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Maths Question 10 – JEE-MAIN 2026

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If a straight line drawn through the point of intersection of the lines 4x+3y−1=0 and 3x+4y−1=0, meets the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is :

First, find the coordinates of the point where the two given lines intersect.

Step 1: Find the point of intersection✦ Active

Solve the system of linear equations 4x+3y−1=0 and 3x+4y−1=0. Subtracting the second equation from the first gives (4x+3y−1)−(3x+4y−1)=0⟹x−y=0⟹x=y. Substituting x=y into 4x+3y−1=0 gives 4x+3x−1=0⟹7x=1⟹x=17. Thus, the point of intersection is A(17,17).

Step 2: Formulate the line equation and intercept relation○ Expand

Let the equation of the line passing through A(17,17) and making intercepts a and b on the axes be xa+yb=1. Since the line passes through A, we substitute its coordinates:

1/7a+1/7b=1⟹17a+17b=1⟹1a+1b=7
💡 Teacher's Secret Hint

Remember that P and Q are the intercepts, so their coordinates are (a,0) and (0,b) respectively.

Step 3: Determine the locus of the midpoint○ Expand

Let (h,k) be the midpoint of the segment PQ. The coordinates of P are (a,0) and Q are (0,b). Using the midpoint formula:

h=a+02⟹a=2h and k=0+b2⟹b=2k

Substitute a=2h and b=2k into the relation 1a+1b=7 from Step 2:

12h+12k=7⟹k+h2hk=7⟹h+k=14hk

Replacing (h,k) with (x,y) to find the locus, we get x+y=14xy, or x+y−14xy=0.

💡 Teacher's Secret Hint

Ensure to replace the locus variables (h,k) with (x,y) at the final step.

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