I'm actually a little ashamed to ask this,

Suppose A and B are not 0. Consider a line "l" whose cartesian equation is $Ax+ By + D = 0$. Suppose that $P_0 = (x_0,y_0)$ does not lie on "l". Show that $n = (A,B)$ is a vector that is perpendicular to "l"

I am trying to do this without the dot-product which makes it even more embarassing. So I know that in order for two lines to be consider perpendicular that either their slopes have to be negative reciprocals of each other or in dot-product form that they equal 0. How can I use that point and the normal "n" to show the vector is perpendicular?

  • $\begingroup$ there is a second part also giving me trouble: Let $Q_X = (x, \frac{-Ax}{B} - \frac{D}{B}$) on point "l". For what value of $x$ is (vector) $P_0Q_x$ parallel to "n"? Now considering that I've been given that two vectors are parallel iff $\frac{v}{||v||} = \frac{w}{||w||}$ where v and w are vectors. I was trying to expand that out and solve for x, but it is turning into a huge mess $\endgroup$ – dc3rd Apr 30 '15 at 22:35

If I'm understanding your question correctly, you want to show that the line Ax + By + D = 0 is perpendicular to the vector (A,B). I don't know what you intend $P_0$ to be used for.

At any rate, since you assume A and B are not zero, you can rewrite this as $y = mx + b$ to see that the slope of the line is -A/B. But the vector (A,B) has slope B/A. So the product of the slopes is -1. (Note that D is irrelevant - changing it translates the line, but doesn't change its slope.)


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