Question

P and Q are the points of trisection of the line segment joining the points \( (3, -7) \) and \( (-5,3) \). If line PQ subtends a right angle at a variable point R, then the locus of R is:

• A. A circle with radius \( \frac{\sqrt{41}}{3} \)
• B. A circle with radius \( \sqrt{409} \)
• C. A pair of straight lines passing through \((-1, -2)\)
• D. A pair of straight lines passing through \( (1, 2) \)

Hint:

For problems involving trisection points and perpendicularity, remember that the locus of a point subtending a right angle to a chord is a circle. Calculate the midpoint and radius of the chord to find the center and radius of the circle.

Answer 1

The Correct Option is A

Step 1: Find the points of trisection, P and Q. Given the points \( A = (3, -7) \) and \( B = (-5, 3) \), we can calculate the coordinates of the trisection points \( P \) and \( Q \) by dividing the segment \( AB \) into three equal parts. The formula for finding the trisection points is as follows: \[ P = \left( \frac{2A + B}{3} \right), \quad Q = \left( \frac{A + 2B}{3} \right) \] Substitute the coordinates of \( A \) and \( B \): \[ P = \left( \frac{2(3, -7) + (-5, 3)}{3} \right) = \left( \frac{(6, -14) + (-5, 3)}{3} \right) = \left( \frac{1, -11}{3} \right) = \left( \frac{1}{3}, -\frac{11}{3} \right) \] \[ Q = \left( \frac{(3, -7) + 2(-5, 3)}{3} \right) = \left( \frac{(3, -7) + (-10, 6)}{3} \right) = \left( \frac{-7, -1}{3} \right) = \left( -\frac{7}{3}, -\frac{1}{3} \right) \] Thus, the coordinates of \( P \) and \( Q \) are: \[ P = \left( \frac{1}{3}, -\frac{11}{3} \right), \quad Q = \left( -\frac{7}{3}, -\frac{1}{3} \right) \] 

Step 2: Use the condition that line PQ subtends a right angle at point R. To find the locus of point \( R \), we use the property that if a line subtends a right angle at a variable point, then the locus of that point is a circle whose diameter is the line segment joining the two points. In this case, the line \( PQ \) subtends a right angle at point \( R \). Therefore, the midpoint of \( PQ \) is the center of the circle, and the radius is half the length of \( PQ \).

Step 3: Calculate the midpoint of PQ. The midpoint \( M \) of the line segment \( PQ \) is given by: \[ M = \left( \frac{\frac{1}{3} + \left(-\frac{7}{3}\right)}{2}, \frac{-\frac{11}{3} + \left(-\frac{1}{3}\right)}{2} \right) \] \[ M = \left( \frac{-\frac{6}{3}}{2}, \frac{-\frac{12}{3}}{2} \right) = \left( -1, -2 \right) \] Thus, the center of the circle is at \( (-1, -2) \).

 Step 4: Calculate the radius of the circle. The radius of the circle is half the length of the segment \( PQ \). The length of \( PQ \) is calculated as the distance between \( P = \left( \frac{1}{3}, -\frac{11}{3} \right) \) and \( Q = \left( -\frac{7}{3}, -\frac{1}{3} \right) \): \[ \text{Length of } PQ = \sqrt{\left( \frac{1}{3} - \left(-\frac{7}{3}\right) \right)^2 + \left( -\frac{11}{3} - \left(-\frac{1}{3}\right) \right)^2} \] \[ = \sqrt{\left( \frac{1}{3} + \frac{7}{3} \right)^2 + \left( -\frac{11}{3} + \frac{1}{3} \right)^2} = \sqrt{\left( \frac{8}{3} \right)^2 + \left( -\frac{10}{3} \right)^2} \] \[ = \sqrt{\frac{64}{9} + \frac{100}{9}} = \sqrt{\frac{164}{9}} = \frac{\sqrt{164}}{3} \] Thus, the radius of the circle is: \[ \text{Radius} = \frac{\frac{\sqrt{164}}{3}}{2} = \frac{\sqrt{164}}{6} \] The square of the radius is: \[ \left( \frac{\sqrt{164}}{6} \right)^2 = \frac{164}{36} = \frac{41}{9} \] 

Step 5: Write the equation of the locus of R. The equation of the circle is: \[ (x + 1)^2 + (y + 2)^2 = \frac{41}{9} \] Thus, the radius is \( \frac{\sqrt{41}}{3} \), and the equation of the locus of \( R \) is a circle with radius \( \frac{\sqrt{41}}{3} \). Thus, the correct answer is \( \boxed{ \frac{\sqrt{41}}{3}} \).