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The slope of a Line

Science > Mathematics > Coordinate Geometry > Straight Lines > The slope of Line

In this article, we shall study to find the slope of line AB, when points A(x1, y1) and B(x2, y2) are given.

Type – I: To Find Slopeof a Line When Two Points on the Line are Given:

Example – 01:

Find the slopes of the lines which pass through the points

  • (2, 5) and (-4, -4)

Let the points be A(2, 5) ≡ (x1, y1) and (-4, -4) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (- 4 – 5)/(- 4 – 2) = (-9)/(-6)

∴  Slope of line AB = 3/2

  • (-2, 3) and (4, -6)

Let the points be A(-2, 3) ≡ (x1, y1) and (4, -6) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (- 6 – 3)/(4  + 2) = (-9)/(6)

∴  Slope of line AB = – 3/2

  • (1, -3) and (-1, -1)

Let the points be A(1, -3) ≡ (x1, y1) and (-1, -1) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (- 1 + 3)/(- 1  – 1) = (2)/(-2)

∴  Slope of line AB = – 1

  • (2, √3) and (3, 2√3)

Let the points be A(2, √3) ≡ (x1, y1) and (3, 2√3) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (2√3 – √3)/(3  – 2) = (√3)/(1)

∴  Slope of line AB = √3

  • (3, 4) and (2, 5)

Let the points be A(3, 4) ≡ (x1, y1) and (2, 5) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (5 – 4)/(2 – 3) = (1)/(-1)

∴  Slope of line AB = – 1

  • (2, 5) and (4, -6)

Let the points be A(2, 5) ≡ (x1, y1) and (4, -6) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (- 6 – 5)/(4  – 2) = (- 11)/(2)

∴  Slope of line AB = – 11/2

  • (-1, 4) and (2, 2)

Let the points be A(-1, 4) ≡ (x1, y1) and (2, 2) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (2 – 4)/(2 + 1) = (-2)/(3)

∴  Slope of line AB = – 2/3

  • (a, 0) and (0, b)

Let the points be A(a, 0) ≡ (x1, y1) and (0, b) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (b – 0)/(0 – a) = (b)/(-a)

∴  Slope of line AB = – b/a

  • (at12, 2at1) and (at22, 2at2)

Let the points be A(at12, 2at1) ≡ (x1, y1) and (at22, 2at2) ≡ (x2, y2)

∴  Slope of line AB = (y2 – y1)/ (x2 – x1) = (2at2 – 2at1)/(at22 – at12)

∴  Slope of line AB = 2a(t2 – t1)/a(t22 – t12)

∴  Slope of line AB = 2(t2 – t1)/(t2 + t1)(t2 – t1)

∴  Slope of line AB = 2)/(t2 + t1)

  • The origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Midpoint of segment AB is M( (0 + 8)/2, (-4, 0)/2) = M(8/2, -4/2) = M(4, – 2)

Now slope of OM = (-2 – 0)/ (4 – 0) = -2/4 = – 1/2

Type – II: Find Value of Arbitrary Constant, When Slope is Given:

Example – 02:

Find the value of ‘k’ if the slope of the line passing through the points

  • (2, 4), (5, k) is 5/3

Let the points be A(2, 4) and B(5, k)

Slope of AB = (k – 4)/(5 – 2) = 5/3

∴  (k – 4)/3 = 5/3

∴  k – 4 = 5

∴  k = 9

  • (2, 5), (k, 3) is 2

Let the points be A(2, 5) and B(k, 3)

Slope of AB = (3 – 5)/(k – 2) = 2

∴  – 2  = 2k – 4

∴  2k = 2

∴  k = 1

  • (k, 2), (-6, 8) is – 5/4

Let the points be A(k, 2) and B(-6, 8)

Slope of AB = (8 – 2)/(-6 – k) = -5/4

∴  6/(-6 – k) = -5/4

∴ 24 = 30 + 5k

∴ 5k = – 6

∴  k = – 6/5

Type – III: To Show Given Points are Collinear:

Example – 03:

By using slopes show that the following points are collinear.

  • A(0,4), B(2,10) and C(3,13)

Given points are A(0,4), B(2,10) and C(3,13)

Slope of line AB = (10 – 4)/(2 – 0) = (6)/(2) = 3 …………. (1)

Slope of line BC = (13 – 10)/(3 – 2) = (3)/(1) = 3 …………. (2)

From equations (1) and (2)

Slope of line AB = Slope of line BC

B is common point

Hence points A, B and C are collinear

  • P(5, 0), Q(10, -3) and R(-5, 6)

Given points are P(5, 0), Q(10, -3) and R(-5, 6)

Slope of line PQ = (-3 – 0)/(10 – 5) = (-3)/(5) = – 3/5 …………. (1)

Slope of line QR = (6 + 3)/(-5 – 10) = (9)/(- 15) = – 3/5 …………. (2)

From equations (1) and (2)

Slope of line PQ = Slope of line QR

Q is common point

Hence points P, Q and R are collinear

  • L(2, 5), M(5, 7) and N(8, 9)

Given points are L(2, 5), M(5, 7) and N(8, 9)

Slope of line LM = (7 – 5)/(5 – 2) = (2)/(3) = 2/3 …………. (1)

