1708Thanks Shubhodip.
got it
i have a doubt in finding Min. value of point MA+MB where M(x,0) and A(3/2,1) and A(3/2,1)
my Confusion is that why We can not take MA+MB>=AB
and answer given is take Image of A at A(3/2,-1). Then find
MA'+MB>=AB
and please give the condition in which we can apply Triangle Inequality
like in angle M,A,C
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6 Answers
36Apart from co-ordinate...
if MAB is a triangle then, MA + MB > AB and in no case can it equal AB.
21You can of course take MA + MB≥AB, but since M can not lie on the segment AB, that will not help to find the minimum, it will be a strict inequality,for exapmle when B≥1, B>-5 is true, but useless.
In order to solve the problem, reflect A, about the X-axis, so that its image is A'. Now MA+ MB= MA'+ MB≥A'B. With equality iff M lies on A'B.(degenerate triangle). Observe that BA' intersects the X-axis at some point, say T. Choose T=M,and we have equality. So the minimum value MA+ MB = |A'B|
21We have learnt that of all the triangles of same height and base ,the perimeter of the isosceles triangle is the least.
So find the eqn. of the line perpendicular bisector of AB. You get M. Then find AM+BM.
I got M≡(5/2,0) and AM+BM=2√2