Answer:
[tex]\tt y = \dfrac{3}{5}x - \dfrac{22}{5}[/tex]
Step-by-step explanation:
To find the equation of the line that passes through the points [tex]\tt (-1, -5)[/tex] and [tex]\tt (4, -2)[/tex], we can use the point-slope form of a linear equation:
[tex]\tt y - y_1 = m(x - x_1)[/tex]
where [tex]\tt (x_1, y_1)[/tex] is a point on the line, and [tex]\tt m[/tex] is the slope.
First, calculate the slope ([tex]\tt m[/tex]) using the formula:
[tex]\tt m = \dfrac{y_2 - y_1}{x_2 - x_1}[/tex]
Let [tex]\tt (x_1, y_1) = (-1, -5)[/tex] and [tex]\tt (x_2, y_2) = (4, -2)[/tex]:
[tex]\tt m = \dfrac{-2 - (-5)}{4 - (-1)}[/tex]
[tex]\tt m = \dfrac{3}{5}[/tex]
Now that we have the slope ([tex]\tt m[/tex]), choose one of the points (let's use [tex]\tt (-1, -5)[/tex]) and substitute the values into the point-slope form:
[tex]\tt y - (-5) = \dfrac{3}{5}(x - (-1))[/tex]
Simplify:
[tex]\tt y + 5 = \dfrac{3}{5}(x + 1)[/tex]
Now, we can rearrange this equation to put it in a more familiar form if needed, such as slope-intercept form ([tex]\tt y = mx + b[/tex]):
[tex]\tt y + 5 = \dfrac{3}{5}x + \dfrac{3}{5}[/tex]
Subtract 5 from both sides:
[tex]\tt y = \dfrac{3}{5}x + \dfrac{3}{5} - 5[/tex]
Combine the constants:
[tex]\tt y = \dfrac{3}{5}x - \dfrac{22}{5}[/tex]
So, the equation of the line passing through [tex]\tt (-1, -5)[/tex] and [tex]\tt (4, -2)[/tex] is [tex]\tt y = \dfrac{3}{5}x - \dfrac{22}{5}[/tex].
