Problem 10 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 7-10: Use slopes to show that the points are vertices of the specified polygon. $$ A(1,4), B(6,-4), C(-15,-6) ; \text { right triangle } $$

Step-by-Step Solution

Answer

The points form a right triangle with a right angle at \( A \).

Step 1: Calculate Slope of AB

To find the slope of the line between two points, use the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). For points \( A(1, 4) \) and \( B(6, -4) \), the slope \( m_{AB} \) is \[ m_{AB} = \frac{-4 - 4}{6 - 1} = \frac{-8}{5} = -\frac{8}{5} \].

Step 2: Calculate Slope of BC

Use the slope formula again for points \( B(6, -4) \) and \( C(-15, -6) \). The slope \( m_{BC} \) is \[ m_{BC} = \frac{-6 + 4}{-15 - 6} = \frac{-2}{-21} = \frac{2}{21} \].

Step 3: Calculate Slope of AC

Calculate the slope for points \( A(1, 4) \) and \( C(-15, -6) \) using the slope formula. The slope \( m_{AC} \) is \[ m_{AC} = \frac{-6 - 4}{-15 - 1} = \frac{-10}{-16} = \frac{5}{8} \].

Step 4: Verify Right Triangle Condition

A right triangle requires two sides to be perpendicular. Perpendicular lines have slopes that are negative reciprocals. Check if any two slopes are negative reciprocals: \( m_{AB} = -\frac{8}{5} \) and \( m_{AC} = \frac{5}{8} \) are negative reciprocals since \(-\frac{8}{5} \times \frac{5}{8} = -1 \). Thus, \( \angle BAC \) is a right angle.

Key Concepts

Slope Calculation

Calculating the slope between two points is fundamental when analyzing the orientation of a line on a graph. It determines how steep the line is and whether it slants upwards or downwards as you move from left to right. To find the slope, you can use the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), where \((x_1, y_1)\) and \((x_2, y_2)\) are coordinates of any two points on the line.

• In this scenario, for line segment AB connecting points A(1,4) and B(6,-4), the slope is calculated as \( m_{AB} = \frac{-4 - 4}{6 - 1} = -\frac{8}{5} \), indicating a downward slant.
• The slope for line segment BC, between B(6,-4) and C(-15,-6), is \( m_{BC} = \frac{-6 + 4}{-15 - 6} = \frac{2}{21} \), a very gentle upward slant.
• Lastly, for line AC connecting A(1,4) and C(-15,-6), the slope is \( m_{AC} = \frac{-6 - 4}{-15 - 1} = \frac{5}{8} \), showing an upward slant as well.

Understanding slope helps determine perpendicularity, an essential aspect of identifying right angles.

Perpendicular Lines

Perpendicular lines are crucial in geometry, especially when determining shapes like right triangles. Two lines or line segments are perpendicular when they intersect at a right angle, forming a 90-degree angle. In terms of their slopes, perpendicular lines have slopes that are negative reciprocals of each other.

• This means if one line has a slope \( m \), the perpendicular line will have a slope of \(-\frac{1}{m} \).
• A negative reciprocal involves flipping the slope fraction and changing its sign. For instance, \(-\frac{8}{5} \) and \( \frac{5}{8} \) are negative reciprocals, crucial to proving triangles as right triangles.

By ensuring two slopes are negative reciprocals, you confirm that the lines are perpendicular, resulting in one right angle.

Negative Reciprocals

Negative reciprocals are a unique relationship between two numbers, often used in geometry to demonstrate perpendicularity of lines. Two numbers are negative reciprocals if their product equals \(-1\). This characteristic guarantees that lines with these slopes will intersect at right angles.

• For example, in our exercise, with slopes \( m_{AB} = -\frac{8}{5} \) and \( m_{AC} = \frac{5}{8} \), their product is \(-1\) \(( -\frac{8}{5} \times \frac{5}{8} = -1)\).
• This verifies that lines AB and AC are perpendicular, forming the desired right triangle, with a right angle at the vertex A.

Recognizing negative reciprocals allows geometrical proofs to establish right angles and verify the nature of right triangles.