JEE Main 2026 — Straight Lines Question with Solution

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

Adding the two equations gives $7x+7y-2=0\Rightarrow x+y=\frac{2}{7}$.

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

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

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

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

$\frac{1}{7a}+\frac{1}{7b}=1\Rightarrow \frac{1}{a}+\frac{1}{b}=7$

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

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

Substituting $a$ and $b$ into the relation $\frac{1}{a}+\frac{1}{b}=7$, we get:

$\frac{1}{2h}+\frac{1}{2k}=7$

$\frac{h+k}{2hk}=7$

$h+k=14hk$

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

$x+y-14xy=0$

Answer: $x+y-14xy=0$