First slide

Equation of line in Straight lines

Question

$A variable line ' L' is drawn through O (0, 0) to meet the lines L_{1} : y - x - 10 = 0 and L_{2} : y - x - 20 = 0 at the$
$points A and B respectively. A point P is taken on ' L^{'} such that \frac{2}{O P} = \frac{1}{O A} + \frac{1}{O B} . Locus of ' P^{'} is$

NA

A

$3 x + 3 y = 40$
B

$3 x + 3 y + 40 = 0$
C

$3 x - 3 y = 40$
D

$3 y - 3 x = 40$

Solution

Let the parametric equation of drawn line is
$\begin{aligned} \frac{x}{\cos θ} = \frac{y}{\sin θ} = r \\ \Rightarrow x = r \cos θ, y = r \sin θ \end{aligned}$
$\begin{aligned} Putting it in ' L_{1}', we get \\ r \sin θ = r \cos θ + 10 \\ \Rightarrow \frac{1}{O A} = \frac{\sin θ - \cos θ}{10} \end{aligned}$
Similarly, putting the general point of drawn line is the equation of L₂, we get
$\frac{1}{O B} = \frac{\sin θ - \cos θ}{20}$
$\begin{array}{l} Let P = (h, k) and O P = r \Rightarrow r \cos θ = h, r \sin θ = k, we have \\ \frac{2}{r} = \frac{\sin θ - \cos θ}{10} + \frac{\sin θ - \cos θ}{20} \\ \Rightarrow 40 = 3 r \sin θ - 3 r \cos θ \\ \Rightarrow 3 y - 3 x = 40 \end{array}$

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