The equation of a line in pointslope 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 xaxis in anticlockwise sense and passes through a given point.
Let the line MN makes an angle θ with the positive direction of xaxis 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 coordinates 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 pointslope 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 xaxis.
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 xaxis. 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 pointslope 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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