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Question

If a variable line drawn through the intersection of the lines $$\dfrac {x}{3} + \dfrac {y}{4} = 1$$ and $$\dfrac {x}{4} + \dfrac {y}{3} = 1$$, meets the coordinate axes at A and B, $$(A\neq B)$$, then the locus of the midpoint of AB is:


A
7xy=6(x+y)
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B
14(x+y297(x+y)+168=0
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C
4(x+y)228(x+y)+49=0
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D
6xy=7(x+y)
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Solution

The correct option is A $$7xy = 6(x + y)$$
$$\dfrac {x}{3} + \dfrac {y}{4} = 1$$ and $$\dfrac {x}{4} + \dfrac {y}{3} = 1$$

These can be simplified as
$$4x+3y=12$$ --(1)
$$3x+4y=12$$ --(2)
Multiply equation (1) by $$3$$ and (2) by $$4,$$
$$12x+9y=36$$
$$12x+16y=48$$
Eliminating $$x,$$ we get
$$y=\dfrac{12}{7}$$
Putting $$y$$ in equation (1), $$x=\dfrac{12}{7}$$
A variable line passes through $$(\dfrac{12}{7},\dfrac{12}{7})$$ and meets coordinate  axis at $$A$$ and $$B,$$ 

Let the $$y-\dfrac{12}{7}=m(x-\dfrac{12}{7})$$

Point A is $$(\dfrac{12}{7}(1-\dfrac{1}{m}),0)$$ 

Point B is $$(0,\dfrac{12}{7}(1-{m}))$$ 

midpoint is $$(\dfrac{6}{7}(1-\dfrac{1}{m}),\dfrac{6}{7}(1-{m}))$$

$$h=\dfrac{6}{7}(1-\dfrac{1}{m})$$

$$k=\dfrac{6}{7}(1-{m})$$  

Eliminate $$m,$$ we get locus as
$$7hk = 6(h+k)$$
$$7xy=6(x+y)$$

Maths

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