00:01
For this example, we have two points that are given, and we want to find the equation of a line that passes through both of them.
00:07
We'll start off by organizing, calling the first point, x1, y1, and likewise denote the second point as x2y2, so that x2 is negative 5, and y2 is 3.
00:20
The next step is that we have to calculate the slope m, and we know by formula, m equals y2 minus y1, in the numerator, divide by x2, minus x1 in the denominator.
00:33
So let's make our substitutions.
00:35
We have four substitutions to make, and that's why we start off by organizing, and now we say y2 is 3, and we see x2 is negative 5.
00:45
Now go to the y1, x1 column, where y1 is a 9, and x1 is a positive 5.
00:56
Let's now simplify what we have.
00:58
In the numerator, we have negative 6, and in the denominator we have negative 10.
01:03
So this is going to be a positive 6 tenths.
01:07
Let's make that a more clear 10...