Published by:
CGP EDU Academic Team
Published on: August 14, 2026
The equation of the locus of the foot of perpendicular drawn from (5,6) on the family of lines (x -2) +
(y -3) = 0 (where
R) is
Text Solution
Verified by ExpertsThe correct answer is:
C
(x - 2) (x - 5) + (y - 3) (y - 6) = 0

Let A = (5,6) and the point of concurrency of the family of lines (x - 2) +
(y -3) = 0 is (2,3) = B and foot of the perpendicular from A tothe family of lines is P = (h, k)
Now, PA is perpendicular to PB
(slope of PA) x (slope of PS) = -1

locus is (x - 2) (x - 5) + (y- 3) (y-6) = 0
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The distance of the point (-1, l) from the line 12 (x + 6) = 5 (y - 2) is
If the straight line drawn through the point and inclined at an angle with the x - axis meets the…
The line 3x + 2y = 24 meets the y-axis at A & the x-axis at B. The perpendicular bisector of AB mee…
Let in coordinates of vertex A is (0, 0) . Equation of the internal angle bisector of is x + y -1…
The sum of all the values of | | such that the lines x + 2y - 3 = 0, 3x - y –1 = 0 and 2 x + y - …
If the line segment joining P (2,3) and Q (5,7) subtends a right angle at R (x, y) and the area of …