To find the area of the triangle formed by the points \( A(-3, 5) \), \( B(3, 7) \), and \( C(2, 3) \) using the method described (calculating the area of the circumscribing rectangle and then subtracting the areas of the right triangles), follow these steps:
Step 1: Find the coordinates of the rectangle
To create a rectangle that circumscribes the triangle defined by points \( A \), \( B \), and \( C \), we need to determine the minimum and maximum x and y coordinates.
- Minimum x-coordinate: \( -3 \) (from point A)
- Maximum x-coordinate: \( 3 \) (from point B)
- Minimum y-coordinate: \( 3 \) (from point C)
- Maximum y-coordinate: \( 7 \) (from point B)
The corners of the rectangle are at:
- Bottom left: \( (-3, 3) \)
- Bottom right: \( (3, 3) \)
- Top left: \( (-3, 7) \)
- Top right: \( (3, 7) \)
Step 2: Calculate the area of the rectangle
The width and height of the rectangle can be calculated as follows:
- Width = \( x_{\text{max}} - x_{\text{min}} = 3 - (-3) = 3 + 3 = 6 \)
- Height = \( y_{\text{max}} - y_{\text{min}} = 7 - 3 = 4 \)
Now we can find the area of the rectangle: \[ \text{Area}_{\text{rectangle}} = \text{Width} \times \text{Height} = 6 \times 4 = 24 \text{ square units} \]
Step 3: Calculate the area of the triangle
We can use the formula for the area of a triangle given by the coordinates of its vertices: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] where \( (x_1, y_1) = (-3, 5) \), \( (x_2, y_2) = (3, 7) \), \( (x_3, y_3) = (2, 3) \).
Substituting the values into the formula: \[ \text{Area} = \frac{1}{2} \left| (-3)(7 - 3) + (3)(3 - 5) + (2)(5 - 7) \right| \] \[ = \frac{1}{2} \left| (-3)(4) + (3)(-2) + (2)(-2) \right| \] \[ = \frac{1}{2} \left| -12 - 6 - 4 \right| \] \[ = \frac{1}{2} \left| -22 \right| = \frac{22}{2} = 11 \text{ square units} \]
Step 4: Calculate the area of the right triangles
The area of the right triangles created in the rectangle can be calculated as follows:
-
Left triangle (bottom left corner):
- Base from \( (-3, 3) \) to \( (-3, 5) \) = 2
- Height from \( (-3, 3) \) to \( (-3, 3) \) to \( (2, 3) \) = 5
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \times 5 = 5 \)
-
Top triangle (top left corner):
- Base from \( (-3, 7) \) to \( (-3, 5) \) = 2
- Height = from \( (-3, 7) \) to \( (3, 7) \) = 3
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \times 3 = 3 \)
-
Right triangle (top right corner):
- Base from \( (3, 7) \) to ( (3, 5) = 2
- Height = from \( (3, 7) \) to ( (3, 3) = 4
- Area = \( \frac{1}{2} \times 2 \times 4 = 4 \)
Step 5: Subtract the areas of the right triangles from the area of the rectangle
\[ \text{Area}{\text{triangles}} = \text{Area}{\text{rectangle}} - \text{Area}_{\text{Triangle}} \] Subtracting from the total area: \[ = 24 - 11 = 13 \text{ total} \]
So the area of the triangle \( ABC \) is found to be 11 square units.
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