### Equation of a Line

Given the graph of the line

In this instance, the graph of a line is given and the question requires finding the equation of the line.

Example

Find the equation of the line above.

Solution:

The equation of a straight line is, y = mx + c

Recall, c is the y intercept (the point at which the graph crosses the y axis), which is the point (0, 2)

That is, c = 2

Now substituting the (x, y) coordinates of any of the other four points on the graph, along with the value for c (2) in the equation, solve for m.

That is, using the point (1, 5), substituting 1 for x, 5 for y, and 2 for c

Yields, y = mx + c

5 = m (1) + 2

5 = m + 2

m = 5 – 2

m = 3

Therefore the equation of the straight line is:

Y = 3x + 2

Given the co-ordinates of two points on the line

If co-ordinates of two points on the line are given, and the question requires using those points to find the equation of the line, use one of the methods below.* *

Example** **

The points, M (-3, -5) and N (5, 3) lie on a straight line L_{2 }. Find the particular equation of the line.

*Method 1*

Substitute the co-ordinates of the points in two different equations, subtracting one equation from the other in solving for m and c.

Using the equation of a straight line:

y = mx + c

Substituting (-3, -5) for x and y respectively

-5 = m (-3) + c

-5 = -3m + c ——- eq (1)

Substituting (5, 3) for x and y respectively

3 = m (5) + c

3 = 5m + c ——– eq (2)

eq (2) – eq (1)

3 – (-5) = 5m – (-3m) + c – c

3 + 5 = 5m + 3m

8 = 8m

m = 1

Substituting m = 1 in eq (1)

-5 = -3(1) + c

-5 = -3 + c

c = -5 + 3

c = -2

Hence the equation of L_{2 }is:

y = mx + c

y = (1) x + (-2)

y = x – 2

*Method 2*

Use the co-ordinates of the two points to find the gradient of the line (m), then using m and one of the points to substitute in the equation of a line to find c and hence the equation of the line L_{2 }.

Gradient, m = y_{2} – y_{1}/ x_{2} – x_{1}

Recall, M (-3, -5) N (5, 3)

m = 3 – (-5)/ 5 – (-3)

m = 3 + 5 / 5 + 3

m = 8 / 8

m = 1

Using, m = 1 and N (5, 3) to substitute in the equation of a line:

y = mx + c

3 = 1 (5) + c

3 = 5 + c

c = 3 – 5

c = -2

Therefore, the equation of L_{2} is:

y = x – 2

Given the gradient and one point on the line

Substitute the given quantities into the equation of a line, solving for c, and hence the equation of the line L_{2}.

Given, m = 1

N (5, 3)

Then, y = mx + c

3 = 1(5) + c

3 = 5 + c

c = -2

Therefore the equation of L_{2} is:

y = x – 2

**Note**: The above is the same as Method 2 (with gradient given).