Hi, I'm having some trouble with some basic algebra (apparently).
First is a problem where I have to find the equation of a line going through a given pair of points in standard form using only integers.
The points are (\(\displaystyle \frac{1}{2}\),\(\displaystyle \frac{1}{4}\)), (\(\displaystyle \frac{-1}{3}\), \(\displaystyle \frac{1}{5}\))
Basically, I found the slope, which was \(\displaystyle \frac{3}{50}\), and my equation was
y = \(\displaystyle \frac{3}{50}x + \frac{1}{5}\)
This gave me:
50y = 3x + 10
50 - 3x = 10
-3x + 50 = 10
3x - 50y = -10
The book, on the other hand, says that the answer is 3x - 50y = -11. I must be missing something really simple, I just can't find out what it is.
The other problem I'm having has to do with perpendicular lines. I'm pretty close, I just have the signs wrong (apparently):
The line perpendicular to x - 2y = 3, and containing (-3,1).
Here's what I have so far:
-2y = -x + 3
2y = x - 3
y = \(\displaystyle \frac{1}{2}x - \frac{3}{2}\)
y - 1 = -2(x + 3)
y - 1 = -2x - 6
y = -2x - 5
-2x + y = -5
2x - y = 5
The book, on the other hand, says the solution is 2x + y = -5. I can't see where I've gone wrong, and I'd appreciate any insight. Thanks.
First is a problem where I have to find the equation of a line going through a given pair of points in standard form using only integers.
The points are (\(\displaystyle \frac{1}{2}\),\(\displaystyle \frac{1}{4}\)), (\(\displaystyle \frac{-1}{3}\), \(\displaystyle \frac{1}{5}\))
Basically, I found the slope, which was \(\displaystyle \frac{3}{50}\), and my equation was
y = \(\displaystyle \frac{3}{50}x + \frac{1}{5}\)
This gave me:
50y = 3x + 10
50 - 3x = 10
-3x + 50 = 10
3x - 50y = -10
The book, on the other hand, says that the answer is 3x - 50y = -11. I must be missing something really simple, I just can't find out what it is.
The other problem I'm having has to do with perpendicular lines. I'm pretty close, I just have the signs wrong (apparently):
The line perpendicular to x - 2y = 3, and containing (-3,1).
Here's what I have so far:
-2y = -x + 3
2y = x - 3
y = \(\displaystyle \frac{1}{2}x - \frac{3}{2}\)
y - 1 = -2(x + 3)
y - 1 = -2x - 6
y = -2x - 5
-2x + y = -5
2x - y = 5
The book, on the other hand, says the solution is 2x + y = -5. I can't see where I've gone wrong, and I'd appreciate any insight. Thanks.