What you have done is correct. Since t is not the same in all three equations, the point does not lie on the line.
Good day,
I working with the following question:
Determine if the point (0, 2, 4) lies on the line with parametric equations:
x = 1 - t;
y = 3 + t;
z = 6 + 2t
My attempt:
I think you have to substitute the respective x, y and z values from the given point into x, y and z and solve. If t is the same in each case, then the point lies on the line. Is that correct?
I get t = 1, t = -1, t = -1 for x, y and z respectively. This means that the point does not lie on the line. However, I feel as though my methods are completely wrong. The only other thing that I could think of is converting the line to it's vector equation and solving that. I'm just not sure.
Another way of thinking about this is: If (0, 2, 4) is on the line then there must exist t such that x= 0= 1- t so t= 1. But then y= 3+ t= 3+ 1= 4, not 2. That is sufficient to say that (0, 2, 4) is not on the line.