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Equation of a Straight Line

L1 is a line ...

Question

$L_{1}$ is a line intersecting $x$ and $y$ axes at $A (a, 0)$ and $B (0, b) .$ $L_{2}$ is a line perpendicular to $L_{1}$ intersecting $x$ and $y$ axes at $C$ and $D$ respectiveley. If the area of the triangle $O C D$ is 4 times the area of triangle $O A B$ , then equation of $A D$ is

x + 2 y = 26

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2 x + y + 2 a = 0

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x - 2 y + 2 b = 0

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2 x - y - 2 a = 0

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Solution

The correct option is D $2 x - y - 2 a = 0$
Equation of $A B : \frac{x}{a} + \frac{y}{b} = 1$
$L_{2} ⊥ L_{1} \Rightarrow \frac{x}{b} - \frac{y}{a} = α$
$\Rightarrow \frac{x}{α b} + \frac{y}{- α a} = 1$
$C : (α b, 0), D : (0, - α a)$
$A r (Δ O C D)$
$= ∣ ∣ ∣ \frac{1}{2} \times O C \times O D ∣ ∣ ∣ = ∣ ∣ ∣ \frac{1}{2} \times (α b) (- α a) ∣ ∣ ∣ = \frac{α^{2}}{2} | a b |$

$A r (Δ O A B)$
$= ∣ ∣ ∣ \frac{1}{2} \times O A \times O B ∣ ∣ ∣ = \frac{1}{2} | a b |$
$\Rightarrow \frac{α^{2}}{2} | a b | = 4 \times \frac{1}{2} | a b | \Rightarrow α = \pm 2$
$A (a, 0), D (0, \mp 2 a)$
Line passing through $A$ and $D$
$\Rightarrow \frac{x}{a} + \frac{y}{\mp 2 a} = 1 \Rightarrow 2 x \mp y - 2 a = 0$

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L1 is a line intersecting x and y axes at A(a,0) and B(0,b). L2 is a line perpendicular to L1 intersecting x and y axes at C and D respectiveley. If the area of the triangle OCD is 4 times the area of triangle OAB, then equation of AD is

$L_{1}$ is a line intersecting $x$ and $y$ axes at $A (a, 0)$ and $B (0, b) .$ $L_{2}$ is a line perpendicular to $L_{1}$ intersecting $x$ and $y$ axes at $C$ and $D$ respectiveley. If the area of the triangle $O C D$ is 4 times the area of triangle $O A B$ , then equation of $A D$ is