4.3 Q-5

Question Statement

Using the slope method, prove that the triangle with vertices $A(6,1)$, $B(2,7)$, and $C(-6,-7)$ is a right triangle.

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Background and Explanation

To solve this, we need to use the concept of the slope of a line. The slope of a line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

For a triangle to be a right triangle, the product of the slopes of two perpendicular sides must be $-1$. If we find two sides of the triangle with slopes whose product equals $-1$, we can conclude that the triangle is a right triangle.

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Solution

Step 1: Calculate the slope of line $\overline{AB}$

The points $A(6,1)$ and $B(2,7)$ are given. Using the slope formula:

$$m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 1}{2 - 6} = \frac{6}{-4} = \frac{-3}{2}$$

So, the slope of line $\overline{AB}$ is $m_1 = \frac{-3}{2}$.

Step 2: Calculate the slope of line $\overline{BC}$

The points $B(2,7)$ and $C(-6,-7)$ are given. Using the slope formula:

$$m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 7}{-6 - 2} = \frac{-14}{-8} = \frac{7}{4}$$

So, the slope of line $\overline{BC}$ is $m_2 = \frac{7}{4}$.

Step 3: Calculate the slope of line $\overline{AC}$

The points $A(6,1)$ and $C(-6,-7)$ are given. Using the slope formula:

$$m_3 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 1}{-6 - 6} = \frac{-8}{-12} = \frac{2}{3}$$

So, the slope of line $\overline{AC}$ is $m_3 = \frac{2}{3}$.

Step 4: Check if the product of the slopes of two lines equals $-1$

Now, we need to check if any two sides of the triangle are perpendicular. To do this, we multiply the slopes of two lines and check if their product is $-1$.

Check the product of $m_1$ and $m_2$:

$$m_1 \times m_2 = \frac{-3}{2} \times \frac{7}{4} = \frac{-21}{8}$$

This is not equal to $-1$, so these lines are not perpendicular.

Next, check the product of $m_1$ and $m_3$:

$$m_1 \times m_3 = \frac{-3}{2} \times \frac{2}{3} = -1$$

Since the product of $m_1$ and $m_3$ equals $-1$, the lines $\overline{AB}$ and $\overline{AC}$ are perpendicular.

Thus, the triangle $\triangle ABC$ has a right angle at $A$, making it a right triangle.

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Key Formulas or Methods Used

• Slope of a line through two points $(x_1, y_1)$ and $(x_2, y_2)$:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

• For perpendicular lines, the product of their slopes is $-1$, i.e., $m_1 \cdot m_2 = -1$.

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Summary of Steps

1. Calculate the slope of line $\overline{AB}$: $m_1 = \frac{-3}{2}$.
2. Calculate the slope of line $\overline{BC}$: $m_2 = \frac{7}{4}$.
3. Calculate the slope of line $\overline{AC}$: $m_3 = \frac{2}{3}$.
4. Check the product of slopes $m_1 \times m_3$. If the result is $-1$, the lines are perpendicular, and the triangle is a right triangle.
5. Conclude that $\triangle ABC$ is a right triangle with a right angle at $A$.