JEE Main 2026 — Mathematics Coordinate Geometry

The point of intersection of the lines $4x+3y-1=0$4x+3y−1=0 and $3x+4y-1=0$3x+4y−1=0 is obtained by solving them simultaneously.

Adding the two equations gives $7x+7y-2=0 \Rightarrow x+y=\dfrac{2}{7}$7x+7y−2=0⇒x+y=72​.

Subtracting the two equations gives $x-y=0 \Rightarrow x=y$x−y=0⇒x=y.

Thus, the point of intersection is $\left(\dfrac{1}{7}, \dfrac{1}{7}\right)$(71​,71​).

Let the equation of the line passing through this point and meeting the coordinate axes at $P$P and $Q$Q be $\dfrac{x}{a} + \dfrac{y}{b} = 1$ax​+by​=1.

Since the line passes through $\left(\dfrac{1}{7}, \dfrac{1}{7}\right)$(71​,71​), we have:

$$\dfrac{1}{7a} + \dfrac{1}{7b} = 1 \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} = 7$$7a1​+7b1​=1⇒a1​+b1​=7

The coordinates of the points where the line meets the axes are $P(a, 0)$P(a,0) and $Q(0, b)$Q(0,b).

Let $(h, k)$(h,k) be the midpoint of $PQ$PQ. Then $h = \dfrac{a}{2}$h=2a​ and $k = \dfrac{b}{2}$k=2b​, which gives $a = 2h$a=2h and $b = 2k$b=2k.

Substituting $a$a and $b$b into the relation $\dfrac{1}{a} + \dfrac{1}{b} = 7$a1​+b1​=7, we get:

$$\dfrac{1}{2h} + \dfrac{1}{2k} = 7$$2h1​+2k1​=7

$$\dfrac{h+k}{2hk} = 7$$2hkh+k​=7

$$h+k = 14hk$$h+k=14hk

Replacing $(h, k)$(h,k) with $(x, y)$(x,y), the locus of the midpoint is:

$$x+y-14xy=0$$x+y−14xy=0

Answer: $x+y-14xy=0$x+y−14xy=0