Algergra
posted by Kay .
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= 1/97/9
Then you use the formula yy1=m(xx1)
so it would be
y6=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!

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=9x30. 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.