The point of intersection of the lines frac{x | vedclass

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(D) Given lines are: $1) \frac{x}{a} + \frac{y}{b} = 1$ $2) \frac{x}{b} + \frac{y}{a} = 1$ Adding the two equations: $(\frac{x}{a} + \frac{x}{b}) + (\

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(D) Given lines are: $1) \frac{x}{a} + \frac{y}{b} = 1$ $2) \frac{x}{b} + \frac{y}{a} = 1$ Adding the two equations: $(\frac{x}{a} + \frac{x}{b}) + (\frac{y}{b} + \frac{y}{a}) = 2$ $x(\frac{a+b}{ab}) + y(\frac{a+b}{ab}) = 2$ $(x + y)(\frac{a+b}{ab}) = 2 \implies x + y = \frac{2ab}{a+b}$ Subtracting the two equations: $(\frac{x}{a} - \frac{x}{b}) + (\frac{y}{b} - \frac{y}{a}) = 0$ $x(\frac{b-a}{ab}) - y(\frac{b-a}{ab}) = 0$ $(x - y)(\frac{b-a}{ab}) = 0 \implies x = y$ Substituting $x = y$ into $x + y = \frac{2ab}{a+b}$,we get $2x = \frac{2ab}{a+b} \implies x = \frac{ab}{a+b}$. Thus,the intersection point is $(\frac{ab}{a+b}, \frac{ab}{a+b})$. Checking option $(a)$: $x - y = \frac{ab}{a+b} - \frac{ab}{a+b} = 0$. (Correct) Checking option $(b)$: $(x + y)(a + b) = (\frac{ab}{a+b} + \frac{ab}{a+b})(a + b) = (\frac{2ab}{a+b})(a + b) = 2ab$. (Correct) Checking option $(c)$: $(lx + my)(a + b) = (l(\frac{ab}{a+b}) + m(\frac{ab}{a+b}))(a + b) = (l+m)(\frac{ab}{a+b})(a+b) = (l+m)ab$. (Correct) Therefore,all options are correct.

The point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{b} + \frac{y}{a} = 1$ lies on the line

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