Answer:
y = - x - 2
Explanation:
The equation of a line perpendicular to PQ in point-slope form is given as:
[tex]y-y_1=-\frac{1}{m}(x-x_1)[/tex]
where:
• m is the ,slope, of the line
,
• (x₁, y₁) is ,any point ,on the line
Get the required point
The point (x₁, y₁) will be the midpoint of PQ with endpoints P(-5, -5) and Q(3, 3) as shown using the midpoint formula.
[tex]\begin{gathered} (x_1,y_1)=(\frac{-5+3}{2},\frac{-5+3}{2}) \\ (x_1,y_1)=(\frac{-2}{2},\frac{-2}{2}) \\ (x_1,y_1)=(-1,-1) \end{gathered}[/tex]
Get the slope of the line passing through the point P(-5, -5) and Q(3, 3) using the slope formula as shown:
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{3-(-5)}{3-(-5)} \\ m=\frac{3+5}{3+5} \\ m=\frac{8}{8} \\ m=1 \end{gathered}[/tex]
Substitute the point (-1, -1) and the slope of PQ which is 1 into the point-slope form of the equation to have:
[tex]\begin{gathered} y-y_1=-\frac{1}{m}(x-x_0) \\ y-(-1)=-\frac{1}{1}(x-(-1)) \\ y+1=-1(x+1) \end{gathered}[/tex]
Expand the parenthesis:
[tex]\begin{gathered} y+1=-x-1 \\ y=-x-1-1 \\ y=-x-2 \end{gathered}[/tex]
Hence, the equation of the line passing through the midpoint and perpendicular to PQ is y = -x - 2
