Question:medium

Let point \( Q \) be the image of point \( P(2,-1) \) in the line \[ 3x+5=4y. \] Find the area of the circle that has the segment \( PQ \) as the diameter.

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If a point is reflected across a line, the distance between the point and its image equals twice the perpendicular distance from the point to the line.
Updated On: May 20, 2026
  • \(9\pi\)
  • \(1.96\pi\)
  • \(36\pi\)
  • \(3\pi\)
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The Correct Option is A

Solution and Explanation

Understanding the Concept: When a point is reflected in a line, the reflecting line becomes the perpendicular bisector of the segment joining the original point and its image. Therefore, the perpendicular distance from the point to the line is exactly half the distance between the point and its image. The equation of the given line is: \[ 3x-4y+5=0 \] The point is: \[ P(2,-1) \] We first find the perpendicular distance of \(P\) from the line.
Step 1: Use the perpendicular distance formula. The perpendicular distance of point \((x_1,y_1)\) from line \[ Ax+By+C=0 \] is given by: \[ d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \] Here, \[ A=3,\quad B=-4,\quad C=5 \] Substituting \(P(2,-1)\): \[ d=\frac{|3(2)-4(-1)+5|}{\sqrt{3^2+(-4)^2}} \] \[ =\frac{|6+4+5|}{\sqrt{9+16}} \] \[ =\frac{15}{5} \] \[ =3 \] Thus, the perpendicular distance from \(P\) to the mirror line is: \[ 3 \]
Step 2: Find the length of \(PQ\). Since the line is the perpendicular bisector, \[ PQ=2\times3=6 \] Therefore, the diameter of the circle is: \[ 6 \] Hence the radius is: \[ r=\frac{6}{2}=3 \]
Step 3: Find the area of the circle. Area of a circle: \[ \pi r^2 \] Substituting \(r=3\): \[ \pi(3)^2 \] \[ =9\pi \] Hence, \[ \boxed{9\pi} \]
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