# Thread: Points in the plane.

1. ## Points in the plane.

I want to find some arbitrary point in the plane

$\displaystyle 2x+4y-2z + 10 =0$

Thank you.

2. Originally Posted by MatteNoob
I want to find some arbitrary point in the plane

$\displaystyle 2x+4y-2z + 10 =0$

Thank you.
Since 3 non-colinear points determine a plane, then try two coordinates and solve or the 3rd coordinate.

2x +4y -2z +10 = 0

Example, try x=1 and y=1,
2(1) +4(1) -2z +10 = 0
-2z = -16
z = 8
Hence, (1,1,8) is a point on the given plane.

3. Thank you very much. :]

4. Hello, MatteNoob!

I want to find some arbitrary point in the plane: .$\displaystyle 2x+4y-2z + 10 \:=\:0$
The easiest points are the intercepts. .You can "eyeball" the answers!

. . $\displaystyle \begin{array}{cccccccc} \text{Let }y = 0,\:z = 0\!: & 2x + 10 &=& 0 & \Rightarrow & x \:=\: \text{-}5 & \Rightarrow & (\text{-}5,0,0) \\ \\[-4mm] \text{Let }x = 0,\:z = 0\!: & 4y + 10&=& 0 & \Rightarrow & y \:=\: \text{-}\frac{5}{2} & \Rightarrow& \left(0,\:\text{-}\frac{5}{2},\:0\right) \\ \\[-4mm] \text{Let }x = 0,\:y=0\!: & \text{-}2z+10 &=& 0 & \Rightarrow & z \:= \:5 & \Rightarrow & (0,\:0,\:5) \end{array}$

5. or...

$\displaystyle \frac{x}{-5}\;+\;\frac{y}{-5/2}\;+\;\frac{z}{5}\;=\;1$

They're just kind of staring at you in this form.

6. Thanks to all of you. I now understand how finding (or checking if) points (are in) the plane. :]