Line AB

𝐴
𝐵
contains point A(−4, 1)
𝐴
(

4
,

1
)
and point B (−1, 3)
𝐵

(

1
,

3
)

1 answer

To find information about the line segment \(AB\) that passes through the points \(A(-4, 1)\) and \(B(-1, 3)\), we can calculate the slope of the line, the equation of the line, and the length of the line segment.

1. Calculate the Slope

The slope \(m\) of the line connecting points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) can be calculated with the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substituting the coordinates of points \(A\) and \(B\): \[ m = \frac{3 - 1}{-1 - (-4)} = \frac{2}{3} \]

2. Write the Equation of the Line

Using the point-slope form of a line equation \(y - y_1 = m(x - x_1)\): \[ y - 1 = \frac{2}{3}(x + 4) \]

Distributing the slope: \[ y - 1 = \frac{2}{3}x + \frac{8}{3} \]

Now, add 1 (or \(\frac{3}{3}\)) to both sides to isolate \(y\): \[ y = \frac{2}{3}x + \frac{8}{3} + \frac{3}{3} \] \[ y = \frac{2}{3}x + \frac{11}{3} \]

3. Calculate the Length of the Segment

The length \(d\) of the line segment can be found using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates of points \(A\) and \(B\): \[ d = \sqrt{(-1 - (-4))^2 + (3 - 1)^2} \] \[ d = \sqrt{(-1 + 4)^2 + (2)^2} = \sqrt{(3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13} \]

Summary

  • The slope of line segment \(AB\) is \(\frac{2}{3}\).
  • The equation of line segment \(AB\) is \(y = \frac{2}{3}x + \frac{11}{3}\).
  • The length of line segment \(AB\) is \(\sqrt{13}\).