So, lets use (2,3), (7,8)

The equation is y = x+1

So, what is "down" ? If you mean lower "x", then just pick an x-value slightly less than the smallest of the two x-values.

The x-values are 2 and 7, the smallest is 2, and "slightly less" would be 1.9

y = x+1 = 1.9+1 = 2.9

So "slightly down" would be (1.9,2.9)

]]>Okay, let the next step be this:

Write the co-ords here on the board, and we'll see what we can do to help you

that was my point with the last post it isnt for a set line it is for an arbitary line.

]]>Write the co-ords here on the board, and we'll see what we can do to help you ]]>

i would caclulate the gradient and the y intercept but it is for an arbitary line.

from the material given i can understand how the maths works how ever i dont see how i can find a point a little down the line unless i know either an x or a y that is on the line near to where i want the new point to be ( if u can understand what i mean, i know it isnt that clear)

]]>Once you have the equation you can calculate any point on the line you want.

]]>I have a line drawn on a grid all i know is start and end co-ords of the line from the start co-ord i need to calculate a point on the line that is just slightly further down the line than the first point. is this possibe?

]]>