First slide

Equation of line in Straight lines

Question

$The equation of line belonging to the family of lines p (2 x + 3 y - 13) + q (x - y + 1) = 0,$
$and which is at maximum distance from the orioin is$

Moderate

A

$4 x + 3 y - 17 = 0$
B

$3 x + 2 y - 12 = 0$
C

$2 x + 3 y - 13 = 0$
D

$x + y - 5 = 0$

Solution

$\begin{array}{l} Given family of lines p (2 x + 3 y - 13) + q (x - y + 1) = 0 \\ \Rightarrow 2 x + 3 y - 13 + \frac{q}{p} (x - y + 1) = 0 \\ \Rightarrow 2 x + 3 y - 13 + λ (x - y + 1) = 0 (∵ Let \frac{p}{q} = λ) \\ \Rightarrow x (2 + λ) + y (3 - λ) + λ - 13 = 0 \dots \dots \dots . . . (1) \\ The distance from origin to (1) is D = \frac{| λ - 13 |}{\sqrt{(2 + λ)^{2} + (3 - λ)^{2}}} \end{array}$
$\begin{array}{l} = \frac{| λ - 13 |}{\sqrt{2 λ^{2} - 2 λ + 13}} \\ {Differentiate with respect to}^{'} λ^{'} \\ \Rightarrow \frac{d D}{d λ} = \frac{\sqrt{2 λ^{2} - 2 λ + 13} - (λ - 13) \frac{(4 λ - 2)}{2 (\sqrt{2 λ^{2} - 2 λ + 13})}}{(2 λ^{2} - 2 λ + 13)} \\ = \frac{2 λ^{2} - 2 λ + 13 - (λ - 13) (2 λ - 1)}{{(2 λ^{2} - 2 λ + 13)}^{\frac{3}{2}}} \end{array}$
For maximum value of D ,
$\begin{array}{l} \frac{d D}{d λ} = 0 \\ \Rightarrow 2 λ^{2} - 2 λ + 13 = (λ - 13) (2 λ - 1) \\ \Rightarrow λ = 0 \\ Substituting λ = 0 in equation (1) we get 2 x + 3 y - 13 = 0 \end{array}$

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