We are given the following equation of a line
[tex]y=-\frac{2}{3}x-3[/tex]Give the equation of the line parallel to line 1 which passes through (5, -9)
Recall that the standard form of the equation of a line in slope-intercept form is given by
[tex]y=mx+b[/tex]Where m is the slope and b is the y-intercept.
Comparing the standard form with the given equation we see that the slope of line 1 is -2/3
Since we are given that the lines are parallel so the slope of the other line must be equal that is -2/3
So, the equation of the other line becomes
[tex]y=-\frac{2}{3}x+b[/tex]To find the value of b, substitute the point (5, -9) into the above equation and solve for b.
[tex]\begin{gathered} -9=-\frac{2}{3}(5)+b \\ -9=-\frac{10}{3}+b \\ -9+\frac{10}{3}=b \\ -\frac{17}{3}=b \\ b=-\frac{17}{3} \end{gathered}[/tex]So, the equation of the other line is
[tex]y=-\frac{2}{3}x-\frac{17}{3}[/tex]Give the equation of the line perpendicular to line 1 which passes through (5, -9)
Since we are given that the lines are perpendicular so the slope of the other line must be negative reciprocal of the given line.
[tex]m_2=-\frac{1}{m_1}=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}[/tex]So, the slope of the other line is 3/2
The equation of the other line becomes
[tex]y=\frac{3}{2}x+b[/tex]To find the value of b, substitute the point (5, -9) into the above equation and solve for b
[tex]\begin{gathered} -9=\frac{3}{2}(5)+b \\ -9=\frac{15}{2}+b \\ -9-\frac{15}{2}=b \\ -\frac{33}{2}=b \\ b=-\frac{33}{2} \end{gathered}[/tex]So, the equation of the other line is
[tex]y=\frac{3}{2}x-\frac{33}{2}[/tex][tex]\begin{gathered} Parallel\colon\; y=-\frac{2}{3}x-\frac{17}{3} \\ Perpendicular\colon\; y=\frac{3}{2}x-\frac{33}{2} \end{gathered}[/tex]