Drawing graph of y = mx + c

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

(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.