Equation of Line: Slope Point Form

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Tangent Functions of Some Angles:

Angle

15° 30° 45° 60° 75° 105° 120° 135° 150°
tanθ 2 – 23 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) = (x1, y1)

Slope of line = m = – 2



By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = -3/2

By slope point form



y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = -3

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)



Slope of line = m = 1/2

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m

By slope point form



y – y1 = m(x – x1)

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) = (x1, y1)

Line is parallel to the x-axis



Slope of line = m = 0

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = tan θ = 1/2

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = – 2

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = – 2

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = – 2

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = – 2

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = – 3

By slope point form

y – y1 = m(x – x1)

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) = (x1, y1)

Slope of line = m = tan θ = tan 30o = 1/3

By slope point form

y – y1 = m(x – x1)

y  – 2 = 1/3(x – 0)

3y  -23 = x

x – 3y + 23 = 0

Ans: The equation of the line is x – 3y + 23 = 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) = (x1, y1)

Slope of line = m = 5

By slope point form

y – y1 = m(x – x1)

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 135o.

Line passes through point (2, -3) = (x1, y1)

Slope of line = m = tan θ = tan 135o = – 1

By slope point form

y – y1 = m(x – x1)

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 60o.

Line passes through point (3, -1) = (x1, y1)

Slope of line = m = tan θ = tan 60o = 3

By slope point form

y – y1 = m(x – x1)

y + 1 = 3(x – 3)

y + 1 =  3 x + 3 3

3 x – y – 1 + 33 = 0

3 x – y – (33 – 1) = 0

Ans: The equation of the line is 3 x – y – (33 – 1) = 0

  • Passing through (-5, 6) and making the inclination of 150o.

Line passes through point (-5, 6) = (x1, y1)

Slope of line = m = tan θ = tan 150o = – 1/3

By slope point form

y – y1 = m(x – x1)

y – 6 = -1/3(x + 5)

3y – 63 =  – x – 5

x + 3y – 6 3  + 5 = 0

x + 3y + (5 – 63) = 0

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

  • Passing through (-2, 3) and making the inclination of 45o.

Line passes through point (-2, 3) = (x1, y1)

Slope of line = m = tan θ = tan 45o = 1

By slope point form

y – y1 = m(x – x1)

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, 23)and inclined with the x-axis at an angle of 75°.

Line passes through point (2, 23) = (x1, y1)

Slope of line = m = tan θ = tan 75o = 2 + 3

By slope point form

y – y1 = m(x – x1)

y – 23 = (2 + 3)(x – 2)

y – 23 = (2 + 3)x – 4 – 23

(2 + 3)x – y – 4 = 0

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

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