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

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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