# Question Video: Finding Unknown Vertices Points of a Triangle Using Its Area Mathematics

If the area of a triangle whose vertices are (ℎ, 0), (6, 0), and (0, 3) is 9 square units, then ℎ =＿. [A] 0 or 12 [B] 0 or −12 [C] −6 or 6 [D] −12 or 12

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### 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.