# Find the equation

1. Dec 29, 2011

### luigihs

Find the equation of the straight line joining points (-1,1) and (2,5)

m = y1 - y2 / x1 - x2

5 - 1 / 2 - (-1) = 4/3
y - 1 = 4/3( x + 1 )
y = 4/3x + 4/3 + 1
y = 4/3x + 5/3< --- this is my attempt solution but I think is wrong! because on the back of my maths book the answer is 3y = 4x + 7 so I don't know what formula I have to use.

2. Dec 29, 2011

### HallsofIvy

I think you mean (y1- y2)/(x1- x2). What you wrote is y1- (y2/x1)- x2.

What do you get if you multiply both sides of y= (4/3)x+ 5/3 by 3?

3. Dec 29, 2011

### luigihs

Why do I have to multiply by 3 ?? I dont remember exactly how to do that somebody teach me that like 3 years ago but is like y - 1 = 4/3 ( x + 1) and I think I have to multiply by 3 each number so I can get rid of the denominator, but I not sure ... y - 1 = 12 ( 3x + 3) <--- like that?

Last edited: Dec 29, 2011
4. Dec 29, 2011

### luigihs

Any ideas?

5. Dec 29, 2011

### SammyS

Staff Emeritus
You have made an error in adding $\displaystyle \frac{4}{3}+1\,.$

The number one is how many thirds?

6. Dec 29, 2011

### luigihs

Oh yes! 4/3 + 1/1 (3) = 4/3 + 3/3 = 7/3 ... But I still having the wrong answer because the answer is 3y = 4x + 7.. and I figure out only the slope formula ( y = mx + b )

7. Dec 29, 2011

### SammyS

Staff Emeritus
Two ways to see that the following two equations are equivalent:
$\displaystyle y=\frac{4}{3}x+\frac{7}{3}\quad\quad \text{(Equation 1)}$

$\displaystyle 3y=4x+7\quad\quad\quad\text{(Equation 2)}$​

One way: (You've been told this before.) Multiply both sides of Equation 1 by 3 . Remember, use the Distributive Law

Second way: Divide [STRIKE]Multiply[/STRIKE] both sides of Equation 2 by 3 .
(Mod note: removed extraneous operation above.)

Last edited by a moderator: Dec 29, 2011
8. Dec 30, 2011

### checkitagain

luigihs,

it should be 4/3 + 1/1(3/3) = 4/3 + 3/3 = 7/3, or

$\dfrac{4}{3} \ = \ \dfrac{1}{1}\bigg(\dfrac{3}{3}\bigg) \ = \ \dfrac{4}{3} + \dfrac{3}{3} \ = \dfrac{7}{3}$

What you had is equivalent to $\dfrac{4}{3} + 3.$

9. Dec 30, 2011

### Redbelly98

Staff Emeritus
In case it wasn't clear from what others have said: you have the correct answer, it is equivalent to the answer given in the book.

10. Dec 30, 2011

### Staff: Mentor

The work above has a typo (marked). Here is the correction:
$\dfrac{4}{3} \ + \ \dfrac{1}{1}\bigg(\dfrac{3}{3}\bigg) \ = \ \dfrac{4}{3} + \dfrac{3}{3} \ = \dfrac{7}{3}$