IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

Harris

Let P=(1,1) and Q=(3,2). The point R on X-axis such that PR+PQ is minimum is

Ans is (5/3,0)

Rates are assured

pannaguma

EDITED.

made a very silly mistake.

iitkgp_bipin

Let any point on x-axis be (x,0).

Apply distance formula :

PR + PQ = {(x-1)²+1²}^1/2 + {(x-3)²+2²}^1/2

PR + PQ = (x²-2x+2)^1/2 + (x²-6x+13)^1/2 = f(x) (let)

At minimum : f '(x) = 0

(1/2)(2x-2).(x²-2x+2)^-1/2 + (1/2)(2x-6)(x²-6x+13)^-1/2 = 0

solve this you'll get (5/3,0) and at this point f '(x) changes its sign from negative to positive.

Hence minima occurs at (5/3,0).

blacklisted_420

take image of say P about the X axis, now the line joining the image and pt Q intersects the X axis in the required point.
image of 1,1 is 1,-1 obviously. so now the line joining 3,2 and 1,-1 has the equation 3x-2y=5 meets X axis at 5/3,0 ...

gr8vipul

NICE APPROACH BY BLACK LISTED
GOOD WRK YAAR !!!!

venkat_tatolu

For these type of problems,

If the P and Q are same side of the line, then finde the image of any

one of P,Q. Say Q' is the image of Q w.r.t the given line(in this problem x-axis)

Then find the equn., of the line PQ' and substitue R(x,0).

We get value of x.

If the points P and Q are either side of the given line,then find the

equn., of PQ and substitute R(x,0).We get value of x.

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