Linear equations (equation of line and perpendicular lines)

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foahchon

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Nov 12, 2006
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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.
 
points[1/2,1/4] and [-1/3,1/5]

y=mx+b substitute point 1, and point 2 for 2 equations
1/4=1/2 m+b
1/5=-1/3m+b subtract 2nd equation from 1st
1/4-1/5 = [1/2+1/3]m
[5-4]/20=5/6 m
6/100 =m

y=mx+b
y=6/100 x+b substitute a point
1/4=[1/2][3/50] +b
25/100-3/100=b

y=3/50 x +11/50
50y=3x+11
3x-50y=-11

-------------------------------------------------------------------------------------------
x-2y=3 perpindicular and contains[-3,1]
y=1/2 x - 3/2 slope = 1/2
a perpindicular line has a slope of -1/m
m of line is -1/[1/2]
m=-2
y=-2x+b but point -3,1 on the line
1=-2[-3]+b
b=-5
y=-2x-5 answer
2x+y=-5
arthur
 
Re: Linear equations (equation of line and perpendicular lin

foahchon said:
Basically, I found the slope, which was \(\displaystyle \frac{3}{50}\), and my equation was

y = \(\displaystyle \frac{3}{50}x + \frac{1}{5}\)
Click to expand...
Slope ok, BUT why 1/5 as y-intercept ? :shock:
 
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