Equation of a Line: Formulas, Forms & Examples

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Equation of a line is an expression that specifies the relationship between coordinate points on a straight line. In the coordinate plane, we know there are infinite points. Consider a line L and an arbitrary point P(x,y) on the XY plane.

The equation of a straight line can be expressed in terms of x and y.
If the equation of the line satisfies the point P(x,y), then the point P is on the line L.
In two-dimensional geometry, a straight line extends to both ends to infinity.
The equation of a line passes through two unique points.
It passes in a uniform direction.
The method is used to determine the different points on a line.
It can be written in the form of point-slope form, slope-intercept form, intercept form and standard form.

Read More: Horizontal and Vertical Lines

Table of Content

Different Forms of Equation of a Line
Equation of a Line Using the Standard Method
Equation of a Line using the Slope-Intercept Method
Things to Remember
Sample Questions

Key Terms: Equation of a line, Straight Line, Slope-Intercept Form, Standard Form, Point Slope Form, Coordinate Points, Intercept Form, Two Point Form

Different Forms of Equation of a Line

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The different methods used to find the equation of line based on the parameters known for a straight line are:

Normal Form

In normal form, the perpendicular line passes through the origin forms the normal method of the line equation. The equation of the line using the normal method is based on the specifications of the length of the normal "p" and the angle formed by the positive x-axis along with the normal $\theta$. The equation is expressed as:

X cos $\theta$ + y sin $\theta$ = P

Equation of line using the normal method

Equation of line using the normal method

Read More: Definite Integral Formula

Intercept Form

In the intercept form, equation of a lines crosses both x-axis and y-axis. Consider a line that x-axis and y-axis at at (a, 0) and (0, b), respectively. Using the point slope form:

y - 0 = -b/a (x - a)
Now, multiplying both sides of the equation by a
ay = -bx + ab
bx + ay = ab
Dividing both sides of the equation by ab,

x/a + y/b = 1

Solved Example of Intercept Form

Example: The equation of a straight line is given by 2x - 4y = 8. Convert this into the intercept form and hence find the intercepts.

Solution:

The equation of given line is:

2x - 4y = 8

Dividing both sides by 8,

2x/8 - 4y/8 = 8/8

x/4 + y/(-2) = 1

This is in the intercept form x/a + y/b = 1.

Read More:

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Slope Intercept Form

The slope-intercept method plays an important role in engineering and mathematics. It is expressed as

y= mx + c

where, m = slope of the line, c = y-intercept of the line. "c” also represents the distance between the point, (0, c) on the y- axis and the origin.

Solved Example of Slope Intercept Form

Example: The cost of a notebook is $12 more than twice the cost of a pen. Represent the situation as an equation of a straight line.

Ans: Assume the cost of pen = $x and the cost of notebook = $y.

According to the question, we have

y = 2x + 12 which is the equation of a straight line.

The required answer is y = 2x + 12

Equation of line using the slope-intercept method

Equation of line using the slope-intercept method

Two-Point Form

The more elaborated explanation of the point-slope method of the equation of a line is the two-point method of a line equation. The slope m= (y₂-y₁)/(x₂-x₁) according to a point-slope form whereas in two-point form, when a line is passing through two points (x₁,y₁) and (x₂, y₂), the line equation is:

(y-y₁) = [(y₂ –y₁)/(x₂-x₁)]* (x-x₁)

Solved Example of Two Point Form

Example: Find the general equation of a line passing through the points (-1, 2) and (3, -3).

Solution: The two points on the straight line are (-1, 2) and (3, -3).

Equation of a line in two-point form:

(y–y₁)/ (y₂–y₁) = (x–x₁) / (x₂–x₁)

Substitute (x₁,y₁)=(−1,2) and (x₂,y₂)=(3,−3)

(y–2)/(−3–2)=(x+1)/(3+1)

(y–2)/(−5)=(x+1)/(4)

4(y–1)=−5(x+1)4

Distribute.

4y–4=−5x–20

Simplify.

5x+4y+16=0

Straight line equation using two-point forms

Straight line equation using two-point forms

Point Slope Form

In the point-slope form, points is used when two points lie on a particular line. It is expressed as

y-y₁ =m (x-x₁)

where, y₁ = coordinate on the y axis, m = slope , x₁= coordinate on the x-axis.

Solved Example of Point Slope Form

Example: Find the equation of a straight line that passes through the points (2, 3) and (-2, 5). Write the equation in standard form.

Solution: To determine the equation of the line, we will use the formula point-slope form.

For this, we first need to find the slope of the line.

