Can some one check my work on making an equation of a line

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February 19th 2008, 05:07 PM
imbadatmath

Can some one check my work on making an equation of a line

1st problem
(6,1) and (4,5)
i first found the slope 5-1=4 over 4-6=-2 so m= -2
i then did y-(5)=2(x-4)
y-5=2x-8
add 5 to both sides
y=2x-3

2nd problem
m=-1/5 (3,4)
so the equation of the line would be y-4=-1/2(x-3) am i done or do i need to do more my teacher really does not teach he goes to his desk and plays games. So could some one please check my work and help me correct it if its wronge thanks for your time.
February 19th 2008, 05:15 PM
bobak

Quote:

Originally Posted by imbadatmath

1st problem
(6,1) and (4,5)
i first found the slope 5-1=4 over 4-6=-2 so m= -2
i then did y-(5)=2(x-4) Your 2 should be -2
y-5=2x-8 This should be y-5=-2x+8
add 5 to both sides
y=2x-3 then you should get y = -2x + 13

There you go
February 19th 2008, 05:20 PM
bobak

Quote:

Originally Posted by imbadatmath

m=-1/5 (3,4)
so the equation of the line would be y-4=-1/2(x-3) I think you made a typo here that should be -1/5. but I am sure that is what you meant.

well what you have written is correct but usually questions are a bit more elbaroate and may ask you to write it as either

$y = mx +c$
$ay + bx + c = 0$ where a,b are c integers.
February 19th 2008, 07:10 PM
imbadatmath

how would i write it in mx+c?
February 19th 2008, 11:50 PM
mr fantastic

Quote:

Originally Posted by imbadatmath

how would i write it in mx+c?

$y - 4 = -\frac{1}{5} (x-3) \Rightarrow y - 4 = -\frac{x}{5} + \frac{3}{5} \Rightarrow y = -\frac{x}{5} + \frac{3}{5} + 4 = -\frac{x}{5} + \frac{3}{5} + \frac{20}{5} = -\frac{x}{5} + \frac{23}{5}$ .

So the mx+c form (that is, slope-intercept form) is $y = -\frac{x}{5} + \frac{23}{5}$ .

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