Find the slope and distance between two points in a coordinate plane. The slope helps determine the steepness of a line, while distance shows how far apart the points are. Enter the coordinates of each point, and this tool will calculate the slope and distance accurately, making it ideal for geometry, engineering, and cartography applications.
Problem: Find the slope and distance between the points \( A(1, 2) \) and \( B(4, 6) \).
Solution: The slope \( m \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 2}{4 - 1} = \frac{4}{3} \]
The distance \( d \) is calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Problem: Find the slope and distance between the points \( C(3, -1) \) and \( D(7, 4) \).
Solution: The slope \( m \) is:
\[ m = \frac{4 - (-1)}{7 - 3} = \frac{5}{4} \]
The distance \( d \) is:
\[ d = \sqrt{(7 - 3)^2 + (4 - (-1))^2} = \sqrt{16 + 25} = \sqrt{41} \]
Problem: Find the slope and distance between the points \( P(0, 0) \) and \( Q(6, 8) \).
Solution: The slope \( m \) is:
\[ m = \frac{8 - 0}{6 - 0} = \frac{8}{6} = \frac{4}{3} \]
The distance \( d \) is:
\[ d = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]
Problem: Find the slope and distance between the points \( A(-2, 3) \) and \( B(4, -1) \).
Solution: The slope \( m \) is:
\[ m = \frac{-1 - 3}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3} \]
The distance \( d \) is:
\[ d = \sqrt{(4 - (-2))^2 + (-1 - 3)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \]
Problem: Find the slope and distance between the points \( E(1, -2) \) and \( F(5, 3) \).
Solution: The slope \( m \) is:
\[ m = \frac{3 - (-2)}{5 - 1} = \frac{5}{4} \]
The distance \( d \) is:
\[ d = \sqrt{(5 - 1)^2 + (3 - (-2))^2} = \sqrt{16 + 25} = \sqrt{41} \]
Problem: Find the slope and distance between the points \( G(3, 4) \) and \( H(6, 10) \).
Solution: The slope \( m \) is:
\[ m = \frac{10 - 4}{6 - 3} = \frac{6}{3} = 2 \]
The distance \( d \) is:
\[ d = \sqrt{(6 - 3)^2 + (10 - 4)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \]
Problem: Find the slope and distance between the points \( I(-1, 4) \) and \( J(4, -2) \).
Solution: The slope \( m \) is:
\[ m = \frac{-2 - 4}{4 - (-1)} = \frac{-6}{5} \]
The distance \( d \) is:
\[ d = \sqrt{(4 - (-1))^2 + (-2 - 4)^2} = \sqrt{25 + 36} = \sqrt{61} \]