Slope = (5-3)/(-2-2) = -2/4 = -1/2

Therefore, the equation of the line passing through (2, 3) and (-2, 5) is y - 3 = (-1/2) (x + 2)

⇒ y - 3 = -x/2 - 1

⇒ y + x/2 = 3 - 1

⇒ x + 2y = 4

The different methods used to find the equations of the line are explained in detail in the following sections.

The video below explains this:

Straight Lines Detailed Video Explanation:

Equation of a Line Using the Standard Method

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The equation of a line using the standard method is expressed as ax+by+c = 0, where a and b are the coefficients, variables are x and y and c is a constant term. The variables x and y represent the points on a line.

Equation of line using the standard method

Read More: Plane Equation of a Plane

Equation of a Line using the Slope-Intercept Method

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The x-intercept "a" and the y-intercept "b" form the line equation in intercept form. (a,o) is the point which cuts the line at the x-axis and (o,b) is the point that cuts the line at the y-axis.

The distance between these points from the origin is represented by the variable a and b.
The distance between the point which cuts the x-axis and the y axis can be determined by the slope-intercept method.

Straight line equation using the slope-intercept method

Straight line equation using the slope-intercept method

Read More: Angle between two lines

Things to Remember

Equation of a line is given by y = mx + c.
It is also known as a linear equation in two variables.
The value of the slope is not affected by the interchange of the coordinates
One has to find the slope/gradient of the line before using the formula.
The equation of a line passing through two points.
Slope (m) = (y₂-y₁)/(x₂-x₁) to find the equation of the line passing through two points.
If a line is perpendicular to ax +by+ c = 0 , the line equation is bx – ay + k = 0

Read More:

Related Articles
Slope Intercept Form Formula	Angle Between Line and Plane	Different Forms of the Equation of Line
Angle Between Two Lines	Point Slope Form Formula	Linear Equations

Sample Questions

Ques: A plane has x-intercepts of 2 units, y-intercepts of 3 units, and z-intercepts of 4 units. Find the equation of the plane. (2 marks)

Ans: xa + yb + zc = 1

Given, a = 2, b= 3 and c = 4

Thus, x/2 + y/3 + z/4 = 1

Or,

6 x + 4 y 3 z = 12

Ques: The equation of a line is given as 3x – 4y + 7 = 0. Find the slope and y-intercept (2 marks)

Ans: 3x – 4y +7 = 0

3x +7 = 4y

4y = 3x+ 7

Y = 3x/4 +7/4

According to slope intercept form of equation,

Y = mx + c

Thus, slope (m) = 34 and y intercept (c ) = 74

Ques: A straight line passes through the point (2,3) and line segments that cross each axis are bisected at this point. Find the equation of this straight line. (3 marks)

Ans: let the equation of the straight line be xa + yb = 1 which meets the x-axes at A(a,0) and y-axes at B(0,b). the coordinates of the midpoint of the line segment AB are (a2, b2). As point (2,3) bisects AB,

a/2 = 2 and b/2 = 3

a = 4 and b= 6

Equation of the straight line is x/4 + y/6 = 1 or 3x + 2y = 12

Ques: A straight line has y intercept of 4 units which is perpendicular to a straight line connecting (4,2) and (2,-3). (2 marks)

Ans: slope of the line = m

m x slope of PQ = -1

m x 2+34-2 = -1

m x 52 = -1

m = -25

Ques: Find the line equation of the line passing through the points (-1,0) and (2,3). (3 marks)

Ans: Comparing the points to the coordinates of (x,y) , we get (x₁,y₁) = (2,3) and (x₂,y₂) = (-1,0)

By substituting,

m = 0-3-1-2

= -3-3

= 1

Putting the value of m and any coordinate,

y – y₁ = m (x – x₁)

y – 0 = 1(x – (-1))

= y = x+1

= y-x = 1

Solving the equation by slope intercept method,

y = x +1

thus, the line equation of the line passing through the points (-1,0) and (2,3) is

y = x +1 or y-x = 1

Ques: A line has an x-intercept of 5 units and a y-intercept of 4 units. Find the equation of the line in standard form. (3 marks)

Ans: x intercept (a) = 5 , y intercept (b) = 4

x/a + y/b = 1

x/5 + x/4 = 1

Converting the equation in standard form:

x/5 + x/4 = 1

(4x+5y)/20 =1

4x +5y = 20

4x + 5y – 20 = 0

The line equation in standard form is 4x + 5 y = 20

Ques: If a line is formed with points (1,3) and a slope of 13. Find the line equation. (3 marks)

Ans: y- y1 = m(x – x1)

y – 3 = 13 (x – 1)

= 3 (y-3) = 1 (x-1)

