We will discuss here about the method of finding the pointslope form of a line.
To find the equation of a straight line passing through a fixed point and having a given slope,
let AB be the line passing through the point (x\(_{1}\), y\(_{1}\)), and let the line be inclined at an angle θ with the positive direction of the xaxis.
Then, tan θ = m = slope.
Let the equation of the line be y = mx + c, ……………. (i)
where m is the slope of the line and c is the yintercept. As A (x\(_{1}\), y\(_{1}\)) is a point on the line AB (x\(_{1}\), y\(_{1}\)) satisfy (i).
Therefore, y\(_{1}\) = mx\(_{1}\) + c ...................... (ii)
Subtracting (ii) from (i)
y – y\(_{1}\) = m(x  x\(_{1}\))
The equation of a line passing through(x\(_{1}\), y\(_{1}\)) and having the slope m is y – y\(_{1}\) = m(x – x\(_{1}\))
For example:
The equation of a line passing through the point (0, 1) and inclined at 30° with the positive direction of the xaxis is y  1 = tan 30° ∙ (x  0) or y  1 = \(\frac{x}{√3}\)
Notes:
(i) Equation of the yaxis:
The yaxis passes through the origin (0,0) and inclined at 90° with the positive direction of the xaxis.
So, the equation of the yaxis is y – 0 = tan 90° ∙ (x – 0)
⟹ y = ∞ ∙ x
⟹ \(\frac{y}{∞}\) = x
⟹ x = 0
The coordinate of any point on the yaxis is (0, k), where k changes from point to point. Thus, the xcoordinate of any point on the yaxis is 0 and so the equation x = 0 is satisfied by the coordinates of any point on the yaxis. Therefore, the equation of the yaxis is x = 0.
(ii) Equation of a line parallel to the yaxis:
Let AB be a line parallel to the yaxis. Let the line be at a distance a from the yaxis. Then, the slope = tan 90° = ∞ and the line passes through the point (a, 0).
Therefore, the equation of AB is y – 0 = tan 90° ∙ (x – a)
or, y cot 90° = x  a
⟹ y × 0 = x  a
⟹ x  a = 0
⟹ x = a
2. Find the equation of the line inclined at 60° with the positive direction of the xaxis and passing through the point (2, 5).
Solution:
The inclination of the line with the positive direction of the xaxis is 60°.
Therefore, the slope of the line = m = tan 60° = √3 and (x\(_{1}\), y\(_{1}\)) = (2, 5).
By the point slope form, the equation of the line is y  y\(_{1}\) = m(x  x\(_{1}\))
Substituting the value we get,
y  5 = √3(x  (2))
or, y  5 = √3(x + 2)
or, y – 5 = √3x + 2√3
or, y = √3x + 2√3 + 5, which is the required equation.
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