Write the equation of the line which is perpendicular to the red line graphed at the point $P(x,2)$.
I started with finding the equation of the red line using $y=a+bx$: $$0=a+b(-6)$$ $$-3=a+b(0)$$ Which gives $a=-3$ and $b=-\frac12$. Therefore the equation of the original line is $y=-3-\frac x2$. Then what should I do?
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Having obtained the line's equation, note that its slope is $-\frac12$. For a line with slope $m$, any perpendicular line will have slope $-\frac1m$, so the line to be found must have slope $-\frac1{-\frac12}=2$.
Furthermore, we are told that the point of intersection has $y=2$. Substituting this into the first line's equation, we get $x=-10$, i.e. $P=(-10,2)$. Then the $c$ in the equation of the perpendicular line $y=mx+c$ satisfies $2=-10\cdot2+c$, from which we get $c=22$.
Therefore the perpendicular line's equation is $y=2x+22$.
Use the fact that, the product of the slopes of this line and the line perpendicular to this gives $-1$. since comparison with $y=mx+c$, gives the slope of this line equals $-\frac12$, the slope of the line perpendicular is $2$.
So, $$y= 2x+c.$$
Now, in the original line to find $x$ such that $y=2$, gives $x=-10$. So, for the new line to pass through this, $(-10,2)$ gives
$$c=22$$
line is $y=2x+22$