Question Video: Finding an Unknown Value By Using Determinants to Calculate the Area of a Triangle | Nagwa Question Video: Finding an Unknown Value By Using Determinants to Calculate the Area of a Triangle | Nagwa

Question Video: Finding an Unknown Value By Using Determinants to Calculate the Area of a Triangle Mathematics • First Year of Secondary School

Fill in the blank: If the area of a triangle whose vertices are (ℎ, 0), (6, 0), and (0, 3) is 9 square units, then ℎ = _.

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Video Transcript

Fill in the blank. If the area of a triangle whose vertices are ℎ, zero; six, zero; and zero, three is nine square units, then ℎ equals what. Is it option (A) zero or negative 12? Option (B) zero or 12. Option (C) negative six or six. Or is it option (D) 12 or negative 12?

In this question, we’re given the coordinates of the vertices of a triangle which contains a single unknown coordinate and the area of a triangle. We want to use this to find the possible values of the unknown. We could do this by using any method for finding the area of a triangle. However, since we are given the vertices, we will do this by using determinants.

We recall that the area of a triangle can be calculated by using the formula one-half times the absolute value of the determinant of the three-by-three matrix 𝑥 sub one, 𝑦 sub one, one, 𝑥 sub two, 𝑦 sub two, one, 𝑥 sub three, 𝑦 sub three, one, where 𝑥 sub 𝑖 and 𝑦 sub 𝑖 are the coordinates of each vertex of the triangle for each value of 𝑖.

We can substitute the coordinates of the triangle and the area being equal to nine into this formula to obtain the following equation. We can solve for ℎ by expanding and isolating ℎ on one side of the equation. We can evaluate the determinant in many ways. The easiest is to note that the second column contains two zeros. So expanding over this column will only have a single nonzero term.

We obtain one-half the absolute value of negative three times ℎ minus six. We can simplify this further by taking the factor of negative three out of the absolute value. This becomes a factor of positive three to give the following. This is equal to the area of the triangle, which is nine. We can solve for ℎ by rearranging. We divide both sides of the equation by three over two. This yields that six is equal to the absolute value of ℎ minus six.

We can then solve this absolute value equation by constructing two equations. Either six equals ℎ minus six or six equals negative one times ℎ minus six. Solving each equation gives us that ℎ is equal to 12 or ℎ equals zero. We can then see that this is given by option (B).

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