### Video Transcript

Find the equation of the straight
line that passes through points one, one and three, four. Give your answer in standard
form.

Equations for straight lines are
usually written in the format π¦ equals ππ₯ plus π, where π is the slope of that
line and π equals the π¦-intercept. That means weβll need both the
slope and the π¦-intercept to write this equation. The formula for the slope between
two points is π¦ two minus π¦ one over π₯ two minus π₯ one. The changes in π¦ over the changes
in π₯. Letβs start there.

We can label our two points: our
first point π₯ one, π¦ one and our second point π₯ two, π¦ two. We plug in the information weβre
given. π¦ two minus π¦ one becomes four
minus one. And π₯ two minus π₯ one becomes
three minus one. Four minus one is three. Three minus one is two. Our slope is three over two. Letβs use slope intercept form to
find what the π¦-intercept is.

π¦ equals three over two π₯ plus
π. Weβre going to take one of our two
points and plug them in for π₯ and π¦. It doesnβt matter which of the two
points we use as long as weβre consistent. We have to use the π₯- and
π¦-coordinate from the same point. If we use π₯ two, π¦ two, four is
equal to three-halves times three plus π. Three-halves times three equals
nine-halves. Bring down the plus π. Bring down the four.

Weβre trying to solve for π which
means we need to move the nine-halves to the other side of the equation. We subtract nine-halves on both
sides. Four minus nine-halves could be
written as eight-halves minus nine-halves. Eight-halves minus nine-halves is
negative one-half, equals π.

Back to the slope intercept form,
we plug in three-halves. And the π¦-intercept, negative
one-half, is plugged in for π. We have to be careful here because
π¦-intercept form is not standard form of a linear equation. A linear equation written in
standard form follows the pattern ππ₯ plus ππ¦ equals π. So now itβs our job to take the
slope intercept form and turn it into standard form to get the constant value by
itself.

I can do that by subtracting
three-halves π₯ from both sides of the equation. On the right, they cancel out
leaving us with negative one-half equals π¦ minus three-halves π₯. Weβll just bring it up, so we have
a little bit more room. Something else we need to know
about π, π, and π is that these values need to be integers. They shouldnβt be fractions. Both our π- and π-values are
given as fractions. But if we multiply both sides of
our equation by two, we can get rid of these fractions.

π¦ times two equals two π¦. Three-halves times two equals
three. Negative one-half times two equals
negative one. The next thing we notice about our
standard form is that weβre adding ππ₯ plus ππ¦. In our equation, we are subtracting
π¦ from π₯. We want π₯ to be leading. And we want it to be positive. We want that π-value to be
positive. So we multiply both sides of our
equation by negative one. Negative one times two π¦ equals
negative two π¦. Negative one times negative three
π₯ equals positive three π₯. Negative one times negative one
equals one. Our last step here will be to
switch our π₯ and π¦.

Now, we have three π₯ minus two π¦
equals one. Standard form of a straight line
that passes through one, one and three, four is three π₯ minus two π¦ equals
one. Remember, what weβre looking for
here is a positive π₯. And π, π, and π must be
integers. They cannot be fractions.