If a straight line drawn through the point of intersection of the lines 4x+3y−1=0 and 3x+4y−1=0, meet the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is:
(A)x+y−7=0
(B)x+y−14xy=0✓ Correct
(C)2x+y+14xy=0
(D)x+2y−14xy=0
Answer: (B)
Solution
Solving 4x+3y=1 and 3x+4y=1 gives the common point (71,71). A line meeting the axes at P(a,0) and Q(0,b) is ax+by=1; passing through (71,71) it satisfies 7a1+7b1=1, that is a1+b1=7. The midpoint (h,k) of PQ has a=2h, b=2k, so
2h1+2k1=7⟹h1+k1=14⟹h+k=14hk.
Replacing (h,k) by (x,y), the locus is x+y−14xy=0.
JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option