

From equations (1) and (3)

Slope of side AB = Slope of side CD

Thus side AB ∥ side CD ………. (5)

Slope of side BC = Slope of side AD

Thus side BC ∥ side AD ………. (6)

From equations (5) and (6)

In □ ABCD

side AB ∥ side CD and side BC ∥ side AD

Thus the pairs of opposite sides are parallel.

Hence □ ABCD is a parallelogram

From equations (5) and (6)

lope of side AC x Slope of side BD 1 x (-1) = -1

∴  AC ⊥ BD

Thus in parallelogram ABCD diagonals are perpendicular.

Hence □ ABCD is a rhombus

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