Question:

The equation of the line passing through the point \( (2, -3) \) and perpendicular to the segment joining the points \( (1, 2) \) and \( (-1, 5) \) is

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For a line perpendicular to another, use the negative reciprocal of the original slope. Then, use the point-slope form to find the equation.
Updated On: Apr 28, 2025
  • \( 2x - 3y - 13 = 0 \)
  • \( 2x - 3y - 7 = 0 \)
  • \( 2x - 3y - 9 = 0 \)
  • \( 2x - 3y - 11 = 0 \)
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The Correct Option is A

Solution and Explanation


The slope of the line joining the points \( (1, 2) \) and \( (-1, 5) \) is: \[ m = \frac{5 - 2}{-1 - 1} = \frac{3}{-2} = -\frac{3}{2}. \] The slope of the line perpendicular to this line will be the negative reciprocal, i.e., \( \frac{2}{3} \). Now, using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \) with the point \( (2, -3) \) and slope \( \frac{2}{3} \): \[ y - (-3) = \frac{2}{3}(x - 2). \] Simplifying: \[ y + 3 = \frac{2}{3}(x - 2) \quad \Rightarrow \quad 3(y + 3) = 2(x - 2) \quad \Rightarrow \quad 3y + 9 = 2x - (4) \] Rearranging: \[ 2x - 3y - 13 = 0. \]
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