An Alternate to the Point-Slope Form

The point-slope form of the equation of a line is given as

y – y₁ = m ( x – x₁) This form is generally used to solve a problem such as:

Find the equation of the line that has slope=2/3 and that contains the point (4,5).

We solve the problem by substituting 2/3 for m, 4 for x₁, and 5 for y₁. This produces the equation y – 5 = (2/3)( x – 4) which we then simplify, by various steps such as y – 5 = (2/3)x – (2/3)4
y = (2/3)x – (8/3) + 5
y = (2/3)x – 8/3 + 15/3
y = (2/3)x + 7/3 to get a final answer in slope-intercept form. Of course, this approach relies on using the point-slope form to begin the solution.

There is another approach to solving this same problem, one that does not use the point-slope form at all. Given the problem

Find the equation of the line that has slope=2/3 and that contains the point (4,5).

we could say that we know the answer will be an equation of the form y = mx + b We just need to find the values for m and b. However, the problem gives us the value for m, namely, m is 2/3. Therefore, the answer must look like y = (2/3)x + b Now we just need to find the value of b. However, the problem also gives us the information that the point (4,5) is on the line. That means that the point (4,5) must make our equation true. Therefore, if we replace x by 4 and y by 5, we must have a true equation. This means that we must have 5 = (2/3)4 + b which we solve for b via steps such as 5 = 8/3 + b
5 – 8/3 = b
15/3 – 8/3 = b
7/3 = b now that we know the value for b, we can write the answer, in slope-intercept form as y = (2/3)x + 7/3

Naturally, both methods produce the same result. The choice of method is open to the problem solver. The point-slope form is taught in virtually every algebra and college algebra class. However, it is not needed to solve these problems. The advantage of this approach is that it uses the basic slope-intercept form and it avoids requiring us to memorize the point-slope form.