Slope of line MN = (9 – 7)/(8 – 5) = (2)/(3) = 2/3 …………. (2)

From equations (1) and (2)

Slope of line LM = Slope of line MN

M is common point

Hence points L, M, and N are collinear

  • D(5, 1), E(1, -1) and F(11, 4)

Given points are D(5, 1), E(1, -1) and F(11, 4)

Slope of line DE = (-1 – 1)/(1 – 5) = (-2)/(-4) = 1/2 …………. (1)

Slope of line EF = (4 + 1)/(11 – 1) = (5)/(10) = 1/2 …………. (2)

From equations (1) and (2)

Slope of line DE = Slope of line EF

E is common point

Hence points D, E, and F are collinear

Type IV: To Find Value of K if Given Points are Collinear:

Example – 04:

If the following points are collinear find value of k using slopes.

  • (-2, -3), (k, 4) and (5, 5)

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

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (4 + 3)/(k + 2) = (5 – 4)/(5 – k)

∴  7/(k + 2) = 1/(5 – k)

∴  7(5 – k) = 1(k + 2)

∴  35 – 7k = k + 2

∴  33 = 8k

∴  k = 33/8

  • (5, k), (-3, 1) and (-7, -2)

Let given points be A(5, k), B(-3, 1) and C(-7, -2)

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (1 – k)/(-3 – 5) = (-2 – 1)/(-7 + 3)

∴  (1 – k)/(-8) = (-3)/(-4)

∴  (1 – k)/(-8) = (3/4)  x (-8)

∴  1 – k = – 6

∴  k = 7

  • (2, 1), (4, 3) and (0, k)

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

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (3 – 1)/(4 – 2) = (k – 3)/(0 – 4)

∴  2/2 = (k – 3)/(- 4)

∴ 1 (-4) = (k – 3)

∴  k – 3  = – 4

∴  k = – 1

  • (-4, 5), (-3, 5) and (-1, k)

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

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (5 – 5)/(-3 + 4) = (k – 5)/(-1 + 3)∴  0/1 = (k – 5)/2

∴ 0 = (k – 5)/2

∴  k – 5  = 0

∴  k = 5

  • (k, 1), (2, -3) and (3, 4)

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

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (-3 – 1)/(2 – k) = (4 + 3)/(3 – 2)

∴  (-4)/(2 – k) = 7/1

∴ -4 = 14 – 7k

∴  7k = 18

∴  k = 18/7

Type V: To Prove Given Condition When Collinearity of Points is Given:

Example – 05:

If three points (h, 0), (a, b) and (0, k) lie on a line, show that a/h + b/k = 1

Solution:

Let A(h, 0), B(a, b) and C(0, k) be the points lying on the line.

∴ Slope of AB = Slope of BC

∴ (b – 0)/(a – h) = (k – b)/(0 – a)

∴ b/(a – h) = (k – b)/(-a)

∴ – ab   = (a – h)(k – b)

∴  ak – kh + hb = 0

∴  ak + hb = kh

Dividing both sides of equation by kh

∴  ak/kh + hb/kh = kh/kh

∴  a/h + b/k = 1  (proved as required)

Example – 08:

A line passes through (x1, y1) and (h, k). If the slope of the line is m, show that k – y1 = m (h – x1).

Line passes through (x1, y1) and (h, k)

Slope of line = (k – y1)/(h – x1) = m

∴  k – y1 = m (h – x1)  (Proved as required)

Example – 09:

Three points P (h, k), Q (x1, y1) and R (x2, y2) lie on a line. Show that (h – x1) (y2 – y1) = (k – y1) (x2 – x1).

As the points lie on a line

Slope of line PQ = Slope of line QR

(k – y1)/(h – x1) = (y2– y1)/(x2 – x1)

∴ (k – y1) (x2 – x1)= (h – x1) (y2 – y1)

∴  (h – x1) (y2 – y1) = (k – y1) (x2 – x1)    (Proved)

Type – VI: Interpretation of Graph:

Example – 10:

Consider the following population and year graph, find the slope of the line AB, and using it, find what will be the population in the year 2010?

slope of Line AB

Solution:

Slope of line AB = (97 – 92)/(1995 -1985) = 5/10 = 1/2

Let the populatiomn in year 2010 be k

Thus point P(2010, k) lies on the line

Slope of AP = Slope of AB

Slope of AP = (k – 92)/(2010 – 1985) = 1/2

∴ (k – 92)/25 = 1/2

∴ 2k – 184 = 25

∴ 2k = 209

∴ k = 104.5

Hence population in 2010 will be 104.5 crore

Example – 11:

In the figure, the time and distance graph of linear motion is given. Two positions of time and distance are recorded as when T = 0, D = 2 and when T = 3, D = 8. Using the concept of slope, find the law of motion, i.e., how distance depends upon time.

Slope of AB = Slope of BC

∴ (8 – 2)/(3 – 0) = (D – 8)(T – 3)

∴ 6/3 = (D – 8)(T – 3)

∴ 2 = (D – 8)(T – 3)

∴ 2(T – 3) = D – 8

∴ 2T – 6 = D – 8

∴ D = 2T – 6 + 8

∴ D = 2T + 2

This relation is known as the law of motion.

Science > Mathematics > Coordinate Geometry > Straight Lines > The slope of a Line

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