Find the point on the line segment joining $P_1(1, 4,-3)$ and $P_2(1, 5,-1)$ that is $2/3$ of the way from $P_1$ to $P_2$

Question

I had used the distance formula $d(p_1,p_2)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ and got $\sqrt5$ and multiply $\sqrt 5$ by $2/3$ I get $1.5$.

Now, how can I get the coordinates $x,y,z$?

Answer 1

HINT

We need the parametric equation for the segment that is

$$P(t)=P_1+t(P_2-P_1)=(1,4,-3)+t(0,1,2)$$

indeed note that

• $P(0)=P_1$
• $P(1)=P_2$

and then take the value $t=\frac23$.

Answer 2

The answer is $$P_1 + (2/3)(P_2 -P_1)$$

You check and see which one is the correct answer.

Answer 3

Hint: Use proportional triangles instead. If a point is $\frac23$ of the way from $P_1$ to $P_2$, then its $x$-coordinate is $\frac23$ of the way between the $x$-coordinates of $P_1$ and $P_2$, and so on.