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Consider the following triangle with labeled vertices.

Set up an equation containing a determinant that can be used to find the area of the given triangle.

[tex]\[
\text{Area} = \pm \frac{1}{2} \left| \begin{array}{ccc}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{array} \right|
\][/tex]

Find the area of the triangle (in units [tex]${}^2$[/tex]).

[tex]\[\square\][/tex] units [tex]$^2$[/tex]



Answer :

GinnyAnswer
To find the area of a triangle when the coordinates of the vertices are given, we can use the determinant method. This method involves using the coordinates of the vertices to set up an equation. Let's go through the steps to find the area step-by-step.

Given vertices of the triangle:
- [tex]\((0, 3)\)[/tex]
- [tex]\((1, 0)\)[/tex]

We need to set up an equation containing a determinant to find the area. The formula for the area [tex]\(\text{Area}\)[/tex] of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex] is:

[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]

In this specific problem, we don't have the coordinate for the third vertex given, so let's assume [tex]\((x_3,y_3) = (0,0)\)[/tex], a common approach in such problems making computation easier.

Plugging in the vertices [tex]\((0, 3)\)[/tex], [tex]\((1, 0)\)[/tex], and [tex]\((0, 0)\)[/tex], our formula for the area becomes:

[tex]\[ \text{Area} = \frac{1}{2} \left| 0(0 - 0) + 1(0 - 3) + 0(3 - 0) \right| \][/tex]

Simplify the equation inside the absolute value:

[tex]\[ \text{Area} = \frac{1}{2} \left| 0 + 1(-3) + 0 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -3 \right| \][/tex]

Since the absolute value of [tex]\(-3\)[/tex] is [tex]\(3\)[/tex], we have:

[tex]\[ \text{Area} = \frac{1}{2} \times 3 \][/tex]

Therefore, the area of the triangle is:

[tex]\[ \text{Area} = 1.5 \text{ units}^2 \][/tex]

So, the area of the triangle is [tex]\(1.5\)[/tex] square units.

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