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

2 Answers
Jul 18, 2018

(-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).

(-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)