IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

Conjurer

In an orthonormal coordinate system(x,y plane),one considers the point a=(12,0), a variable point b on the line y = x and a variable point c on the line x=24.If triangle abc has to have minimum perimeter.Then the coordinates of b are:

kishore.subramanian.b

A:(12,0) B:(x,x) C:(24,y)

Area of triangle = 1/2 (12(x-y)+x(y-0)+24(0-y))

A = 6x-6y+xy/2-12x

Partial differentiate w.r.t x

A/

x=y/2-6 =0

y=3

Partial differentiate w.r.t y

A/

y=x/2-6 =0

x=3

Hence A:(12,0) B:(3,3) C:(24,3)

Rate if correct......

Conjurer

Answer is (9,9)

sboosy

the fixed pt is 12,0 (A)

another pt C lies on x=24

let us reduce distance CA by placing C at perpendicular to A

that is C becomes 24,0

now the other pt B lies on y=x

Consider the perpendicular to y=x ...which passes through A

the pt will be nothing but 6,6

now although this is shortest from A ..from C it is lon

Consider the perpendicular to y=x which passes through C

the pt will be nothing but 12,12

now although this is shortest to C it will be far from A

so the pt required is going to be midpt of 6,6 and 12,12

lying on the line y=x..because as we have seen the sum of the distance of

this line from the 2 pts A and C are going to increase on either side of this pt

and this pt as it turns out to be is

9,9

kishore.subramanian.b

is there any other method to solve the question.........Can't be my method be used.....where have i gone wrong..........please tell......

akhil_o

@ kishore subramaniam...u have minimised AREA not perimeter as the qn requires

kishore.subramanian.b

ops....thanks akhil

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