Equation of Line Joining (-1, 1) & (2, 5): 3y = 4x + 7

[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.

 

[HallsofIvy]

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



m = y1 - y2 / x1 - x2

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

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

 

[luigihs]

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?

Why do I have to multiply by 3 ?? I don't 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?

 

[luigihs]

Any ideas?

 

[SammyS]

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.

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

The number one is how many thirds?

 

[luigihs]

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

The number one is how many thirds?

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 )

 

[SammyS]

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 )

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.)

 

[checkitagain]

Oh yes! &gt; &gt; 4/3 + 1/1 (3) &lt; &lt; = 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 )

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.

 

[Redbelly98]

Oh yes! 4/3 + 1/1 (3) = 4/3 + 3/3 = 7/3 ...[/RIGHT] 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 )

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.

 

[Mark44]

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}

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}

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