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#### Tangent Functions of Some Angles:

Angle |
15° | 30° | 45° | 60° | 75° | 105° | 120° | 135° | 150° |

tanθ | 2 – 2√3 | 1/√3 | 1 | √3 | 2 + √3 | -(2 + √3) | – √3 | – 1 |
– 1/√3 |

#### Example – 01:

- Find the slope of lines whose inclinations are
**45°**

Given θ = 45°

∴ The slope of a line = m = tan θ = tan 45° = 1

**60°**

Given θ = 60°

∴ The slope of a line = m = tan θ = tan 60° = √3

**30°**

Given θ = 30°

∴ The slope of a line = m = tan θ = tan 30° = 1/√3

**120°**

Given θ = 120°

∴ The slope of a line = m = tan θ = tan 120°

∴ m = tan (90° + 30°) = – cot 30° = – √3

**(3π/4)**^{c}

Given θ = (3π/4)^{c}

∴ The slope of a line = m = tan θ = tan (3π/4)^{c}

∴ m = tan (π/2 + π/4)^{c} = – tan (π/4)^{c} = – 1/√2

**(5π/6)**^{c}

Given θ = (5π/6)^{c}

∴ The slope of a line = m = tan θ = tan (5π/6)^{c}

∴ m = tan (π – π/6)^{c} = – tan (π/6)^{c} = – 1/√3

**105°**

Given θ = 105°

∴ The slope of a line = m = tan θ = tan 105° = tan (60° + 45°)

- The line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise.

The angle made by line with positive direction of the y-axis = – 30° (anticlockwise)

The angle made by line with positive direction of the x-axis = 90° – 30° = 60°

Now θ = 60°

∴ The slope of a line = m = tan θ = tan 60° = √3

#### Example – 02:

- Find the inclinations of lines whose slopes are
**1**

The slope of line = m = tan θ =1

∴ θ = tan^{-1}(1) = 45° or (π/4)^{c}

**– 1**

The slope of line = m = tan θ = -1

∴ θ = tan^{-1}(-1) = 135° or (3π/4)^{c}

**√3**

The slope of line = m = tan θ =√3

∴ θ = tan^{-1}(√3) = 60° or (π/3)^{c}

**– √3**

The slope of line = m = tan θ = –√3

∴ θ = tan^{-1}(- √3) = 120° or (2π/3)^{c}

**1/√3**

The slope of line = m = tan θ =1/√3

∴ θ = tan^{-1}(1/√3) = 30° or (π/6)^{c}

**– 1/√3**

The slope of line = m = tan θ = – 1/√3

∴ θ = tan^{-1}(- 1/√3) = 150° or (5π/6)^{c}

#### Example – 03:

- Find the slopes of the lines which pass through the points
- (2, 5) and (-4, -4)

Let the points be A(2, 5) ≡ (x_{1}, y_{1}) and (-4, -4) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (- 4 – 5)/(- 4 – 2) = (-9)/(-6)

∴ Slope of line AB = 3/2

- (-2, 3) and (4, -6)

Let the points be A(-2, 3) ≡ (x_{1}, y_{1}) and (4, -6) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (- 6 – 3)/(4 + 2) = (-9)/(6)

∴ Slope of line AB = – 3/2

- (1, -3) and (-1, -1)

Let the points be A(1, -3) ≡ (x_{1}, y_{1}) and (-1, -1) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (- 1 + 3)/(- 1 – 1) = (2)/(-2)

∴ Slope of line AB = – 1

- (2, √3) and (3, 2√3)

Let the points be A(2, √3) ≡ (x_{1}, y_{1}) and (3, 2√3) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (2√3 – √3)/(3 – 2) = (√3)/(1)

∴ Slope of line AB = √3

- (3, 4) and (2, 5)

Let the points be A(3, 4) ≡ (x_{1}, y_{1}) and (2, 5) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (5 – 4)/(2 – 3) = (1)/(-1)

∴ Slope of line AB = – 1

- (2, 5) and (4, -6)

Let the points be A(2, 5) ≡ (x_{1}, y_{1}) and (4, -6) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (- 6 – 5)/(4 – 2) = (- 11)/(2)

∴ Slope of line AB = – 11/2

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

Let the points be A(-1, 4) ≡ (x_{1}, y_{1}) and (2, 2) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (2 – 4)/(2 + 1) = (-2)/(3)

∴ Slope of line AB = – 2/3

- (a, 0) and (0, b)

Let the points be A(a, 0) ≡ (x_{1}, y_{1}) and (0, b) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (b – 0)/(0 – a) = (b)/(-a)

∴ Slope of line AB = – b/a

- (at
_{1}^{2}, 2at_{1}) and (at_{2}^{2}, 2at_{2})

Let the points be A(at_{1}^{2}, 2at_{1}) ≡ (x_{1}, y_{1}) and (at_{2}^{2}, 2at_{2}) ≡ (x_{2}, y_{2})

∴ Slope of line AB = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (2at_{2} – 2at_{1})/(at_{2}^{2} – at_{1}^{2})

∴ Slope of line AB = 2a(t_{2} – t_{1})/a(t_{2}^{2} – t_{1}^{2})

∴ Slope of line AB = 2(t_{2} – t_{1})/(t_{2} + t_{1})(t_{2} – t_{1})

∴ Slope of line AB = 2)/(t_{2} + t_{1})

- 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, M(-4, 0)/2) = M(8/2, -4/2) = M(4, – 2)

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

#### Example – 04:

- 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

#### Example – 05:

- 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

#### Example – 06:

- 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

#### Example – 07:

- 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:

- line passes through (x
_{1}, y_{1}) and (h, k). If slope of the line is m, show that k – y_{1}= m (h – x_{1}).

Line passes through (x_{1}, y_{1}) and (h, k)

Slope of line = (k – y_{1})/(h – x_{1}) = m

∴ k – y_{1} = m (h – x_{1}) (Proved as required)

#### Example – 09:

- 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?

**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

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).

Example 5 In Fig 10.9, time and

distance graph of a 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 law of motion, i.e., how distance

depends upon time.

Science > Mathematics > Straight Lines > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |