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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 equation of following lines
**passing through (3, -2) with slope -2**

Line passes through point (3, -2) = (x_{1}, y_{1})

Slope of line = m = – 2

By slope point form

y – y_{1} = m(x – x_{1})

y + 2 = – 2(x – 3)

y + 2 = – 2x + 6

2x + y – 4 = 0

**Ans:** The equation of the line is 2x + y – 4 = 0

**Passing through (1, 2) with slope – 3/2**

Line passes through point (1, 2) = (x_{1}, y_{1})

Slope of line = m = -3/2

By slope point form

y – y_{1} = m(x – x_{1})

y – 2 = -3/2(x – 1)

2y – 4 = -3x + 3

3x + 2y – 7 = 0

**Ans:** The equation of the line is 3x + 2y – 8 = 0

**Passing through (6, 2) with slope – 3**

Line passes through point (6, 2) = (x_{1}, y_{1})

Slope of line = m = -3

By slope point form

y – y_{1} = m(x – x_{1})

y – 2 = -3 (x – 6)

y – 2 = -3x + 18

3x + y – 20 = 0

**Ans:** The equation of the line is 3x + y – 20 = 0

**Passing through the point (– 4, 3) with slope 1/2.**

Line passes through point (-4, 3) = (x_{1}, y_{1})

Slope of line = m = 1/2

By slope point form

y – y_{1} = m(x – x_{1})

y – 3 = 1/2(x + 4)

2y – 6 = x + 4

x – 2y + 10 = 0

**Ans:** The equation of the line is x – 2y + 10 = 0

**Passing through (0, 0) with slope m.**

Line passes through point (0, 0) = (x_{1}, y_{1})

Slope of line = m

By slope point form

y – y_{1} = m(x – x_{1})

y – 0 = m(x – 0)

y = mx

**Ans:** The equation of the line is y = mx

**Passing through (-4, – 3) and parallel to the x-axis**

Line passes through point (-4, -3) = (x_{1}, y_{1})

Line is parallel to the x-axis

Slope of line = m = 0

By slope point form

y – y_{1} = m(x – x_{1})

y + 3 = 0(x + 4)

y + 3 = 0

**Ans:** The equation of the line is y + 3 = 0

**Having x-intercept 4 and tan θ = 1/2, where θ is inclination of line.**

x- intercept of line is 4

Line passes through point (4, 0) = (x_{1}, y_{1})

Slope of line = m = tan θ = 1/2

By slope point form

y – y_{1} = m(x – x_{1})

y – 0 = 1/2(x – 4)

2y = x – 4

x – 2y – 4 = 0

**Ans:** The equation of the line is x – 2y – 4 = 0

**Having y – intercept 2 and slope – 2**

y- intercept of line is 2

Line passes through point (0, 2) = (x_{1}, y_{1})

Slope of line = m = – 2

By slope point form

y – y_{1} = m(x – x_{1})

y – 2 = -2(x – 0)

y – 2 = -2x

2x + y – 2 = 0

**Ans:** The equation of the line is 2x + y – 2 = 0

**Having y – intercept 5 and slope – 2**

y- intercept of line is 5

Line passes through point (0, 5) = (x_{1}, y_{1})

Slope of line = m = – 2

By slope point form

y – y_{1} = m(x – x_{1})

y – 5 = -2(x – 0)

y – 5 = -2x

2x + y – 5 = 0

**Ans:** The equation of the line is 2x + y – 5 = 0

**Having y – intercept -4 and slope – 1/3**

y- intercept of line is -4

Line passes through point (0, -4) = (x_{1}, y_{1})

Slope of line = m = – 2

By slope point form

y – y_{1} = m(x – x_{1})

y + 4 = -1/3(x – 0)

3y + 12 = – x

x + 3y + 12 = 0

**Ans:** The equation of the line is x + 3y + 12 = 0

**Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.**

x- intercept of line is – 3

Line passes through point (-3, 0) = (x_{1}, y_{1})

Slope of line = m = – 2

By slope point form

y – y_{1} = m(x – x_{1})

y – 0 = -2(x + 3)

y = -2x – 6

2x + y + 6 = 0

**Ans:** The equation of the line is 2x + y + 6 = 0

**Intersecting the x-axis at a distance of 4 units to the left of origin with slope –3.**

x- intercept of line is – 4

Line passes through point (-4, 0) = (x_{1}, y_{1})

Slope of line = m = – 3

By slope point form

y – y_{1} = m(x – x_{1})

y – 0 = -3(x + 4)

y = -3x – 12

3x + y + 12 = 0

**Ans:** The equation of the line is 3x + y + 12 = 0

**Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis**

y- intercept of line is + 2

Line passes through point (0, 2) = (x_{1}, y_{1})

Slope of line = m = tan θ = tan 30^{o} = 1/√3

By slope point form

y – y_{1} = m(x – x_{1})

y – 2 = 1/√3(x – 0)

√3y -2√3 = x

x – √3y + 2√3 = 0

**Ans:** The equation of the line is x – √3y + 2√3 = 0

**It cuts intercept of length 3 on y axis and having slope 5.**

y- intercept of line is 3

Line passes through point (0, 3) = (x_{1}, y_{1})

Slope of line = m = 5

By slope point form

y – y_{1} = m(x – x_{1})

y – 3 = 5(x – 0)

5x – y + 3 = 0

**Ans:** The equation of the line is 5x – y + 3 = 0

**Passing through (2, -3) and making the inclination of 135**^{o}.

Line passes through point (2, -3) = (x_{1}, y_{1})

Slope of line = m = tan θ = tan 135^{o} = – 1

By slope point form

y – y_{1} = m(x – x_{1})

y + 3 = -1(x – 2)

y + 3 = – x + 2

x + y + 1 = 0

**Ans:** The equation of the line is x + y + 1 = 0

**Passing through (3, -1) and making inthe clination of 60**^{o}.

Line passes through point (3, -1) = (x_{1}, y_{1})

Slope of line = m = tan θ = tan 60^{o} = √3

By slope point form

y – y_{1} = m(x – x_{1})

y + 1 = √3(x – 3)

y + 1 = √3 x + 3 √3

√3 x – y – 1 + 3√3 = 0

√3 x – y – (3√3 – 1) = 0

**Ans:** The equation of the line is √3 x – y – (3√3 – 1) = 0

**Passing through (-5, 6) and making the inclination of 150**^{o}.

Line passes through point (-5, 6) = (x_{1}, y_{1})

Slope of line = m = tan θ = tan 150^{o} = – 1/√3

By slope point form

y – y_{1} = m(x – x_{1})

y – 6 = -1/√3(x + 5)

√3y – 6√3 = – x – 5

x + √3y – 6 √3 + 5 = 0

x + √3y + (5 – 6√3) = 0

**Ans:** The equation of the line is x + √3y + (5 – 6 √3) = 0

**Passing through (-2, 3) and making the inclination of 45**^{o}.

Line passes through point (-2, 3) = (x_{1}, y_{1})

Slope of line = m = tan θ = tan 45^{o} = 1

By slope point form

y – y_{1} = m(x – x_{1})

y – 3 = 1(x + 2)

y – 3 = x + 2

x – y + 5 = 0

**Ans:** The equation of the line is root x – y + 5 = 0

**Passing through (2, 2√3)and inclined with the x-axis at an angle of 75°.**

Line passes through point (2, 2√3) = (x_{1}, y_{1})

Slope of line = m = tan θ = tan 75^{o} = 2 + √3

By slope point form

y – y_{1} = m(x – x_{1})

y – 2√3 = (2 + √3)(x – 2)

y – 2√3 = (2 + √3)x – 4 – 2√3

(2 + √3)x – y – 4 = 0

**Ans:** The equation of the line is root (2 + √3)x – y – 4 = 0

Science > Mathematics > Straight Lines > You are Here |

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Mathematics |