To find the image of the point \( P(1, 0, 0) \) with respect to the line given by the equations:
\[
\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8}
\]
we can follow these steps:
### Step 1: Parametrize the line
Let \( \lambda \) be the parameter. From the line equations, we can express the coordinates \( (x, y, z) \) in terms of \( \lambda \):
\[
x = 2\lambda + 1
\]
\[
y = -3\lambda - 1
\]
\[
z = 8\lambda - 10
\]
### Step 2: Find a point on the line
We can denote a point on the line as \( Q(2\lambda + 1, -3\lambda - 1, 8\lambda - 10) \).
### Step 3: Find the direction ratios of the line
The direction ratios of the line can be derived from the coefficients of \( \lambda \):
\[
\text{Direction ratios} = (2, -3, 8)
\]
### Step 4: Find the direction ratios of line PQ
Let \( R(x, y, z) \) be the image of point \( P \). The direction ratios of the line \( PQ \) can be expressed as:
\[
PQ = (2\lambda + 1 - 1, -3\lambda - 1 - 0, 8\lambda - 10 - 0) = (2\lambda, -3\lambda - 1, 8\lambda - 10)
\]
### Step 5: Set up the perpendicularity condition
Since \( PQ \) is perpendicular to the line, the dot product of the direction ratios of \( PQ \) and the line must equal zero:
\[
(2, -3, 8) \cdot (2\lambda, -3\lambda - 1, 8\lambda - 10) = 0
\]
This gives us the equation:
\[
2(2\lambda) + (-3)(-3\lambda - 1) + 8(8\lambda - 10) = 0
\]
### Step 6: Simplify the equation
Expanding this, we have:
\[
4\lambda + 9\lambda + 3 + 64\lambda - 80 = 0
\]
Combining like terms:
\[
77\lambda - 77 = 0
\]
### Step 7: Solve for \( \lambda \)
From this, we find:
\[
\lambda = 1
\]
### Step 8: Substitute \( \lambda \) back to find point \( Q \)
Substituting \( \lambda = 1 \) back into the equations for \( Q \):
\[
Q = (2(1) + 1, -3(1) - 1, 8(1) - 10) = (3, -4, -2)
\]
### Step 9: Find the midpoint of \( P \) and \( R \)
Let \( R(x, y, z) \) be the image of point \( P \). The midpoint \( M \) of \( P \) and \( R \) is given by:
\[
M = \left( \frac{1 + x}{2}, \frac{0 + y}{2}, \frac{0 + z}{2} \right) = Q
\]
Setting this equal to \( Q(3, -4, -2) \):
1. For \( x \):
\[
\frac{1 + x}{2} = 3 \implies 1 + x = 6 \implies x = 5
\]
2. For \( y \):
\[
\frac{0 + y}{2} = -4 \implies y = -8
\]
3. For \( z \):
\[
\frac{0 + z}{2} = -2 \implies z = -4
\]
### Final Answer
Thus, the image of the point \( P(1, 0, 0) \) with respect to the line is:
\[
R(5, -8, -4)
\]
