Q: A variable line is drawn through O, to cut two fixed straight lines L_{1} and L_{2} in A_{1} and A_{2}, respectively. A point A is taken on the variable line such that $\displaystyle \frac{m+n}{OA} = \frac{m}{OA_1} + \frac{n}{OA_2} $ . Show that the locus of A is a straight line passing through the point of intersection of L_{1} and L_{2} , where O is being the origin.

Sol. Let the variable line passing through the origin is

$\displaystyle \frac{x}{cos\theta} = \frac{y}{sin\theta} = r_i $ ….(i)

Let the equation of the line L_{1} is p_{1} x + q_{1} y = 1 …. (ii)

Equation of the line L_{2} is p_{2} x + q_{2} y = 1 …. (iii)

Let the variable line intersects the line (ii) at A_{1} and (iii) at A_{2}

Let OA_{1} = r_{1}

Then A_{1} = (r_{1} cosθ , r_{1} sinθ)

A_{1} lies on L_{1}

$\displaystyle r_1 = OA_1 = \frac{1}{p_1 cos\theta + q_1 sin\theta}$

$\displaystyle r_2 = OA_2 = \frac{1}{p_2 cos\theta + q_2 sin\theta}$

Let OA = r

Let co-ordinate of A are (h, k)

(h, k) = (r cosθ , r sinθ)

$\displaystyle \frac{m+n}{OA} = \frac{m}{OA_1} + \frac{n}{OA_2} $

$\displaystyle \frac{m+n}{r} = \frac{m}{r_1} + \frac{n}{r_2} $

m + n = m (p_{1} rcosθ + q_{1} rsinθ ) + n(p_{2} rcosθ + q_{2} rsinθ)

(p_{1} h + q_{1} k – 1) + (n/m)(p_{2} h + q_{2} k – 1) = 0

Therefore, locus of A is (p_{1} x + q_{1} y -1) + (n/m)(p_{2} x + q_{2} y -1) = 0

L_{1} + λ L_{2} = 0 where λ = n/m

This is the equation of the line passing through the intersection of L_{1} and L_{2}.

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