Question:
P is a point on \( x + y + 5 = 0 \), whose perpendicular distance from \( 2x + 3y + 3 = 0 \) is \( \sqrt{13} \), then the coordinates of P are:
P is a point on \( x + y + 5 = 0 \), whose perpendicular distance from \( 2x + 3y + 3 = 0 \) is \( \sqrt{13} \), then the coordinates of P are:
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- The perpendicular distance formula is useful for solving constraints on points lying on lines.
- Always solve for \( y \) in terms of \( x \) when given a line equation to simplify calculations.
Updated On: May 8, 2025
- \( (20, -25) \)
- \( (1, -6) \)
- \( (-6,1) \)
- \( (\sqrt{13}, -5 - \sqrt{13}) \)
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The Correct Option is B
Solution and Explanation
Step 1: Equation of the line containing point \( P \). We are given the equation \( x + y + 5 = 0 \), which simplifies to \( y = -x - 5 \).
Step 2: Perpendicular distance formula. The formula for the perpendicular distance from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is: \[ {Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \] Substituting the values, we get: \[ \frac{|2x_1 + 3y_1 + 3|}{\sqrt{2^2 + 3^2}} = \sqrt{13}. \] Solving this gives two possibilities for the equation.
Step 3: Solve for the coordinates of \( P \). We solve the system of equations for both cases and find that the coordinates of \( P \) are \( (1, -6) \).
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