Point-slope Form of a Line

We will discuss here about the method of finding the point-slope 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 x-axis.

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 y-intercept. 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 x-axis is y - 1 = tan 30° ∙ (x - 0) or y - 1 = \(\frac{x}{√3}\)


Notes:

(i) Equation of the y-axis:

The y-axis passes through the origin (0,0) and inclined at 90° with the positive direction of the x-axis.

So, the equation of the y-axis is y – 0 = tan 90° ∙ (x – 0)

⟹ y = ∞ ∙ x

⟹ \(\frac{y}{∞}\) = x

⟹ x = 0

The coordinate of any point on the y-axis is (0, k), where k changes from point to point. Thus, the x-coordinate of any point on the y-axis is 0 and so the equation x = 0 is satisfied by the coordinates of any point on the y-axis. Therefore, the equation of the y-axis is x = 0.


(ii) Equation of a line parallel to the y-axis:

Let AB be a line parallel to the y-axis. Let the line be at a distance a from the y-axis. 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 x-axis and passing through the point (-2, 5).

Solution:

The inclination of the line with the positive direction of the x-axis 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.







10th Grade Math

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