What are the intercepts of the plane `4x+3y-12z+6=0` on the axes?

Answer

To find the intercepts of the plane given by the equation \(4x + 3y - 12z + 6 = 0\) on the axes, we will follow these steps:

### Step 1: Find the x-intercept
To find the x-intercept, we set \(y = 0\) and \(z = 0\) in the equation of the plane.

\[ 4x + 3(0) - 12(0) + 6 = 0 \]

This simplifies to:

\[ 4x + 6 = 0 \]

Now, solve for \(x\):

\[ 4x = -6 \implies x = -\frac{6}{4} = -\frac{3}{2} \]

Thus, the x-intercept is \((- \frac{3}{2}, 0, 0)\).

### Step 2: Find the y-intercept
Next, we find the y-intercept by setting \(x = 0\) and \(z = 0\) in the equation of the plane.

\[ 4(0) + 3y - 12(0) + 6 = 0 \]

This simplifies to:

\[ 3y + 6 = 0 \]

Now, solve for \(y\):

\[ 3y = -6 \implies y = -\frac{6}{3} = -2 \]

Thus, the y-intercept is \((0, -2, 0)\).

### Step 3: Find the z-intercept
Finally, we find the z-intercept by setting \(x = 0\) and \(y = 0\) in the equation of the plane.

\[ 4(0) + 3(0) - 12z + 6 = 0 \]

This simplifies to:

\[ -12z + 6 = 0 \]

Now, solve for \(z\):

\[ -12z = -6 \implies z = \frac{6}{12} = \frac{1}{2} \]

Thus, the z-intercept is \((0, 0, \frac{1}{2})\).

### Summary of Intercepts
- x-intercept: \((- \frac{3}{2}, 0, 0)\)
- y-intercept: \((0, -2, 0)\)
- z-intercept: \((0, 0, \frac{1}{2})\)