JEE Main 2026 Apr 8 Shift 2, Mathematics Q10: Points and Straight Lines

JEE Main2026Apr 8, Shift 2Mathematics

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

A
$x + y - 7 = 0$
B
$x + y - 14 x y = 0$
C
$2 x + y + 14 x y = 0$
D
$x + 2 y - 14 x y = 0$

Solution

Adding the two given lines: $7 x + 7 y = 2$ , so the intersection point is $(\frac{1}{7}, \frac{1}{7})$ .

Let the variable line meet the axes at $P (2 h, 0)$ and $Q (0, 2 k)$ , so that $(h, k)$ is the midpoint. The intercept form is $\frac{x}{2 h} + \frac{y}{2 k} = 1$ .

This line passes through $(\frac{1}{7}, \frac{1}{7})$ :

$\frac{1}{14 h} + \frac{1}{14 k} = 1 ⟹ \frac{1}{h} + \frac{1}{k} = 14$

Replacing $(h, k)$ with $(x, y)$ : $\frac{1}{x} + \frac{1}{y} = 14$ , i.e. $x + y - 14 x y = 0$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q10: Points and Straight Lines

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