Science > Mathematics > Coordinate Geometry > Straight Lines > Slope Point Form
In this article, we shall study to find the equation of a straight line using slope point form.
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 -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) = (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 + 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 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 – 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 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, 2√3)and inclined with the x-axis at an angle of 75°.
Line passes through point (2, 2√3) = (x1, y1)
Slope of line = m = tan θ = tan 75o = 2 + √3
By slope point form
y – y1 = m(x – x1)
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