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Before we begin, remember that two lines are enough to define a plane. Therefore, we will need two lines for each plane in order to graph them. 2x+3y-z=12 & (I) 2x-4y+z=8 & (II) Let's graph each plane separately.
x= 0
Zero Property of Multiplication
Identity Property of Addition
| y-intercept | z-intercept | |
|---|---|---|
| Substitute | 3y- 0=12 | 3( 0)-z=12 |
| Calculate | y= 4 | z= -12 |
| Point | ( 0, 4, 0) | ( 0, 0, -12) |
Next, we can plot the intercepts on the coordinate space and draw the line through them.
y= 0
Zero Property of Multiplication
Identity Property of Addition
| x-intercept | z-intercept | |
|---|---|---|
| Substitute | 2x- 0=12 | 2( 0)-z=12 |
| Calculate | x= 6 | z= -12 |
| Point | ( 6, 0, 0) | ( 0, 0, -12) |
Now that we know the intercepts, let's graph the second line!
Finally, we have two lines to graph a plane.
We will graph Plane (II) in the same way as we graphed Plane (I). Let's first find two lines such that one passes through x= 0 and the second through y= 0.
| Substitution | Resulting Equation | |
|---|---|---|
| x=0 | 2( 0)-4y+z=8 | -4y+z=8 |
| y=0 | 2x-4( 0)+z=8 | 2x+z=8 |
Next, we will find the intercepts of each line to graph them.
| -4y+z=8 | 2x+z=8 | |||
|---|---|---|---|---|
| Intercept | y-intercept | z-intercept | x-intercept | z-intercept |
| Substitution | -4y+ 0=8 | -4( 0)+z=8 | 2x+ 0=8 | 2( 0)+z=8 |
| Calculation | y= - 2 | z= 8 | x= 4 | z= 8 |
| Point | ( 0, - 2, 0) | ( 0, 0, 8) | ( 4, 0, 0) | ( 0, 0, 8) |
Now that we know the intercepts of both lines, we can graph them.
Now we can graph the plane that contains the lines.