Determine if the triangle with vertices X(4, −3), Y(9, 1), and Z(0, 2) is a right triangle. Can

you think of more than one way to prove it is/isn’t a right triangle?

To determine if the triangle with vertices \(X(4, -3)\), \(Y(9, 1)\), and \(Z(0, 2)\) is a right triangle, we can use two different methods:

1. **Using the Distance Formula and Pythagorean Theorem:**
2. **Using the Slope Method to check for Perpendicularity:**

### Method 1: Using the Distance Formula and Pythagorean Theorem

First, we calculate the lengths of all three sides using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

#### Step 1: Calculate \(XY\)
\[ X(4, -3) \quad Y(9, 1) \]
\[ XY = \sqrt{(9 - 4)^2 + (1 - (-3))^2} \]
\[ XY = \sqrt{5^2 + 4^2} \]
\[ XY = \sqrt{25 + 16} \]
\[ XY = \sqrt{41} \]

#### Step 2: Calculate \(YZ\)
\[ Y(9, 1) \quad Z(0, 2) \]
\[ YZ = \sqrt{(0 - 9)^2 + (2 - 1)^2} \]
\[ YZ = \sqrt{(-9)^2 + 1^2} \]
\[ YZ = \sqrt{81 + 1} \]
\[ YZ = \sqrt{82} \]

#### Step 3: Calculate \(ZX\)
\[ Z(0, 2) \quad X(4, -3) \]
\[ ZX = \sqrt{(4 - 0)^2 + (-3 - 2)^2} \]
\[ ZX = \sqrt{4^2 + (-5)^2} \]
\[ ZX = \sqrt{16 + 25} \]
\[ ZX = \sqrt{41} \]

Next, we check if the Pythagorean theorem holds:
\[ a^2 + b^2 = c^2 \]

Let's assume \( XY \) and \( ZX \) are the legs, and \( YZ \) is the hypotenuse:
\[ (\sqrt{41})^2 + (\sqrt{41})^2 = (\sqrt{82})^2 \]
\[ 41 + 41 = 82 \]
\[ 82 = 82 \]

Since the Pythagorean theorem holds, the triangle \( XYZ \) is a right triangle.

### Method 2: Using the Slope Method to Check for Perpendicularity

We can find the slopes of the lines \( XY \), \( YZ \), and \( ZX \) and check if any two are negative reciprocals (which indicates perpendicularity).

#### Step 1: Calculate the slope of \( XY \)
\[ X(4, -3) \quad Y(9, 1) \]
\[ m_{XY} = \frac{1 - (-3)}{9 - 4} \]
\[ m_{XY} = \frac{4}{5} \]

#### Step 2: Calculate the slope of \( YZ \)
\[ Y(9, 1) \quad Z(0, 2) \]
\[ m_{YZ} = \frac{2 - 1}{0 - 9} \]
\[ m_{YZ} = \frac{1}{-9} \]
\[ m_{YZ} = -\frac{1}{9} \]

#### Step 3: Calculate the slope of \( ZX \)
\[ Z(0, 2) \quad X(4, -3) \]
\[ m_{ZX} = \frac{-3 - 2}{4 - 0} \]
\[ m_{ZX} = \frac{-5}{4} \]

To check for perpendicularity:
- \( m_{XY} \cdot m_{YZ} = \frac{4}{5} \cdot -\frac{1}{9} = -\frac{4}{45} \) (not equal to -1)
- \( m_{YZ} \cdot m_{ZX} = -\frac{1}{9} \cdot -\frac{5}{4} = \frac{5}{36} \) (not equal to -1)
- \( m_{ZX} \cdot m_{XY} = -\frac{5}{4} \cdot \frac{4}{5} = -1 \) (equal to -1)

Since the product of the slopes of \( ZX \) and \( XY \) is \(-1\), they are perpendicular, confirming that \( XYZ \) is a right triangle.

### Conclusion

Using both the Pythagorean theorem and the slope method, we have confirmed that the triangle with vertices \(X(4, -3)\), \(Y(9, 1)\), and \(Z(0, 2)\) is indeed a right triangle.

Would you like any further details or have any questions about the solution?

---

Here are 8 related questions for further practice:

1. How do you find the area of a right triangle given its side lengths?
2. How can you determine if a triangle is isosceles using its vertices?
3. What is the equation of the line passing through \(X\) and \(Y\)?
4. How do you find the circumcenter of a triangle given its vertices?
5. What is the equation of the circle circumscribed around triangle \(XYZ\)?
6. How do you find the orthocenter of a triangle?
7. Can you determine if three points form an equilateral triangle? How?
8. How do you find the centroid of a triangle given its vertices?

**Tip:** When checking for right angles using slopes, remember that the slopes of perpendicular lines multiply to \(-1\).

Learn how to determine if triangle XYZ with vertices X(4, -3), Y(9, 1), and Z(0, 2) is a right triangle using the distance formula and Pythagorean theorem. Detailed step-by-step solution and verification using both methods provided. Suitable for high school students studying coordinate geometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation