The point of intersection of the line joining the points (-3, 4,-8) and (5,-6, 4) with the XY plane is

Options

A. `(7/3,8/3,0)`
B. `(7/3,-8/3,0)`
C. `(-7/3,8/3,0)`
D. `(7/3,8/3,0)`

Answer

To find the point of intersection of the line joining the points (-3, 4, -8) and (5, -6, 4) with the XY plane, we can follow these steps:

### Step 1: Determine the Direction Ratios
First, we need to find the direction ratios of the line joining the two points. The direction ratios can be calculated as follows:

Let the first point be \( P_1(-3, 4, -8) \) and the second point be \( P_2(5, -6, 4) \).

The direction ratios \( (a, b, c) \) can be calculated as:
- \( a = x_2 - x_1 = 5 - (-3) = 5 + 3 = 8 \)
- \( b = y_2 - y_1 = -6 - 4 = -10 \)
- \( c = z_2 - z_1 = 4 - (-8) = 4 + 8 = 12 \)

Thus, the direction ratios are \( (8, -10, 12) \).

### Step 2: Write the Equation of the Line
Using the point-direction form of the line, we can write the equation of the line passing through point \( P_1(-3, 4, -8) \) with direction ratios \( (8, -10, 12) \):

\[ \frac{x + 3}{8} = \frac{y - 4}{-10} = \frac{z + 8}{12} = k \]

### Step 3: Set \( z = 0 \) for the XY Plane
Since we are looking for the intersection with the XY plane, we set \( z = 0 \):

\[ \frac{0 + 8}{12} = k \implies \frac{8}{12} = k \implies k = \frac{2}{3} \]

### Step 4: Find \( x \) and \( y \)
Now, we can substitute \( k \) back into the equations to find \( x \) and \( y \):

1. For \( x \):
\[ \frac{x + 3}{8} = \frac{2}{3} \]
Cross-multiplying gives:
\[ 3(x + 3) = 16 \implies 3x + 9 = 16 \implies 3x = 16 - 9 \implies 3x = 7 \implies x = \frac{7}{3} \]

2. For \( y \):
\[ \frac{y - 4}{-10} = \frac{2}{3} \]
Cross-multiplying gives:
\[ 3(y - 4) = -20 \implies 3y - 12 = -20 \implies 3y = -20 + 12 \implies 3y = -8 \implies y = -\frac{8}{3} \]

### Step 5: Final Coordinates
Thus, the coordinates of the point of intersection with the XY plane are:
\[ \left( \frac{7}{3}, -\frac{8}{3}, 0 \right) \]

### Conclusion
The point of intersection of the line with the XY plane is \( \left( \frac{7}{3}, -\frac{8}{3}, 0 \right) \).

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