To determine the relationship among the points A(1, -3), B(3, 4), and C(4, 7), we will analyze the slopes and distances between these points.
### Step 1: Calculate the slopes of the line segments AB, BC, and AC.
1. **Slope of AB**:
- Formula: \( \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \)
- Points: A(1, -3) and B(3, 4)
- Calculation:
\[
\text{slope of AB} = \frac{4 - (-3)}{3 - 1} = \frac{4 + 3}{3 - 1} = \frac{7}{2}
\]
2. **Slope of BC**:
- Points: B(3, 4) and C(4, 7)
- Calculation:
\[
\text{slope of BC} = \frac{7 - 4}{4 - 3} = \frac{3}{1} = 3
\]
3. **Slope of AC**:
- Points: A(1, -3) and C(4, 7)
- Calculation:
\[
\text{slope of AC} = \frac{7 - (-3)}{4 - 1} = \frac{7 + 3}{4 - 1} = \frac{10}{3}
\]
### Step 2: Analyze the slopes.
- The slopes calculated are:
- Slope of AB: \( \frac{7}{2} \)
- Slope of BC: \( 3 \)
- Slope of AC: \( \frac{10}{3} \)
Since all three slopes are different, the points A, B, and C do not lie on the same straight line.
### Step 3: Calculate the distances between the points.
1. **Distance AB**:
- Formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
- Calculation:
\[
d_{AB} = \sqrt{(3 - 1)^2 + (4 - (-3))^2} = \sqrt{(2)^2 + (7)^2} = \sqrt{4 + 49} = \sqrt{53}
\]
2. **Distance BC**:
- Calculation:
\[
d_{BC} = \sqrt{(4 - 3)^2 + (7 - 4)^2} = \sqrt{(1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]
3. **Distance AC**:
- Calculation:
\[
d_{AC} = \sqrt{(4 - 1)^2 + (7 - (-3))^2} = \sqrt{(3)^2 + (10)^2} = \sqrt{9 + 100} = \sqrt{109}
\]
### Step 4: Analyze the distances.
- The distances calculated are:
- Distance AB: \( \sqrt{53} \)
- Distance BC: \( \sqrt{10} \)
- Distance AC: \( \sqrt{109} \)
Since all three distances are different, the triangle formed by points A, B, and C is not equilateral.
### Step 5: Check for right angle triangle.
To check if it is a right triangle, we can use the slopes. For a right triangle, the product of the slopes of two sides should equal -1.
- Product of slopes AB and BC:
\[
\frac{7}{2} \times 3 = \frac{21}{2} \quad \text{(not equal to -1)}
\]
- Product of slopes AB and AC:
\[
\frac{7}{2} \times \frac{10}{3} = \frac{70}{6} \quad \text{(not equal to -1)}
\]
- Product of slopes BC and AC:
\[
3 \times \frac{10}{3} = 10 \quad \text{(not equal to -1)}
\]
Since none of the products equal -1, the triangle is not a right triangle.
### Conclusion:
The points A(1, -3), B(3, 4), and C(4, 7) form a triangle, but it is neither equilateral nor a right triangle.
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