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Examples of Slope Intercept Form

We saw earlier the slope intercept form of a straight line. Here’s a quick recap of it –

The straight line y = mx + c has a gradient or slope of m, and the y-intercept of c.

Now let us look at some more examples of slope intercept form of a straight line.

 

Example 1: Find the x and y intercepts of the line y = 3x + 5, and sketch the line.

Given y = 3x + 5, the y-intercept is c = 5 (we can also get this by putting x = 0 in the line).

To get the x-intercept, put y = 0

0 = 3x + 5, so x = –5/3    (-5/3, 0).slope_intercept_form_1

To sketch the line, mark the points –

x-intercept (-5/3, 0) and y-intercept (0, 3), and join the line.

 

Example 2: Find the x and y intercepts of the line 4x + y = 2, and sketch the line.

Given 4x + y = 2, we rearrange it to get it in the form y = mx + c. So

y = -4x + 2

gradient = m = -4 (so the line is downwards)

the y-intercept is c = 2 (we can also get this by putting x = 0 in 4x + y = 2).

To get the x-intercept, put y = 0 in 4x + y = 2

We get 4x = 2 or x = 1/2slope_intercept_form_2

So we have obtained the x and y-intercepts – (5/3, 0) and (0, 2) respectively.

Mark these points on the graph, and plot the line.

 

Example 3: Write the equation of the line from each of the diagram.

3a. y-intercept = c = 2  slope_intercept_form_3a

gradient = m = {rise}/{run}

=  {2~-~0}/{0~-~3}

=  –2/3

Equation of the line is y = mx + c. Substituting the values of m and c here we get

y = –2/3x + 2

 

3b. y-intercept = c = 2 slope_intercept_form_3b

To calculate the gradient, consider any two points on the line. Here let us take (2, 8) and (3, 11).

gradient = m = {y_2~-~y_1}/{x_2~-~x_1}

=  {11~-~8}/{3~-~2}

=  3

So the equation of the line is y = 3x + 2

 

Example 4: Check if the following lines are parallel.

4a. y = 3x + 2 and y = 3x – 5.

Here both lines have gradient, m = 3. Hence the two lines are parallel.

 

4b. y = 5x – 2 and y = 2x – 5.

Here the gradient of the first line m1 = 5, and the gradient of the second line m2 = 2

Since m_1~<>~m_2, the lines are not parallel.

 

4c. 2x + y + 5 = 0 and 2x + y – 6 = 0.

Converting the lines to gradient intercept form, viz. y = mx + c, we get

2x + y + 5 = 0, or y = -2x – 5

So gradient of first line m1 = -2

2x + y – 6 = 0, or y = -2x + 6

And gradient of second line m2 = -2.

Since m1 = m2, the two lines are parallel.

 

Example 5: Find the slope of the line AB where A is (1, 3) and B (4, 9), and write the equation of the line in gradient intercept form.

Slope of line AB = {y_2~-~y_1}/{x_2~-~x_1} = m

A: x1 = 1, y1 = 3

B: x2 = 4, y2 = 9

m = {9~-~3}/{4~-~1}  =  6/3  =  2

With m = 2, we have the equation of the line in gradient intercept form as y = 2x + c.

To find c, we substitute one of the given points say A (1, 3)

When x = 1, y = 3. Hence

3 = 2 x 1 + c

3 = 2 + c

c = 1

Now the equation of the line AB is y = 2x + 1

 

Example 6: Show that points A (-1, -3), B (3, 5) and C (8, 15) are all collinear i.e. they all line in a straight line.

To show a set of points to be collinear, we find the gradients using the points. If all the gradients are the same, then the points can be said to be on the same line i.e. they are collinear.

Let us look at the gradients of the lines AB and BC.

Gradient of AB = {y_2~-~y_1}/{x_2~-~x_1}

=  {5~-~(-3)}/{3~-~(-1)}

=  8/4

=  2

Gradient of BC = {y_2~-~y_1}/{x_2~-~x_1}

=  {15~-~5}/{8~-~3}

=  10/5

=  2

Since the gradients of AB and BC are same, the points A, B and C are collinear, i.e. they lie on the same line. (Note: The point B is common to the two lines).

 

Example 7: Find the equation of a line that makes an angle of 60° with the x-axis, and has a y-intercept of -4.

Given y-intercept = c = -4, and theta = 60°.

m = tan theta = tan 60°  =  sqrt{3}

So the equation of the line in gradient intercept form (y = mx + c) is

y = sqrt{3}x – 4

 

Example 8: Find the equation of a line with y-intercept 3, and an angle of inclination of 120° with the x-axis.

Given y-intercept = c = 3, and theta = 120°.

m = tan theta = tan 120°  =  –sqrt{3}

So the equation of the line in gradient intercept form (y = mx + c) is

y = –sqrt{3}x + 3