# 3 Points in a line!

by sutupidmath
Tags: line, points
 P: 1,635 I encountered a problem where it asked us to show whether 3 points belong to a straight line in 3D. I have worked such problems in 2D before, but never in 3D, so i tried to carry out the same idea here as well. I understand that to do such a thing as a way one could start by first deriving the eq. of such a line that connects two points,(using some vector algebra i would think), and then test whether the coordinates of the third point satisfy the equations, however i am interested to know whether the following method would work as well. In fact it worked for 2 or 3 cases that i tested, but i don't know how would i show that it is valid in general. I started by assuming that since we are working in 3D, we are gonna have slopes with respect to the x-axis and y-axis as well. Say we have three points in space: $$P_1(x_1,y_1,z_1),P_2(x_2,y_2,z_2),P_3(x_3,y_3,z_3)$$ Now the slopes with respect to x-axis and y-axis respectively would be: (i) for the segment(line) $$\bar{P_1P_2}$$ $$\frac{z_2-z_1}{x_2-x_1}, \frac{z_2-z_1}{y_2-y_1}$$ (ii) for $$\bar {P_1P_3}$$ $$\frac{z_3-z_1}{x_3-x_1},\frac{z_3-z_1}{y_3-y_1}$$ (iii) for $$\bar {P_2P_3}$$ $$\frac{z_3-z_2}{x_3-x_2},\frac{z_3-z_2}{y_3-y_2}$$ Now, in order for these points to lie on a straight line, their slopes w.r.t x and y have to be equal. Also, after we set them equal, i managed with some elementary algebraic manipulations to come with a single relation, which worked for those same cases i tested: $$\frac{z_2-z_1}{z_3-z_1}=\frac{y_2-y_1}{y_3-y_1}=\frac{x_2-x_1}{x_3-x_1}$$ THis last expression, indeed, turned out to be not the only one, with some slightly different manipulatoins i could come up with simmilar relations which seemed to work as well. So any feedback? THnx in advance.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 The line through the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $x= x_1+ (x_2-x_1)t$ $y= y_1+ (y_2-y_1)t$ $z= z_1+ (z_2-z_1)t$ If t= 0, that gives $(x_1, y_1, z_1)$ and when t= 1, $(x_2, y_2, z_2)$. Since a line is determined by two points that is all you need. Now, to determine whether $(x_3, y_3, z_3)$ is on that line, try to solve: $x_3= x_1+ (x_2-x_1)t$ $y_3= y_1+ (y_2-y_1)t$ $z_3= z_1+ (z_2-z_1)t$ See if that can be solved for a single value of t. Solve, say, the first equation for t and the see if it also satisfies the other two equations. If so the point is on the line, if not, the point is not on the line.