If a straight line drawn through the point of intersection of the lines 4x + 3y – 1 = 0 and 3x + 4y – 1 = 0, meets the coordinate axes at the points P and Q, then the locus of the midpoint of PQ is: [2026]

Question

If a straight line drawn through the point of intersection of the lines 4x + 3y – 1 = 0 and 3x + 4y – 1 = 0, meets the coordinate axes at the points P and Q, then the locus of the midpoint of PQ is: [2026]

Smart Achievers · Accepted Answer

(2)

$L_{1} : 4 x + 3 y - 1 = 0$

$L_{2} : 3 x + 4 y - 1 = 0$

$L : L_{1} + λ L_{2} = 0$

$\Rightarrow (4 x + 3 y - 1) + λ (3 x + 4 y - 1) = 0$

$\Rightarrow (4 + 3 λ) x + (3 + 4 λ) y - (1 + λ) = 0$

Point of intersection with x-axis:

$P (\frac{λ + 1}{3 λ + 4}, 0)$

Point of intersection with y-axis:

$Q (0, \frac{λ + 1}{3 + 4 λ})$

Let mid-point of PQ be (h, k)

$h = \frac{\frac{λ + 1}{3 λ + 4} + 0}{2} = \frac{λ + 1}{6 λ + 8}$ ... (i)

$k = \frac{0 + \frac{λ + 1}{3 + 4 λ}}{2} = \frac{λ + 1}{8 λ + 6}$ ... (ii)

From (i), $λ = \frac{1 - 8 h}{6 h - 1}$

Put ' $λ$ ' in eq (ii), we get

$k = \frac{\frac{1 - 8 h}{6 h - 1} + 1}{8 (\frac{1 - 8 h}{6 h - 1}) + 6}$

$\Rightarrow h + k - 14 h k = 0$

$∴$ Locus is x + y – 14xy = 0

Solution

Q.
If a straight line drawn through the point of intersection of the lines 4x + 3y – 1 = 0 and 3x + 4y – 1 = 0, meets the coordinate axes at the points P and Q, then the locus of the midpoint of PQ is: [2026]

1 x + y – 7 = 0

2 x + y – 14xy = 0

3 2x + y + 14xy = 0

4 x + 2y – 14xy = 0

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