Chapter 1: Problem 20
Prove by means of slopes that the three points \(A(3,1), B(6,0)\), and \(C(4,4)\) are the vertices of a right triangle, and find the area of the triangle.
Short Answer
Expert verified
The points form a right triangle, and the area of the triangle is 6 square units.
Step by step solution
01
- Calculate the slope of AB
To find the slope of the line passing through points A(3,1) and B(6,0), use the slope formula: \[\text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 1}{6 - 3} = \frac{-1}{3}\]
02
- Calculate the slope of BC
Next, find the slope of the line passing through points B(6,0) and C(4,4): \[\text{slope of } BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{4 - 6} = \frac{4}{-2} = -2\]
03
- Calculate the slope of CA
Then, calculate the slope of the line passing through points C(4,4) and A(3,1): \[\text{slope of } CA = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 4}{3 - 4} = \frac{-3}{-1} = 3\]
04
- Check for perpendicular slopes
For the triangle to be a right triangle, one pair of slopes should be negative reciprocals, indicating perpendicular lines. Check for the pair \(\frac{-1}{3}\) and 3: \(\frac{-1}{3} \times 3 = -1\). Since their product is -1, they are perpendicular.
05
- Calculate the area of the right triangle
Use the formula for the area of a triangle: \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\]. Here, the base is the vertical height from point C to line AB and the horizontal distance from point A to B: Base (horizontal distance AB) = 3 units, Height (vertical distance from C to horizontal line through B) = 4 units. \[\text{Area} = \frac{1}{2} \times 3 \times 4 = 6\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slopes
The slope of a line indicates its steepness and direction. It is calculated as:
First, let’s calculate the slope of line AB:
\( \text{slope of } AB = \frac{0 - 1}{6 - 3} = \frac{-1}{3} \)
Next, let’s find the slope of line BC:
\( \text{slope of } BC = \frac{4 - 0}{4 - 6} = \frac{4}{-2} = -2 \)
Finally, compute the slope of line CA:
\( \text{slope of } CA = \frac{1 - 4}{3 - 4} = \frac{-3}{-1} = 3 \)
For a right triangle, two lines must be perpendicular. Perpendicular lines have slopes that are negative reciprocals. For lines AB and CA, the slopes are \(\frac{-1}{3} \) and 3, and their product is -1, indicating they are perpendicular.
- slope = \frac{y_2 - y_1}{x_2 - x_1}
First, let’s calculate the slope of line AB:
\( \text{slope of } AB = \frac{0 - 1}{6 - 3} = \frac{-1}{3} \)
Next, let’s find the slope of line BC:
\( \text{slope of } BC = \frac{4 - 0}{4 - 6} = \frac{4}{-2} = -2 \)
Finally, compute the slope of line CA:
\( \text{slope of } CA = \frac{1 - 4}{3 - 4} = \frac{-3}{-1} = 3 \)
For a right triangle, two lines must be perpendicular. Perpendicular lines have slopes that are negative reciprocals. For lines AB and CA, the slopes are \(\frac{-1}{3} \) and 3, and their product is -1, indicating they are perpendicular.
Coordinate Geometry
Coordinate geometry helps to determine points' positions and relationships on a plane using coordinates, like (x, y).
It can also prove geometric properties algebraically. In this problem, we utilize point coordinates to prove a triangle is right-angled.
The three points A (3,1), B (6,0), and C (4,4) form a triangle. By finding the slopes of the lines between these points, we prove perpendicularly and right angles.
To summarize:
It can also prove geometric properties algebraically. In this problem, we utilize point coordinates to prove a triangle is right-angled.
The three points A (3,1), B (6,0), and C (4,4) form a triangle. By finding the slopes of the lines between these points, we prove perpendicularly and right angles.
To summarize:
- Analyze both x-coordinates and y-coordinates changes to find slopes.
- Confirm perpendicular lines by checking if slopes are negative reciprocals.
Area Calculation
The area of a triangle can be determined using the formula:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
For our triangle, we used points A, B, and C.
We identified AB as the horizontal base and the vertical distance from C to the line through A and B as the height.
Measurements:
\( \frac{1}{2} \times 3 \times 4 = 6 \text{ square units} \)
This area calculation shows the size of the right triangle directly based on the coordinates.
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
For our triangle, we used points A, B, and C.
We identified AB as the horizontal base and the vertical distance from C to the line through A and B as the height.
Measurements:
- Base (distance AB ) = 3 units, because from (3,1) to (6,0), the horizontal change is 3 units.
- Height (vertical distance from C to horizontal line through B at y = 0 ) = 4 units, measured from (4,4) to y=0.
\( \frac{1}{2} \times 3 \times 4 = 6 \text{ square units} \)
This area calculation shows the size of the right triangle directly based on the coordinates.