Other Answers (6)
Relevance
Let me give you TWO methods, in detail:
FIRST METHOD:
The equation is y = m x + b
The line passes through (8,9) and (10,3) when both of the following equations hold:
9 = 8 m + b
3 = 10 m + b
Substract those two equations from each other to eliminate b. You obtain: 12 = 18 m, or m = 2/3.
Plug that value of m into either equation to obtain the value of b, namely (using the first equation): 9 = 8 (2/3) + b, so:
b = 916/3 = 11/3
If you prefer, you can also obtain the value of b directly by eliminating m between the two equations (multiply the first by 5 and the second by 4 and add them up). You obtain: 4512 = 9 b, so b = 11/3, indeed.
SECOND METHOD (more efficient):
At a slightly more advanced level, we may obtain the equation of the line directly by stating that (x+8, y9) is proportional to (10+8,39). If you know what a determinant is, you may just write that the relevant 2 by 2 determinant is zero. Otherwise, just write that the proportionality means:
(x+8)(39) = (y9)(10+8)
Just simplify that:
12 (x+8) = 18 (y9) or 2(x+8)+3(y9) = 0
This boils down to 2x+3y=11
You may keep this equation as is or divide by 3 to put it in the form
y = (2/3)x+(11/3) or y = (112x) / 3
RECOMMENDATIONS:
I advise you to get familiar with the second method ASAP and, also, to acquire a taste for the nicer form of the equation of a line where the coefficient of y need not be 1. Here: 2x+3y=11.
Last step: Check that the equation obtained holds for both points:
2(8) + 3(9) = 11
2(10) + 3(3) = 11
Trimming all the fat, the entire computation merely consists in simplifying a single line, as shown in the link below:Source(s):

First, you have to figure out what the slope is. The formula for doing that is this: y2y1/x2x1 = the slope of the line. OK?
So, lets use the first point as (x1, y1) and the second point as (x2, y2).
Considering that, we'd have 39/10(8) = 12/18 = 2/3 = slope
Now, use the point/slope forumula to get the equation you need.
Here it is: yy1 = m(xx1)
The first y and x stay just that way, as y and x. The y1 and x1 are the coordinates of whichever point given you choose to work with. I'll choose point #2.
y  9 = m(x  (8))
y  9 = 2/3(x + 8)
now multiply the right side out:
y  9 = 2/3 x  16/3
The form you want to put this in is y = mx + b, I know this because you mentioned b in your question. b indicates the y intercept of the line between these 2 points.
So, now you have to add 9 to both sides of the equation above to get it in the form of y = mx + b:
y 9 +9 = 2/3 x  16/3 + 9
y = 2/3 x  16/3 + 27/3
y = 2/3 x + 27/3  16/3
y = 2/3 x + 11/3 Done!
Source(s):
BS math, teacher 
By letting p1=( 8, 9 ) and p2 = ( 10, 3 ), we have the slope of
m = change y / change x = 2/3
Using p1 ( 8, 9 ) in the pointslope formula, we have:
( y  9 ) = 2/3 ( x  8 )
solving for y = 2/3 x + 11/3 ( y  intercept is b = 11/3 when x =0 )
Using p2 ( 10, 3 ) in the pointslope formula yields us once again....
( y + 3 ) = 2/3 ( x  10 ) expanding and collecting the like terms we obtain
y = 2/3 x + 11/3 ( yintercept b = 11/3 when x =0 )
multiplying both sides of the equation by 3, we have:
3y = 2x + 11
therefore, the equation of the line that passes through the
points ( 8, 9 ) and ( 10, 3 ) is:
2x +3y  11 = 0Source(s):

First find the slope 'm'
m =( y" y') /( x"x') = 39 / 10(8) = 12 / 18 = 2/3
so the equation in slopeintercept form will be
y = 2x/3 + b
now substitute x=8 and y=9
9 = 2x8 /3 +b
9 = 16/3 +b
b = 916/3
b = (2716)/3
b = 11/3
so the equation of the line is y = 2x/3 +11/3
or , 3y +2x = 11 
First find m, or the slope.
m = (y2  y1) / (x2  x1)
m = (3  9) / (10  (8) )
m = 12 / 18
m = 2/3
Point slope form:
(y  y1) = m (x  x1)
Pick an ordered pair. I'll pick (8, 9)
(y  9) = (2/3) (x  8)
(y  9) = (2/3) (x + 8)
y  9 = (2/3)x  16/3
y = (2/3)x  16/3 + 9
y = (2/3)x + 11/3 
first find the slope: rise over run y2y1/ x2x1
y9= 2/3 ( x (8) )
y9= 2/3x 16/3
add 9 to get y by itself
y= 2/3x + 11/3
What is the equation of the line that passes through (8,9) and (10,3)?
I am confused how to find "b". The answer is in pointslope form but i don't know how to get there. HELP!!! thx
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