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[wiredgamer]

I don't know if this is the right spot for this but I would think this is an admin issue but not for the forum. On the main site in the Point slope form page in algebra I think the first more practice problem is wrong. it has the problem:
Your point is (-1,5). The slope is 1/2. Create the equation that describes this line in point-slope form. Try working it out on your own. The answer is: y+1=1/2∗(x−5).
the x,y are backward I think? shouldn't the -1 be the x and the 5 is y? so the problem should be:
Y-5=1/2 * (x+1)
Unless im just slow or not understanding which is possible I haven't done this kind of math in years.

 

[Deleted member 4993]

I don't know if this is the right spot for this but I would think this is an admin issue but not for the forum. On the main site in the Point slope form page in algebra I think the first more practice problem is wrong. it has the problem:
Your point is (-1,5). The slope is 1/2. Create the equation that describes this line in point-slope form. Try working it out on your own. The answer is: y+1=1/2∗(x−5).
the x,y are backward I think? shouldn't the -1 be the x and the 5 is y? so the problem should be:
Y-5=1/2 * (x+1)
Unless im just slow or not understanding which is possible I haven't done this kind of math in years.

You are correct.

 

[Steven G]

I don't know if this is the right spot for this but I would think this is an admin issue but not for the forum. On the main site in the Point slope form page in algebra I think the first more practice problem is wrong. it has the problem:
Your point is (-1,5). The slope is 1/2. Create the equation that describes this line in point-slope form. Try working it out on your own. The answer is: y+1=1/2∗(x−5).
the x,y are backward I think? shouldn't the -1 be the x and the 5 is y? so the problem should be:
Y-5=1/2 * (x+1)
Unless im just slow or not understanding which is possible I haven't done this kind of math in years.

You are correct and thanks for telling us.
I want you to know that questions like this can be check and found to be wrong.
If in fact the equation of the line was y+1=1/2∗(x−5), then the point (-1,5) should be on the line, that means (-1, 5) should satisfy the equation.
Lets see if it works:
(5) + 1 \(\displaystyle \stackrel{?}{=}\) (1/2)((-1)-5)
6 \(\displaystyle \stackrel{?}{=}\)(1/2)(-6)
6 \(\displaystyle \neq\) -3

So either I made a mistake in my check and/or the equation is wrong.