In this chapter we will learn to draw graph of y = mx + c step by step.

Understand that the the equation y = mx + c is a **linear equation** so it’s **graph will always be a straight line**.

To draw the graph of above equation,** we need two data**, the **slope of the line** and **intersection of line at y axis**.

We have already discussed in previous lesson, the formula for above data;

(a) slope of y = mx + c is given by “m”

(b) y intercept of above equation is given by “c “

Now using the above two data points we can plot any given graph.

Let us understand the process with the help of example.

**Example 01**

Plot graph of y = x + 5

**Solution**

To draw the graph, follow the below steps;**(i)** **Compare the above equation with y = mx + c**

Here ,m = 1 and y intercept = 5

**(ii)** **y intercept = 5**, tells that the line intersect the y axis at (0, 5)

Locate point (0, 5) on the graph.

**(iii)** Now we have to find the **angle of line with respect to horizontal axis.**

Slope of line (m) = 1

tan\theta \ =\ 1\\\ \\ \theta \ =\ 45\ degree

Hence, the line makes 45 degree angle from horizontal axis.**(iv)** **Plot 45 degree angle from point (0,5)**

Hence, the above black line represents the equation y = x + 5

**Example 02**

Plot graph of following equation;

\mathtt{y\ -\sqrt{3} x+4=0}

**Solution**

Follow the below steps;

(i) **Represent equation in form of y = mx + c**

\mathtt{y\ =\ \sqrt{3} x\ -\ 4}

Here, m = \mathtt{\sqrt{3}}

y intercept (c) = -4

**(ii)** y intercept (c) = -4, tells that the **line intersect the y axis at (0, -4)**

Locating the point (0,-4) on the graph paper.

**(iii)** Now let’s **find the angle of line with respect to horizontal axis**.

\mathtt{m\ =\ \sqrt{3}}\\\ \\ \mathtt{tan\theta \ =\ \sqrt{3}}\\\ \\ \mathtt{\theta \ =\ 60\ degree}

Hence, the line makes **60 degree angle with horizontal axis**.

**(iv)** From point (0, -4), **draw a line making 60 degree horizontal angle**.

The above black line represents the given equation.

**Example 03**

Plot the graph of below equation;

\mathtt{3y\ +\sqrt{3} x-2=0} **Solution**

First arrange the equation in form of y = mx + c

\mathtt{3y\ +\sqrt{3} x-2=0}\\\ \\ \mathtt{3y\ =-\sqrt{3} x+2}\\\ \\ \mathtt{y\ =\ \frac{-\sqrt{3}}{3} x\ +\ \frac{2}{3}}\\\ \\ \mathtt{y\ =\ \frac{-1}{\sqrt{3}} x+\ \frac{2}{3}}

Here, m = \mathtt{\frac{-1}{\sqrt{3}}} and c = 2/3

**(i) Here y intercept (c) = 2/3**

It means that the line intersect y axis at point (0, 2/3).

Locating point (0, 2/3) on the graph.

**(ii)** Now **find the angle made by line with horizontal axis**.

\mathtt{m=\frac{-1}{\sqrt{3}}}\\\ \\ \mathtt{tan\theta =\frac{-1}{\sqrt{3}}}\\\ \\ \mathtt{\theta \ =\ -30\ degree}

So the line makes -30 degree with horizontal axis.

The above black line represents the given equation.