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**

**Straight Line****Slope of a Straight Line****Slope of a Line through Two Given Points****Collinearity of Three Points****Equation of a Line Parallel to x-axis****Equation of a Line Parallel to y-axis****Slope-intercept Form****Point-slope Form****Straight line in Two-point Form****Straight Line in Intercept Form****Straight Line in Normal Form****General Form into Slope-intercept Form****General Form into Intercept Form****General Form into Normal Form****Point of Intersection of Two Lines****Concurrency of Three Lines****Angle between Two Straight Lines****Condition of Parallelism of Lines****Equation of a Line Parallel to a Line****Condition of Perpendicularity of Two Lines****Equation of a Line Perpendicular to a Line****Identical Straight Lines****Position of a Point Relative to a Line****Distance of a Point from a Straight Line****Equations of the Bisectors of the Angles between Two Straight Lines****Bisector of the Angle which Contains the Origin****Straight Line Formulae****Problems on Straight Lines****Word Problems on Straight Lines****Problems on Slope and Intercept**

**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.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.