Answer:
[tex]y = -\frac{3}{2}x+15[/tex]
Step-by-step explanation:
Given:
Given point P(6, 6)
The equation of the line.
[tex]y = \frac{2}{3}x[/tex]
We need to find the equation of the line perpendicular to the given line that contains P
Solution:
The equation of the line.
[tex]y = \frac{2}{3}x[/tex]
Now, we compare the given equation by standard form [tex]y = mx +c[/tex]
So, slope of the line [tex]m_{1} = \frac{2}{3}[/tex], and
y-intercept [tex]c=0[/tex]
We know that the slope of the perpendicular line [tex]m_{1}\times m_{2} = -1[/tex]
[tex]m_{2}=-\frac{1}{m_{1}}[/tex]
[tex]m_{2}=-\frac{1}{\frac{2}{3} }[/tex]
[tex]m_{2}=-\frac{3}{2}[/tex]
So, the slope of the perpendicular line [tex]m_{2}=-\frac{3}{2}[/tex]
From the above statement, line passes through the point P(6, 6).
Using slope intercept formula to know y-intercept.
[tex]y=mx+c[/tex]
Substitute point [tex]P(x_{1}, y_{1})=P(6, 6)[/tex] and [tex]m = m_{2}=-\frac{3}{2}[/tex]
[tex]6=-\frac{3}{2}\times 6 +c[/tex]
[tex]6=-3\times 3 +c[/tex]
[tex]c=6+9[/tex]
[tex]c=15[/tex]
So, the y-intercept of the perpendicular line [tex]c=15[/tex]
Using point slope formula.
[tex]y=mx+c[/tex]
Substitute [tex]m = m_{2}=-\frac{3}{2}[/tex] and [tex]c=15[/tex] in above equation.
[tex]y = -\frac{3}{2}x+15[/tex]
Therefore: the equation of the perpendicular line [tex]y = -\frac{3}{2}x+15[/tex]