Question

The image of point \(P(x, y, z)\) with respect to the line: \[ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}, \] is \(P'(1, 0, 7)\). Find the coordinates of point \(P\).

Hint:

To find the coordinates of a point \(P\) whose image with respect to a line is given, use the parametric equation of the line and the midpoint formula. Simplify systematically to solve for the unknowns.

Answer 1

Step 1: Parametric equation of the line. The given line can be expressed in parametric form as: \[ x = t, \quad y = 1 + 2t, \quad z = 2 + 3t, \] where \(t\) is a parameter. 

Step 2: Midpoint of \(P\) and \(P'\). The image \(P'(1, 0, 7)\) of \(P(x, y, z)\) with respect to the given line implies that the midpoint \(M\) of \(P\) and \(P'\) lies on the line. 
The coordinates of the midpoint \(M\) are: \[ M = \left(\frac{x + 1}{2}, \frac{y + 0}{2}, \frac{z + 7}{2}\right). \] 

Step 3: Condition for \(M\) lying on the line. Since \(M\) lies on the line, its coordinates must satisfy the parametric equations of the line. Thus: \[ \frac{x + 1}{2} = t, \quad \frac{y}{2} = 1 + 2t, \quad \frac{z + 7}{2} = 2 + 3t. \] 

Step 4: Solve for \(t\). From the first equation: \[ t = \frac{x + 1}{2}. \] Substitute \(t\) into the second equation: \[ \frac{y}{2} = 1 + 2\left(\frac{x + 1}{2}\right). \] 

Simplify: \[ \frac{y}{2} = 1 + x + 1 \implies \frac{y}{2} = x + 2 \implies y = 2x + 4. \tag{1} \] 

Substitute \(t\) into the third equation: \[ \frac{z + 7}{2} = 2 + 3\left(\frac{x + 1}{2}\right). \] 

Simplify: \[ \frac{z + 7}{2} = 2 + \frac{3x + 3}{2} \implies z + 7 = 4 + 3x + 3 \implies z = 3x. \tag{2} \] 

Step 5: Use the coordinates of \(P'\) to find \(P\). From the midpoint condition: \[ \frac{x + 1}{2} = t, \quad \frac{y + 0}{2} = 1 + 2t, \quad \frac{z + 7}{2} = 2 + 3t. \] 

Substitute \(t = \frac{x + 1}{2}\): \[ \frac{y + 0}{2} = 1 + 2\left(\frac{x + 1}{2}\right). \] 

Simplify: \[ \frac{y}{2} = 1 + x + 1 \implies y = 2x + 4. \tag{3} \] Similarly: \[ \frac{z + 7}{2} = 2 + 3\left(\frac{x + 1}{2}\right). \] Simplify: \[ \frac{z + 7}{2} = 2 + \frac{3x + 3}{2} \implies z = 3x. \tag{4} \] 

Step 6: Solve for \(P\). We now solve for the coordinates of \(P(x, y, z)\). 
Recall the midpoint conditions and verify: From the midpoint: \[ M = \left(\frac{x + 1}{2}, \frac{y}{2}, \frac{z + 7}{2}\right), \] and since \(M\) lies on the line, its coordinates must satisfy: \[ x = t, \quad y = 1 + 2t, \quad z = 2 + 3t. \] 
Using the given image \(P'(1, 0, 7)\) and midpoint relationships: \[ \frac{x + 1}{2} = t, \quad \frac{y}{2} = 1 + 2t, \quad \frac{z + 7}{2} = 2 + 3t. \] 

Substitute \(t = \frac{x + 1}{2}\) into the second and third equations: 1. 

From the second equation: \[ \frac{y}{2} = 1 + 2\left(\frac{x + 1}{2}\right). \] Simplify: \[ \frac{y}{2} = 1 + x + 1 \implies y = 2x + 4. \] 2. From the third equation: \[ \frac{z + 7}{2} = 2 + 3\left(\frac{x + 1}{2}\right). \] 

Simplify: \[ \frac{z + 7}{2} = 2 + \frac{3x + 3}{2} \implies z + 7 = 4 + 3x + 3 \implies z = 3x. \] Thus, the coordinates of \(P(x, y, z)\) are determined as: \[ P(x, y, z) = (0, 4, 0). \] 

Final Answer: The coordinates of \(P\) are: \[ P(x, y, z) = (0, 4, 0). \]