About Area of a Triangle Using Determinants:
In some cases, it is faster to find the area of a triangle given three vertices using determinants. This is done by labeling all the points as (x1, y1), (x2, y2), and (x3, y3). We then use this to set up a matrix where each row contains a given x as the first entry, a given y as the second entry, and a 1 as the final entry. The determinant is found and then multiplied by (+/-) 1/2. This will give us the area for our given triangle in square units.
Test Objectives
- Demonstrate the ability to find the determinant of a matrix
- Demonstrate the ability to find the area of a triangle
#1:
Instructions: find the area.
$$a)\hspace{.2em}(3,-1), (2,5), (9,-1)$$
$$b)\hspace{.2em}(6,2), (3,5), (-8, -2)$$
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#2:
Instructions: find the area.
$$a)\hspace{.2em}(-1,-2), (-3,-7), (6,9)$$
$$b)\hspace{.2em}(-1, -10), (4,5), (8,17)$$
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#3:
Instructions: find the area.
$$a)\hspace{.2em}(-2,-4), (3, -1), (9,9)$$
$$b)\hspace{.2em}(6, -7), (10, 11), (20, -40)$$
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#4:
Instructions: find the area.
$$a)\hspace{.2em}(-13, -15), (7,0), (-1,-1)$$
$$b)\hspace{.2em}(-2,-1), (3,5), (9,17)$$
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#5:
Instructions: find the area.
$$a)\hspace{.2em}(-5,-7), (-2, -9), (12, 20)$$
$$b)\hspace{.2em}(-10,-15), (-2, -20), (4, 7)$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}18\hspace{.2em}square \hspace{.2em}units$$
$$b)\hspace{.2em}27\hspace{.2em}square \hspace{.2em}units$$
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#2:
Solutions:
$$a)\hspace{.2em}\frac{13}{2}\hspace{.2em}square \hspace{.2em}units$$
$$b)\hspace{.2em}Collinear$$
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#3:
Solutions:
$$a)\hspace{.2em}16\hspace{.2em}square \hspace{.2em}units$$
$$b)\hspace{.2em}192\hspace{.2em}square \hspace{.2em}units$$
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#4:
Solutions:
$$a)\hspace{.2em}50\hspace{.2em}square \hspace{.2em}units$$
$$b)\hspace{.2em}12\hspace{.2em}square \hspace{.2em}units$$
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#5:
Solutions:
$$a)\hspace{.2em}\frac{115}{2}\hspace{.2em}square \hspace{.2em}units$$
$$b)\hspace{.2em}123\hspace{.2em}square \hspace{.2em}units$$
