Almost ashamed to ask this question.

[RadiationX]

When we want to find the equation of a line we can use:
Y-Y_1 = m(X-X_1) or Y = mX + B

My question is aren't both equations essentially the same?

For equation 1. If I know two points that the line passes through just by plugging and chugging won't that lead to what the Y intercept is?

For example I have these two points that a line passes through: (2,12) & (6,0).
by looking at the graph I know that the line has a Y intercept, even though it is not drawn. When I use equation one my constant (B) the Y intercept is 0 even though it should not be. What the heck am I doing wrong?

 

[cristo]

Yes, they are the same equation. I'm not sure what you're doing wrong, but if we use your first equation, put one point in (say (6,0)), then y-0=m(x-6) => y=m(x-6). Then use the second point to find m; 12=m(2-6) => m=-3; thus y=-3(x-6)=-3x+18 is the equation for the line.

 

[dontdisturbmycircles]

Yes they are the same, the second one is a special case where Y_{1} is the y intercept (B) and thus X_{1} is 0 so you get Y=mX+Y_{1}

edit: woops, hi cristo =-).

 

[cristo]

edit: woops, hi cristo =-).

Haha; that gets me back for butting into your thread earlier!

 

[RadiationX]

Yes, they are the same equation. I'm not sure what you're doing wrong, but if we use your first equation, put one point in (say (6,0)), then y-0=m(x-6) => y=m(x-6). Then use the second point to find m; 12=m(2-6) => m=-3; thus y=-3(x-6)=-3x+18 is the equation for the line.

You have hit upon the point of my confusion exactly cristo! Jeez, so simple but I have not used this equation since precalc so i forgot how it worked! the point slope form of the line is used to find the slope of the line. It is the variable that we are solving for.