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Let the equation of any line passing through the point of intersection of the given lines be
(x+2y-1)+a×(2x-y-1)=0
Reducing the equation to its intercept form…
(x×(1+2a)/(1+a))+(y×(2-a)/(1+a))=1
Therefore coordinates of A and B,where this line meets the coordinate axes respectively…
A=((1+a)/(1+2a),0) on x axis and,
B=(0,(1+a)/(2-a)) on y axis…
Mid point of AB=((1+a)/(2+4a),(1+a)/(4-2a)
Now we find the locus of this point by eliminating ‘a’ between the two expressions
x=(1+a)/(2+4a) &
y=(1+a)/(4-2a)
y=x/(10x-3)……..required locus …
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