The equation of a line in point-slope form we will learn how to find the equation of the straight line which is inclined at a given angle to the positive direction of x-axis in anticlockwise sense and passes through a given point.
Let the line MN makes an angle θ with the positive direction of x-axis in anticlockwise sense and passes through the point Q (x\(_{1}\), y\(_{1}\)). We have to find the equation of the line MN.
Let P (x, y) be any point on the line MN. But Q (x\(_{1}\), y\(_{1}\)) is also a point on the same line. Therefore, the slope of the line MN = \(\frac{y - y_{1}}{x - x_{1}}\)
Again, the line MN makes an angle θ
with the positive direction of the axis of x; hence, the slope of the
line = tan θ = m (say).
Therefore, \(\frac{y - y_{1}}{x - x_{1}}\) = m
⇒ y - y\(_{1}\) = m (x - x\(_{1}\))
The above equation y - y\(_{1}\) = m (x - x\(_{1}\)) is satisfied by the co-ordinates of any point P lying on the line MN.
Therefore, y - y\(_{1}\) = m (x - x\(_{1}\)) represent the equation of the straight line AB.
Solved examples to find the equation of a line in point-slope form:
1. Find the equation of a straight line passing through (-9, 5) and inclined at an angle of 120° with the positive direction of x-axis.
Solution:
First find the slope of the line:
Here slope of the line (m) = tan 120° = tan (90° + 30°) = cot 30° = √3.
Given point (x\(_{1}\), y\(_{1}\)) ≡ (-9, 5)
Therefore, x\(_{1}\) = -9 and y\(_{1}\) = 5
We know that the equation of a straight line passes through a given point (x\(_{1}\), y\(_{1}\)) and has the slope ‘m’ is y - y\(_{1}\) = m (x - x\(_{1}\)).
Therefore, the required equation of the straight lien is y - y\(_{1}\) = m (x - x\(_{1}\))
⇒ y - 5 = √3{x - (-9)}
⇒ y - 5 = √3(x + 9)
⇒ y - 5 = √3x + 9√3
⇒ √3x + 9√3 = y - 5
⇒ √3x - y + 9√3 + 5 = 0
2. A straight line passes through the point (2, -3) and makes an angle 135° with the positive direction of the x-axis. Find the equation of the straight line.
Solution:
The required line makes an angle 135° with the positive direction of the axis of x.
Therefore, the slope of the required line = m= tan 135° = tan (90° + 45°) = - cot 45° = -1.
Again, the required line passes through the point (2, -3).
We know that the equation of a straight line passes through a given point (x\(_{1}\), y\(_{1}\)) and has the slope ‘m’ is y - y\(_{1}\) = m (x - x\(_{1}\)).
Therefore, the equation of the required straight line is y - (-3) = -1(x -2)
⇒ y + 3 = -x + 2
⇒ x + y + 1 = 0
Notes:
(i) The equation of a straight line of the form y - y\(_{1}\) = m (x - x\(_{1}\)) is called its point-slope form.
(ii) The equation of the line y - y\(_{1}\) = m (x - x\(_{1}\)) is sometimes expressed in the following form:
y - y\(_{1}\) = m(x - x\(_{1}\))
We know that m = tan θ = \(\frac{sin θ}{cos θ}\)
⇒ y - y\(_{1}\) = \(\frac{sin θ}{cos θ}\) (x - x\(_{1}\))
⇒ \(\frac{x - x_{1}}{cos θ}\) = \(\frac{y - y_{1}}{sinθ}\) = r, Where r = \(\sqrt{(x - x_{1})^{2} + (y - y_{1})^{2}}\) i.e., the distance between the points (x, y) and (x1, y1).
The equation of a straight line as \(\frac{x - x_{1}}{cos θ}\) = \(\frac{y - y_{1}}{sinθ}\) = r is called its Symmetrical form.
● The Straight Line
11 and 12 Grade Math
From Point-slope Form to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Apr 22, 24 05:19 PM
Apr 22, 24 01:35 PM
Apr 21, 24 10:57 AM
Apr 20, 24 05:39 PM
Apr 20, 24 05:29 PM