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The lines CD and EF are perpendicular with points $C(1,2)$, $D(3,-4)$, $E(-2,5)$, and $F(k,4)$. Find the value of the constant $k$.

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Forgot how easy "algebraic geometry" could be. Hmm... –  Pedro Tamaroff Sep 22 '13 at 18:32
    
Please, show your work, what you know, what you have tried, &c. People won't do your homework for you here. It is expected you do something yourself. –  Pedro Tamaroff Sep 22 '13 at 18:33
    
Hello, and welcome to Math SE. Please let us know what steps you've taken to solve the problem and where you had difficulties. Also, if this is a homework question, please use the corresponding tag. –  Jonathan Y. Sep 22 '13 at 18:35

2 Answers 2

Hint: $(2, -6)\cdot(k+2,-1)=0$.

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so far, I think I have gradient of line EF which = -1/3. –  Ed Prince Sep 22 '13 at 18:45
    
I would find it useful for a nudge in the right direction, as I'm not sure where to start. It is homework –  Ed Prince Sep 22 '13 at 18:46
    
Two vectors are perpendicular iff their dot product (inner product) is $0$. –  njguliyev Sep 22 '13 at 18:47

The OP presumably does not know about vectors or dot products. Here's the appropriate hint: What do you know about the slopes of perpendicular lines? Now can you do the problem?

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The product of the gradients = -1. If a gradient of a line was 1/2, the perpendicular gradient = -2/1 = -2. –  Ed Prince Sep 22 '13 at 18:59
    
Yes. Good! So, you were right above, the line $EF$ has slope (gradient) $-1/3$. I missed that, sorry. So now use the definition of the gradient ... if you know $E$ and $F$ are on the line, find the gradient and solve for $k$. –  Ted Shifrin Sep 22 '13 at 19:01
    
tried plugging into y = mx + c and end up with 4 = -1/3k + 3 1/3 –  Ed Prince Sep 22 '13 at 21:06
    
Your $3\ 1/3$ is wrong. It should be $13/3$ (better not to use mixed numbers). If you then solve for $k$, what do you get? But you don't need that. Just compute the slope using only the coordinates of $E$ and $F$. –  Ted Shifrin Sep 22 '13 at 21:43
    
How do you compute the slope? Struggling to get my head around it. –  Ed Prince Sep 23 '13 at 16:10

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