M megan0430 New member Joined Aug 23, 2006 Messages 30 Aug 29, 2006 #1 Prove that if y=mx+b, where m and b are constants, then m is the slope. Write a reason to justify each step in the proof. I'm not 100% sure what the problem is asking me to do, so please help :]
Prove that if y=mx+b, where m and b are constants, then m is the slope. Write a reason to justify each step in the proof. I'm not 100% sure what the problem is asking me to do, so please help :]
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Aug 29, 2006 #2 you could start with: \(\displaystyle m=\frac{y-y_{1}}{x-x_{1}}\) A little algbra creates the point-slope form: \(\displaystyle y-y_{1}=m(x-x_{1})\) You could rewrite it as: \(\displaystyle y=mx-mx_{1}+y_{1}\) \(\displaystyle y_{1}-mx_{1}=b=y-intercept\) So, you have slope-intercept form: \(\displaystyle y=mx+b\)
you could start with: \(\displaystyle m=\frac{y-y_{1}}{x-x_{1}}\) A little algbra creates the point-slope form: \(\displaystyle y-y_{1}=m(x-x_{1})\) You could rewrite it as: \(\displaystyle y=mx-mx_{1}+y_{1}\) \(\displaystyle y_{1}-mx_{1}=b=y-intercept\) So, you have slope-intercept form: \(\displaystyle y=mx+b\)
