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A line is a one-dimensional figure formed by a series of points. We can describe a line with an equation if we know any two points that lie on the line. The steepness of a line is defined as the slope of the line. Learn the various methods of framing an equation for a line with the help of some solved examples....Read MoreRead Less

A one-dimensional figure with length but no width is known as a line. A line is made up of a series of points that are infinitely extended in opposite directions. Two points in a two-dimensional plane determine a line. Collinear points are two points that are located on the same line.

The slope of a line indicates the line’s steepness and direction.

Any two distinct points on a line can be used to calculate the slope of the line. The slope of a line formula determines the ratio of “vertical change” to “horizontal change” between two points on a line.

In mathematics, the slope of a line is the change in the \(y-\)coordinate for a change in the \(x-\)coordinate.

The net change in the \(y-\)coordinate is denoted by \( \Delta y \) and the net change in the \(x-\) coordinate is denoted by \( \Delta x \). So, the change in the y-coordinate for a change in the \(x-\) coordinate is expressed as

m = \(\frac{Change~ in~ y~ coordinates}{Change~ in~ x~ coordinates}\)= \(\frac{\Delta y}{\Delta x}\)

The letter \(‘m’\) is commonly used to symbolize the slope.

In point-slope form, a linear equation written as \( y-y_1=m(x-x_1) \). The slope of the line is \( m \), and it passes through the point \( (x_1,y_1) \) .

**Equation: **\( y-y_1=m(x-x_1) \)

Assume \( P_1(x_1, y_1)\) is a fixed point on a non-vertical line with m as its slope.

Consider \( P(x,y)\) to be any point on the line L.

The ratio of the difference of the coordinates to the difference of x-coordinates is the slope of the line passing through the points \( (x_1, y_1) \) \( (x, y) \)

\( m= \frac{(y-y_1)}{(x-x_1)} \)

\( \Rightarrow~y-y_1=m(x-x_1) \ldots(i)\)

As a point, \( p_1(x_1,y_1)\) , which includes all points \( (x,y)\) on L, satisfies the equation I and no other point in the plane does.

As a result, the equation I will be without a doubt the equation for the given line L.

Thus, if and only if, the point \((x,y)\) lies on the line with slope m passing through the fixed point \((x_1,y_1)\) its coordinates satisfy the equation.

\( y-y_1=m(x-x_1) \)

As a result, this is the line equation’s point-slope form.

We can follow the steps below to solve point slope form for a given straight line and find the equation of the given line.

Step 1: Write down the slope of the straight line, ‘m’, and the coordinates \( (x_1, y_1)\) of a given point on the line.

Step 2: In the point slope formula, substitute the given values: \(y-y_1=m(x-x_1)\) .

Step 3: Simplify to get the line equation in the standard form.

**Example 1:**

Write an equation for a line that passes through the point (-2,1) and has a slope of \( \frac{2}{3}\) in the point-slope form.

**Solution:**

According to the point-slope formula we know that,

Equation of line is:

\(y-y_1= m(x-x_1)\) **Write the point slope form**

\(y-1=\frac{2}{3}\left [ x-(-2) \right ]\) **Substitute \( \dfrac{2}{3} \) **** for ****m, -2**** for \(x_1\)****, and 1 for \(y_1\)****.**

\( y-1=\dfrac{2}{3}(x+2)\) **Simplify**

So, \(y-1=\dfrac{2}{3}(x+2)\) is the equation.

**Example 2:**

Write an equation for the line that passes through the point (-4,2) and has a slope of \( \frac{1}{3}\) in the point-slope form.

**Solution:**

According to the point-slope formula we know that,

Equation of line is:

\( y-y_1=m(x-x_1)\) **Write the point slope form**

\( y-2 =\dfrac{1}{3}\left [ x-(-4) \right ]\) **Substitute \( \dfrac{1}{3} \) **** for ****m, -4**** for \( x_1\)****, and 2 for \( y_1\)**

\( y-2=\frac{1}{3}(x+4)\) **Simplify**

So, \(y-2=\frac{1}{3}(x+4)\) is the equation.

**Example 3:**

Write an equation for the line that passes through the point (1,2) and has a slope of \( \frac{3}{5}\) in the point-slope form.

**Solution:**

According to the point-slope formula we know that,

Equation of line is \( y-y_1=m(x-x_1)\)

**\( y-2=\dfrac{3}{5}\left [ x-(1) \right ]\) Substitute\( \dfrac{3}{5} \) **** for ****m, 1**** for\( x_1 \)****, and 2 for \( y_1 \)**

\( y-2=\dfrac{3}{5}(x-1) \) **Simplify**

So, \( y-2=\dfrac{3}{5}(x-1) \) is the equation.

Frequently Asked Questions on Equation of a Line:

\( y-y_1=m(x-x_1)\) is the equation of a line passing through the point \( (x_1, y_1) \) with slope \( m \).

This is the equation for the point slope form.

The formula for the point-slope form of an equation is used to find the equation of a straight line given a point on it and the slope of the line.