= 3y – 9 = x – 1

= 3y – x = 8

Expressing it in slope intercept method,

3y – x = 8

3y = 8+x

Y = 13x + 83

Thus, the equation of line is 3y – x = 8

Or,

Y = 13 x + 83

Ques: Prove that the points (11,4), (1,-1), and (5,1) are collinear. Find the line equation of the line which passes through these points. (3 marks)

Ans: let the given points be:

R = (11,4)

Q = (1,-1)

P = (5,1)

The equation of the line passing through the points P and Q

Y – 1 = -1-11-5 (x-5)

Y – 1 = -2-4 (x-5)

Y – 1 = 12 (x -5)

2(y-1) = (x-5)

2y – 2 = x- 5

x- 2y – 3 = 0

Thus, the point R (11,4) lies on the straight line whose equation is x-2y-3 = 0.

Ques: The equation of a line is given as 2x – 5y + 7 = 0. Find the slope and y-intercept (2 marks)

Ans: 2x – 5y +7 = 0

2x +7 = 5y

5y = 2x+ 7

5y = 3x/5 +7/5

According to slope intercept form of equation,

5y = mx + c

Thus, slope (m) = 3/5 and y intercept (c ) = 7/5

Ques: A straight line passes through the point (1,3) and line segments that cross each axis are bisected at this point. Find the equation of this straight line. (3 marks)

Ans: let the equation of the straight line be xa + yb = 1 which meets the x-axes at A(a,0) and y-axes at B(0,b). the coordinates of the midpoint of the line segment AB are (a₂, b₂₎. As point (1,3) bisects AB,

a/2 = 1 and b/2 = 3

a = 2 and b= 6

Equation of the straight line is x/2 + y/6 = 1 or 3x + y = 6

Ques: The equation of a line is given as 4x – 5y + 9 = 0. Find the slope and y-intercept (2 marks)

Ans: 4x – 5y +9 = 0

4x +9 = 5y

5y = 4x+ 9

5y = 4x/5 +9/5

According to slope intercept form of equation,

5y = mx + c

Thus, slope (m) = 4/9 and y intercept (c ) = 9/5

Ques: Find the equation of a straight line that passes through the points (2, 3) and (-2, 4). Write the equation in standard form. (3 marks)

Ans: To determine the equation of the line, we will use the formula point-slope form.

For this, we first need to find the slope of the line.

Slope = (4-3)/(-2-2) = -1/4

Therefore, the equation of the line passing through (2, 3) and (-2, 4) is y - 4 = (-1/4) (x + 2)

⇒ y - 4 = -x/4 - 2/4

⇒ y + x/4 = 4 - 2/4

⇒ x + 4y = 14

Ques: A plane has x-intercepts of 5 units, y-intercepts of 3 units, and z-intercepts of 4 units. Find the equation of the plane. (2 marks)

Ans: xa + yb + zc = 1

Given, a = 5, b= 3 and c = 4

Thus, x/5 + y/3 + z/4 = 1

Or,

12 x + 20 y + 15 z = 60

Ques: A straight line passes through the point (1,4) and line segments that cross each axis are bisected at this point. Find the equation of this straight line. (3 marks)

a/2 = 1 and b/2 = 4

a = 2 and b= 8

Equation of the straight line is x/2 + y/8 = 1 or 4x + y = 8

Ques: The cost of a notebook is $9 more than twice the cost of a pen. Represent the situation as an equation of a straight line. (2 marks)

Ans: Assume the cost of pen = $x and the cost of notebook = $y.

According to the question, we have

y = 2x + 9 which is the equation of a straight line.

The required answer is y = 2x + 9

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Equation of a Line: Formulas, Forms & Examples

Different Forms of Equation of a Line

Normal Form

Intercept Form

Solved Example of Intercept Form

Slope Intercept Form

Solved Example of Slope Intercept Form

Two-Point Form

Solved Example of Two Point Form

Point Slope Form

Solved Example of Point Slope Form

Straight Lines Detailed Video Explanation:

Equation of a Line Using the Standard Method

Equation of a Line using the Slope-Intercept Method

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

3.
Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

4.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

5.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

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Equation of a Line: Formulas, Forms & Examples

Different Forms of Equation of a Line

Normal Form

Intercept Form

Solved Example of Intercept Form

Slope Intercept Form

Solved Example of Slope Intercept Form

Two-Point Form

Solved Example of Two Point Form

Point Slope Form

Solved Example of Point Slope Form

Straight Lines Detailed Video Explanation:

Equation of a Line Using the Standard Method

Equation of a Line using the Slope-Intercept Method

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.Evaluate:\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

3.Find:Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

4. Find: If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

5.Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

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1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

3.
Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

4.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

5.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).