Math Lesson 14.3.2 - How to find the Equation of a Line?

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Welcome to our Math lesson on How to find the Equation of a Line? , this is the second lesson of our suite of math lessons covering the topic of Equation of Linear Graphs, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

How to find the Equation of a Line?

We defined and explained equation of a line in previous tutorials. However, here we will prove the accuracy of this equation using the gradient concept. Thus, given that any shift along the graph line brings a change in coordinates - which we denote as Δx and Δy respectively - we can write the final coordinates (x, y) of a point that starts moving from the point with coordinates (x₀, y₀) and shifts by Δx and Δy along the line as

and

We can express the above equations as

and

From the definition of the gradient m, we know that

Hence,

Expanding the right side and collecting the like terms yields

Expressing y₀ - mx₀ by a single letter (we usually take it as n), yields the simplified equation of a line

which we have taken as granted in the previous tutorials.

Look at the figure below for a better understanding of the above situation.

Since the constant n is

we can find the formulas for the x- and y-intercepts by rearranging the above equation to isolate x₀ and y₀. This only requires us to consider the upper-right part of the XY coordinates system. Thus, the x-intercept of a line for which the corresponding y-coordinate is zero (y₀ = 0) is

and the y-intercept of the same line for which the corresponding x-coordinate is zero (x₀ = 0) is

Example 2

1. Without plotting the graph, find the x- and y-intercepts of the line
   
   y = 2x - 5
2. Then, plot the graph to confirm the result.

Solution 2

1. We have m = 2 and n = -5. Given the formulas for calculating the x- and y-intercepts of the line
   x_int = -n/m and y_int = n
   we obtain after substituting the above values:
   x_int = -(-5)/2
   = 5/2
   = 2.5
   and
   y_int = -5
   This means the intercepts are at points P(0, -5) and Q(2.5, 0)
2. The graph of a line is obtained by connecting two known points of the graph and extending the straight line obtained beyond these two points. We can take two random values for the variable x; for example x = 1 and x = 4 and find the corresponding values by substituting them in the line's equation. Thus, for x = 1, we have
   y(1) = 2 ∙ 1 - 5
   = 2 - 5
   = -3
   and for x = 4, we have
   y(4) = 2 ∙ 4 - 5
   = 8 - 5
   = 3
   Therefore, the graph passes through the two points A(1, -3) and B(4, 3). The graph is obtained by connecting these two points and extending the line beyond them, as shown in the figure below. As you see, all the four points identified in (a) belong to the line y = 2x - 5. This means the solution is correct.

Remark! You don't need to remember the formulas for the x-and y-intercepts given above in theory. All you need to do is to substitute y = 0 in the line's equation to find the y-intercept and x = 0 to find the y-intercept. Let's explain this through an example.

Example 3

1. Without plotting any graph or using any formula find the intercepts of the line
   y = -4x + 1
   with the x- and y-axes.
2. Use the formulas of the intercepts to confirm the results obtained in (a).
3. Reconfirm the results obtained in (a) and (b) by plotting the graph of the given line.

Solution 3

1. Taking x = 0 in the original equation for calculating the y-intercept yields
   y = -4x + 1
   y = -4 ∙ 0 + 1
   y = 0 + 1
   y_int = 1
   Hence, the point P(0, 1) is the y-intercept of the given line.
   Likewise, taking y = 0 in the original equation for calculating the x-intercept yields
   y = -4x + 1
   0 = -4x + 1
   4x = 1
   x = 1/4
   or
   x_int = 0.25
   Hence, the point Q(0.25, 0) is the x-intercept of the given line.
2. Since the line's equation is
   y = -4x + 1
   we have m = -4 and n = 1.
   From theory, we know that
   x_int = -n/m and y_int = n
   Substituting the known values therefore yields
   x_int = -1/-4
   x_int = 1/4
   or
   x_int = 0.25
   and
   y_int = n
   y_int = 1
   Since the intercepts have the coordinates P(0, y_int) and Q(y_int, 0) we again obtain the results obtained in (a), i.e. P(0, 1) and Q(0.25, 0) for the intercepts of the given line with the axes.
3. Let's plot the graph by connecting two points of the graph and extending them beyond the ends of that segment. This is done by taking two random x-coordinates and finding the corresponding y-values. Thus, taking x = -1 and x = 1 yields for the corresponding y-coordinates:
   y(-1) = -4 ∙ (-1) + 1
   = 4 + 1
   = 5
   and
   y(1) = -4 ∙ 1 + 1
   = -4 + 1
   = -3
   Hence, points A(-1, 5) and B(1, -3) are points of the graph so we can draw the graph's line based on these two points as shown in the figure below.

More Equation of Linear Graphs Lessons and Learning Resources

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