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

• Feb 19th 2008, 05:07 PM
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
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.
• Feb 19th 2008, 05:15 PM
bobak
Quote:

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
y=2x-3 then you should get y = -2x + 13

There you go
• Feb 19th 2008, 05:20 PM
bobak
Quote:

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

$\displaystyle y = mx +c$
$\displaystyle ay + bx + c = 0$ where a,b are c integers.
• Feb 19th 2008, 07:10 PM
$\displaystyle y - 4 = -\frac{1}{5} (x-3) \Rightarrow y - 4 = -\frac{x}{5} + \frac{3}{5} \Rightarrow y = -\frac{x}{5} + \frac{3}{5} + 4 = -\frac{x}{5} + \frac{3}{5} + \frac{20}{5} = -\frac{x}{5} + \frac{23}{5}$.
So the mx+c form (that is, slope-intercept form) is $\displaystyle y = -\frac{x}{5} + \frac{23}{5}$.