Question

At what point does the line `(x+4)/2 = (y-6)/-1 = (z+2)/4` meet the coordinate planes?

Answer 1

`(-3, 5.5, 0)`

Explanation:

While an untraditional definition of a line, we can try and break down what it is meaning.

In order to really define this line, I'm going to introduce a new variable `t`. `t` is equal to each of the things above, i.e.
`t = (x+4)/2`, `t = (y-6)/-1`, `t = (z+2)/4`. If you were to solve for (x,y,z) it is clear how this would be a single parameter to define a 3D object, i.e. a 1D curve. In this case, since each of these relationships is linear, the whole function will be linear and it will be a line in 3D. Therefore, there is only one crossing of the coordinate plane.

The "coordinate plane" refers to the traditional x-y plane. This has the property that z = 0. Hence, we can plug that in to find `t`:
`(0+2)/4 = t = 1/2`.
Therefore,
`1/2 = (x+4)/2 implies x+4 = 1 implies x = -3`
`1/2 = (y-6)/-1 implies 2y-12 = -1 implies y = 11/2`

Therefore, it intercepts the plane at `(-3, 11/2)`.

Answer 2

`(-3, 11/2, 0), \ (0, 4, 6)` & `(8, 0, 22)`

Explanation:

Given equation of line:

`\frac{x+4}{2}=\frac{y-6}{-1}=\frac{z+2}{4}`

Let `\frac{x+4}{2}=\frac{y-6}{-1}=\frac{z+2}{4}=t`

`x=2t-4, \ y=-t+6` & `z=4t-2`

Thus, there is a general point on the given line `(2t-4, -t+6, 4t-2)`

Now, the line will intersect the XY-plane at the point where `z=0` hence

`4t-2=0\implies t=1/2`

hence, the point where the line intersects the XY-plane at the point

`(2(1/2)-4, -1/2+6, 4(1/2)-2)\equiv(-3, 11/2, 0)`

Now, the line will intersect the YZ-plane at the point where `x=0` hence

`2t-4=0\implies t=2`

hence, the point where the line intersects the YZ-plane at the point

`(2(2)-4, -2+6, 4(2)-2)\equiv(0, 4, 6)`

Now, the line will intersect the ZX-plane at the point where `y=0` hence

`-t+6=0\implies t=6`

hence, the point where the line intersects the ZX-plane at the point

`(2(6)-4, -6+6, 4(6)-2)\equiv(8, 0, 22)`