3 co-ordinates given. Equation of perpendicular line.

Question

Sorry, I appreciate that this isn't exactly a deep and meaningful math question, but I've gotten stuck and would appreciate the aid. The question is as follows:

Point $A$ has coordinates $(-6, -9)$.

Point $B$ has coordinates $(-1, 1)$.

Point $C$ has coordinates $(-2, -2)$.

Find an equation of the line that passes through point $C$ and is perpendicular to $AB$. Give your equation in the form $ax + by = c$, where $a$ and $b$ are integers.

I would appreciate all help! Thanks, Josh.

Answer 1

$(-1,1) - (-6,-9) = (5, 10)$ This can be used directly to find the coefficients $a,b$ that you need.

Or you can do some extra work...

The equation for line AB:

$10(x+1) - 5(y-1) = 0\\ 10x - 5y = -15\\ 2x - y = -3 $

A line perpendicular to AB will have a slope that is equal to the negative reciprocal. Or in standard form.

$x + 2y = c$

Plug $(-2,-2)$ for $(x,y)$ to solve for $c.$

Answer 2

HINT

angular coefficient line AB: $$m_{AB}=\frac{-9-1}{-6+1}=2$$

angular coefficient line perpendicular to AB: $$m=-\frac{1}{m_{AB}}=-\frac12$$

line through point C perpendicular to AB: $$y-(-2)=m(x-(-2))\implies y+2=-\frac12 x-1\\\implies y+\frac12x+3=0\implies 2y+x+6=0$$

Answer 3

Recall that the equation $ax+by=c$ describes a line perpendicular to $(a,b)$, passing through a certain point to be determined. In your case, $\vec{AB} = (5,10)$, so $5x+10y=c$. To find out $c$, plug in the coordinates of the point $C$ and you're done.