Respuesta :
Answer:
[tex]y=-\frac{5}{3}x-8\frac{1}{3}[/tex]Explanation:
A line equidistant from the points (-11, -8) and (-1, -2) will be a line perpendicular to the given line.
Step 1: Find the midpoint of the line
[tex]\begin{gathered} M(x,y)=\mleft(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\mright) \\ =\mleft(\dfrac{-11+(-1)}{2},\dfrac{-8+(-2)}{2}\mright) \\ =(-\frac{12}{2},-\frac{10}{2}) \\ =(-6,-5) \end{gathered}[/tex]Step 2: Find the slope of the line:
[tex]\begin{gathered} \text{Slope},m=\frac{Change\text{ in y-axis}}{Change\text{ in x-axis}} \\ =\frac{-2-(-8)_{}}{-1-(-11)} \\ =\frac{-2+8}{-1+11} \\ =\frac{6}{10} \\ m=\frac{3}{5} \end{gathered}[/tex]• Two lines are perpendicular if the product of their slopes is -1.
Since the new line is perpendicular, its slope will be:
[tex]n=-\frac{5}{3}[/tex]Step 3: Find the equation of the line
We are then to find the equation of a line passing through (-6,-5) with a slope of -5/3.
Using the point-slope form:
[tex]y-y_1=m(x-x_1)[/tex]We have:
[tex]\begin{gathered} y-(-5)=-\frac{5}{3}(x-(-2)) \\ y+5=-\frac{5}{3}(x+2) \\ y=-\frac{5}{3}(x+2)-5 \\ y=-\frac{5}{3}x-\frac{10}{3}-5 \\ y=-\frac{5}{3}x-8\frac{1}{3} \end{gathered}[/tex]The equation of the line equidistant from the points (given in slope-intercept form) is:
[tex]y=-\frac{5}{3}x-8\frac{1}{3}[/tex]