### Video Transcript

If the area of a triangle whose
vertices are ℎ, zero; six, zero; and zero, three is nine square units, then ℎ is
equal to (A) zero or 12, (B) zero or negative 12, (C) negative six or six, (D)
negative 12 or 12.

We’re given that a triangle has an
area of nine square units. And we’re given the three
coordinates of its vertices. We need to use this information to
find the possible values of ℎ. It’s worth pointing out here one
method to do this is to plot all three of these points on a diagram and then just
calculate the possible areas. And this definitely works and can
be used to get the correct answer. However, we’re going to do this by
using determinants.

Recall, we can find the area 𝐴 of
a triangle given its vertices by using the following formula. 𝐴 is equal to one-half multiplied
by the absolute value of the determinant of a three-by-three matrix, where each row
in our matrix is the coordinate pair with an extra value of one. And it doesn’t matter which order
we choose our points. So we’ll just order them from left
to right. So by using the vertices given to
us in the question, we get the area of our triangle is equal to one-half times the
absolute value of the determinant of the following three-by-three matrix.

The next thing we’re going to need
to do is find the value of this determinant. There’s a few different ways of
doing this. We can see that one of the columns
of our matrix has two values of zero. So we’ll find this determinant by
expanding over the second column. So we need to do this by finding
the matrix minors. Remember, our first two values are
zero. So we don’t need to calculate
these. We only need to do the final one
with the coefficient of three. And remember, we multiply this by
negative one to the power of three plus two because it’s in row three, column
two.

Now, remember, we need to find the
determinant of the matrix minor we get by removing the third row and second
column. We can see this is the two-by-two
matrix ℎ, one, six, one. So we’ve shown the area of our
triangle is equal to one-half times the absolute value of negative one to the power
of three plus two times three multiplied by the determinant of the two-by-two matrix
ℎ, one, six, one.

Now, all we need to do is evaluate
this expression. First, negative one to the power of
three plus two is just negative one. Next, if we evaluate the
determinant of our two-by-two matrix, we see it’s equal to ℎ minus six. So the area of our triangle is
one-half times the absolute value of negative three times ℎ minus six. And we can simplify this
further. We can take the negative three
outside of our absolute value. And remember, this means it becomes
positive three. So we get three over two times the
absolute value of ℎ minus six.

Remember, in the question, we’re
told the area of our triangle is nine. So, in fact, this expression must
be equal to nine. So we just need to solve an
equation involving the absolute value symbol. We’ll start by dividing both sides
of our equation through by three over two. Nine divided by three over two is
six. We get six is equal to the absolute
value of ℎ minus six.

Remember, to solve equations
involving the absolute value symbol, we need to consider the positive and negative
solution for our absolute value. And by taking the positive and
negative, we get two equations we need to solve. Six is equal to ℎ minus six, and
six is equal to negative one times ℎ minus six. And these are both linear
equations, and we can solve them. We get ℎ is equal to 12 or ℎ is
equal to zero. And this is option (A). Therefore, we were able to show
that ℎ is equal to zero or 12.

And it’s also worth pointing out we
could check our answer by substituting our values of ℎ into this coordinate pair,
plotting these points, and then calculating the area of the triangle and seeing that
we get nine. And this can be a good way to check
that we got the correct answer.