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
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