If a straight line drawn through the…

Single correct+4 / -1Answer matches NTA’s final key

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:

$x + y - 7 = 0$
$x + y - 14xy = 0$
$2x + y + 14xy = 0$
$x + 2y - 14xy = 0$

Solution

Subtracting the two equations gives $x = y$ , and then $7x = 1$ : the lines meet at $\left(\frac17, \frac17\right)$ .

Let the midpoint of $PQ$ be $(h, k)$ . Then $P = (2h, 0)$ , $Q = (0, 2k)$ and the line is $\frac{x}{2h} + \frac{y}{2k} = 1$ . It passes through $\left(\frac17, \frac17\right)$ :

$\frac{1}{14h} + \frac{1}{14k} = 1 \;\Rightarrow\; h + k = 14hk.$

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

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Co-ordinate Geometry

Solution

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