Published by:
CGP EDU Academic Team
Published on: August 12, 2026
Given two intersecting straight lines inclined at an angle α , intersecting at the origin, MA and MB are perpendicular from variable point M (2t –1, t + 2) on the two given intersecting straight lines meeting at N, then the locus of N is a -
Text Solution
Verified by ExpertsThe correct answer is:
B
Lines are x = 0, y = mx
A (2t –1, 0)
solving the lines y = mx and
(y – (t–2)) = – 
(x – (2t –1)) we get
Let N(h, k) lies on BN, AN
equation of line BN ≡ k = 
AN ≡ k = m(h – (2t –1))
using value of t, t = 
in BN
2mh – (4 + 2m 2 + m) k + 5m 3 = 0
which is straight line.
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
Equation of a straight line on which length of perpendicular from the origin is four units and the …
If we reduce to the form , then the value of p is
The algebraic sum of the perpendicular distances from the points (2,0), (0, 2) and (1, 1) to a vari…
The line L has intercepts a and b on the co-ordinate axes. Keeping the origin fixed, the co-ordinat…
Let 2x–3y =0 be a given line and P (sin θ θ , 0) and Q (0, cos θ θ ) be the two points. Then P and …
If the line ax + by = 1 passes through point of intersection of y = x tan α + p sec α , y sin(30 0 …