Question

What is the equation to the straight line joining the origin to the point of intersection of the lines \(\dfrac{x}{a}+\dfrac{y}{b}=1\) and \(\dfrac{x}{b} + \dfrac{y}{a}=1 \ ?\)
1. x + y = 0
2. x + y + 1 = 0
3. x - y = 0
4. x + y + 2 = 0

Answer 1

Correct Answer - Option 3 : x - y = 0

Concept:

Equation of family of lines passing through the intersection of two lines S₁ = 0 and S₂ = 0 is given by S₁ + λS₂ = 0

Calculations:

Given, the equation of lines are \(\dfrac{x}{a}+\dfrac{y}{b}=1\) and \(\dfrac{x}{b} + \dfrac{y}{a}=1\)

The equation to the straight line joining the origin to the point of intersection of the lines \(\dfrac{x}{a}+\dfrac{y}{b}=1\) and \(\dfrac{x}{b} + \dfrac{y}{a}=1 \) is 

\((\dfrac{x}{a}+\dfrac{y}{b}-1) + λ (\dfrac{x}{b} + \dfrac{y}{a}-1) = 0\)    ....(1)

which is passing through the origin.

⇒ \((\dfrac{0}{a}+\dfrac{0}{b}-1) - λ (\dfrac{0}{b} + \dfrac{0}{a}-1) = 0\)

⇒ \(λ = -1\)

Equation of line becomes, 

⇒ \((\dfrac{x}{a}+\dfrac{y}{b}-1) +(-1) (\dfrac{x}{b} + \dfrac{y}{a}-1) = 0\)

⇒ \((\dfrac{x}{a}+\dfrac{y}{b}-1)-(\dfrac{x}{b} + \dfrac{y}{a}-1) = 0\)

⇒ \(\dfrac{x}{a}+\dfrac{y}{b}-1 - \dfrac{x}{b} - \dfrac{y}{a}+1 = 0\)

⇒ \(\rm x(\dfrac{1}{a}-\dfrac{1}{b})- y(\dfrac{1}{a} - \dfrac{1}{b}) = 0\)

⇒ x - y = 0

Hence, the equation to the straight line joining the origin to the point of intersection of the lines \(\dfrac{x}{a}+\dfrac{y}{b}=1\) and \(\dfrac{x}{b} + \dfrac{y}{a}=1 \) is x - y = 0.