Question:
If the image of the point \( (3, 8) \) in the line \( x + 3y = 7 \) is \( (\alpha, \beta) \), then \( \alpha + \beta = \)
If the image of the point \( (3, 8) \) in the line \( x + 3y = 7 \) is \( (\alpha, \beta) \), then \( \alpha + \beta = \)
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Use the formula for the image of a point to reflect it over the given line. The image of the point is equidistant from the line.
Updated On: May 13, 2025
- \( -1 \)
- \( 3 \)
- \( -5 \)
- \( -9 \)
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The Correct Option is C
Solution and Explanation
Step 1: Use the formula for the image of a point.
The formula for the image of a point \( (x, y) \) with respect to a line \( ax + by + c = 0 \) is: \[ x' = x - \frac{2a(ax + by + c)}{a^2 + b^2}, \quad y' = y - \frac{2b(ax + by + c)}{a^2 + b^2} \]
Step 2: Substitute the given values.
For the line \( x + 3y - 7 = 0 \), \( a = 1 \), \( b = 3 \), and \( c = -7 \). The point is \( (3, 8) \). \[ x' = 3 - \frac{2(1)(20)}{1^2 + 3^2} = -1, \quad y' = 8 - \frac{2(3)(20)}{1^2 + 3^2} = -4 \]
Step 3: Find \( \alpha + \beta \).
\[ \alpha + \beta = -1 + (-4) = -5 \] Final Answer: \[ \boxed{-5} \]
The formula for the image of a point \( (x, y) \) with respect to a line \( ax + by + c = 0 \) is: \[ x' = x - \frac{2a(ax + by + c)}{a^2 + b^2}, \quad y' = y - \frac{2b(ax + by + c)}{a^2 + b^2} \]
Step 2: Substitute the given values.
For the line \( x + 3y - 7 = 0 \), \( a = 1 \), \( b = 3 \), and \( c = -7 \). The point is \( (3, 8) \). \[ x' = 3 - \frac{2(1)(20)}{1^2 + 3^2} = -1, \quad y' = 8 - \frac{2(3)(20)}{1^2 + 3^2} = -4 \]
Step 3: Find \( \alpha + \beta \).
\[ \alpha + \beta = -1 + (-4) = -5 \] Final Answer: \[ \boxed{-5} \]
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