### Video Transcript

A straight line is defined by the
equation π¦ equals ππ₯ plus π. Given that the point π₯ one, π¦ one
lies on the line, find an expression for π in terms of π, π₯ one, and π¦ one. Given also that the point π₯ two,
π¦ two lies on the line and π is the slope of the line, find an expression for π
in terms of π¦ one, π¦ two, π₯ one, and π₯ two. By substituting in for π and
factoring out π, find the formula for the equation of the line. And finally, substitute in your
expression for π to complete your formula.

So, letβs start at the
beginning. We can model a linear equation with
the line π¦ equals ππ₯ plus π. If we know that the point π₯ one,
π¦ one lies on this line, we can find an expression for π in terms of π, π₯ one,
and π¦ one. If we want to find an expression
for π in these terms, π will be on one side of the equation and the other side
will only contain π, π₯ one, and π¦ one as those are the terms we want to know π
in. If π₯ one, π¦ one lies on this
line, then we can say that π¦ one is equal to π times π₯ one plus π.

And now, what we want is to
rearrange the equation so that π is by itself. And that means weβll subtract ππ₯
one from both sides of the equation, which gives us π¦ one minus π times π₯ one
equals π. But more commonly, weβll put the π
on the left so that we have π equals π¦ one minus π times π₯ one. And we have our first expression,
π¦ one minus π times π₯ one.

We also know that the point π₯ two,
π¦ two lies on this line and π is the slope of the line. We wanna find an expression for π
in terms of π¦ one, π¦ two, π₯ one, and π₯ two. And similarly, if we want π in
terms of these values, π needs to be by itself. And the other side of the equation
will only include π¦ one, π¦ two, π₯ one, and π₯ two. From part one, we know that π
equals π¦ one minus π times π₯ one. And since π₯ two, π¦ two also lies
on this line, we can plug in π₯ two and π¦ two into this equation. Since these are at the same lines,
their π-values will be the same. And that means we can take π¦ one
minus π times π₯ one and set it equal to π¦ two minus π times π₯ two.

Remember, our goal is to get π by
itself. So, we add π times π₯ two to both
sides of the equation. And then, weβll subtract π¦ one
from both sides of the equation to move that π¦ one value to the other side. On the left, weβll have π times π₯
two minus π times π₯ one since the π¦βs cancel out. And on the other side, weβll have
π¦ two minus π¦ one since the ππ₯ two cancels out.

To get π by itself, we see that we
have two π-values that we can factor out, which will give us π times π₯ two minus
π₯ one equals π¦ two minus π¦ one. From there to finally get π by
itself, weβll divide both sides of the equation by π₯ two minus π₯ one. On the left, they cancel out, and
we have π equals π¦ two minus π¦ one over π₯ two minus π₯ one.

This completes part two. So, we clear some room for part
three.

By substituting in for π and
factoring out π, find the formula for the equation of the line.

This time, we start back with the
definition of a straight line π¦ equals ππ₯ plus π. And in place of π, we plug in π¦
one minus π times π₯ one and everything else remains the same. Since weβre only adding and
subtracting each of the terms, we can take away the parentheses. Our instructions say we need to
factor out π. To make that easier to see, we can
put the order π¦ one plus ππ₯ minus π times π₯ one. And now, weβre able to factor out
the π so that we have π¦ equals π¦ one plus π times π₯ minus π₯ one.

Itβs also good to keep the π¦βs on
the same side of the equation. And we can do that here by
subtracting π¦ one from both sides, which gives us π¦ minus π¦ one equals π times
π₯ minus π₯ one. This is our equation for the line
from step three.

In our final step, we want to
substitute in what we found for π into the equation from step three. Beginning here, π¦ minus π¦ one
equals π times π₯ minus π₯ one. We take our value for π, π¦ two
minus π¦ one over π₯ two minus π₯ one, and plug that in. Which gives us π¦ minus π¦ one
equals π¦ two minus π¦ one over π₯ two minus π₯ one times π₯ minus π₯ one. And this completes our formula.

For any linear equation, if you
know two points that fall on that line, you can plug those values in for π₯ one, π¦
one; π₯ two, π¦ two and solve for the equation.