Posted by **Kay** on Friday, September 26, 2008 at 10:56pm.

I am stumped! I have been working on these problems for the longest time and I can not figure out what I am doing wrong. Some how when I get to the last two steps I mess up, and I am beyond frustrated.

The directions say:

Write an equation of the line containing the given point and perpendicular to the given line.

1. (-5,9) 5x=6y+7

2. (3,5) 4x+y=9

3. (3,-5) 6x+5y=7

To show you where I am messing up I am going to give you one I have already tried to do, then maybe someone can tell me where I am going wrong. I have a test on this tomorrow and I don't get it at ALL!

(4,6) x+9y=7

y= -1/9-7/9

Then you use the formula y-y1=m(x-x1)

so it would be

y-6=-1/9x+4

Then here is where I am having problems, when I try to multiply the -1/9 by 4 and then add 6 to it, I come up with the wrong answer.

I got y= -1/9x+10/9. The right answer is -1/9x+58/9. The answers all have to be in the formula of y=mx+b

Thanks for your help!

- Algergra -
**bobpursley**, Friday, September 26, 2008 at 11:16pm
Start here: you have the general equation correct..

y= -x/9 -7/9

Now you want a line perpendicular to this line, so it has the slope 9 (negative reciprocal of -1/9

So the equation of the perpendicular is

y=9x + b.

Put in the point 4,6

6=9*4+b so b= -30

so the answer is y=9x-30. That is the equation of a line perpendicular and through the point 4,6. Your stated right answer is wrong.

One more:

(-5,9) 5x=6y+7

y=5/6 x -7/6

so the perpendicular line will be slope -6/5, or

y=-6x/5 +b

put in the point (-5,9)

9=30/5 + b

or b=3

y=-6x/5 + 3 is the equation of the line perpendicular containing the point.

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