# Application of the Concept of Slope of a Line

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#### Example – 01:

• Find the value of k so that the line joining (3, k) and (2, 7) is parallel to the line through (-1, 4) and (0, 6).
• Solution:

Let A(3, k), B(2, 7), and C(-1, 4), D(0, 6) be the given points

Given line AB ∥ line CD

∴  Slope of line AB = Slope of line CD

∴   (7 – k)/(2 – 3) = (6 – 4)/(0 + 1)

∴   (7 – k)/(-1) = 2/1

∴   (7 – k) = 2

∴   k = 7 + 2

∴   k = 9

#### Example – 02:

• A(0, 9), B(1, 11), C(3, 13) and D(7, k) are the four points if AB ⊥ CD then find k

Let A(0, 9), B(1, 11), and C(3, 13), D(7, k) be the given points

Slope of AB = (11 – 9)/(1 – 0) = 2/1 = 2

Slope of CD = (k – 13)/(7 – 3) = (k – 13)/4

Given line AB ⊥ line CD

∴  Slope of line AB x Slope of line CD = -1

∴   2  x  ((k – 13)/4) = – 1

∴   (k – 13)/2 = – 1

∴   k – 13 = – 2

∴   k  = – 2 + 13

∴   k = 11

#### Example – 03:

• The line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
• Solution:

Let A(-2, 6), B(4, 8), and C(8, 12), D(x , 24) be the given points

Slope of AB = (8 – 6)/(4 + 2) = 2/6 = 1/3

Slope of CD = (24 – 12)/(x – 8) = 12/(x – 8)

Given line AB ⊥ line CD

∴  Slope of line AB x Slope of line CD = -1

∴   (1/3  x  (12/(x – 8)) = – 1

∴   12/(x – 8)= – 3

∴   12 = – 3x + 24

∴  3x  = 12

∴   x = 4

#### Example – 04:

• If A(6, 4) and B(2, 12) are two given points then find the slope of the line (i) parallel to AB and (ii) perpendicular to AB

Given A(6, 4) and B(2, 12)

Slope of AB = (12 – 4)/(2 – 6) = (8)/(- 4) = – 2

Let l be the line parallel to line AB

As lines are parallel their slopes are equal

∴  Slope of line l = slope of line AB = – 2

Let m be the line perpendicular to line AB

As lines are perpendicular to each other, product of their slopes is -1

∴  Slope of line m =  – 1/slope of line AB = – (1/- 2) = 1/2

The slope of line parallel to AB is – 2 and that of the line perpendicular to AB is 1/2

#### Example – 05:

• Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right-angled triangle.
• Solution:

Let A(4, 4), B(3, 5), and C(-1, -1) be the given points

Slope of AB = (5 – 4)/(3 – 4) = (1)/(-1) = – 1  ………….. (1)

Slope of BC = (-1 – 5)/(-1 – 3) = (-6)/(-4) = 3/2  ………….. (2)

Slope of AC = (-1 – 4)/(-1 – 4) = (-5)/(-5) = 1  ………….. (3)

From equations (1) and (3) we have

Slope of AB x Slope of AC = (-1) x (1) = -1

Thus AB ⊥  AC

Hence ΔABC is right angled triangle right angled at A

∴   The points (4, 4), (3, 5) and (–1, –1) are the vertices of a right-angled triangle.

#### Example – 06:

• Without using the Pythagoras theorem, show that the points (7, 10), (-2, 5) and (3, –4) are the vertices of a right-angled triangle.
• Solution:

Let A(7, 10), B(-2, 5), and C(3, -4) be the given points

Slope of AB = (5 – 10)/(-2 – 7) = (-5)/(-9) = 5/9  ………….. (1)

Slope of BC = (-4 – 5)/(3 + 2) = (-9)/(5) = – 9/5  ………….. (2)

Slope of AC = (-4 – 10)/(3 – 7) = (-14)/(-4) = 7/2  ………….. (3)

From equations (1) and (2) we have

Slope of AB x Slope of BC = (5/9) x (- 9/5) = -1

Thus AB ⊥  BC

Hence ΔABC is right angled triangle right angled at B

∴   The points (7, 10), (-2, 5), and (3, -4) are the vertices of a right-angled triangle.

#### Example – 07:

• 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 – 08:

• 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 – 09:

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

Slope 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

#### Example – 10:

• 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 – 11:

• A(2, 3), B(-2, 1) and C(4, -3) are the vertices of ΔABC. Find the slope of (i) side AB (ii) altitude through A (iii) median through A and (iv) perpendicular bisector of AB.
• Solution:
• Slope of side AB: Slope of side AB = (1 – 3)/(-2 – 2) = (-2)/(-4) = 1/2

• Slope of Altitude Through A: Slope of BC = (-3 -1)/(4 + 2) = (-4)/(6) = – 2/3

Now AD ⊥ side BC

∴  Slope of AD = 3/2

• Median Through A: M is midpoint of side BC, By midpoint formula

M ≡ ((-2 + 4)/2, (1 -3)/2) = M(2/2, -2/2) = M(1, -1)

End points of median are A(2, 3) and M (1, -1)

Slope of median through A = (-1 – 3)/(1 – 2) = (-4)/(-1) = 4

• Perpenicular bisector of AB: Slope of side AB = (1 – 3)/(-2 – 2) = (-2)/(-4) = 1/2

line l is perpendicular bisector of side AB

line ⊥ side AB

Slope of line l = – 2

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