Let P be #(5,3)# and a point R on #y=x# and Q on x-axis are such that #PQ + QR + RP# is minimum. The the coordinates of Q are?

A)#(17/4,0)#
B)#(17,0)#
C)#(17/2,0)#
D)#(3,0)#

2 Answers
Cesareo R.
Sep 28, 2017

See below.

Explanation:

#P = (5,3)#
#R = (x_1,x_1)#
#Q = (x_2,0)#

#PQ = (5-x_2,3)#
#QR = (x_2-x_1,-x_1)#
#RP = (x_1-5, x_1-3)#

and now

#f(x_1,x_2)=abs(PQ)+abs(QR)+abs(RP) = sqrt((5-x_2)^2+3^2)+sqrt((x_2-x_1)^2+x_1^2)+sqrt((x_1-5)^2+(x_1-3) ^2))#

Now the determination of #min f(x_1,x_2)# furnishes
#x_1 =17/5, x_2 =17/4# and #f(17/5,17/4)=8.246#

so the answer is option #A#

CW
Sep 29, 2017

A) #(17/4,0)#

Explanation:

Another approach to solve this problem is given below.
enter image source here
See Fig 1.
The rule for a reflection of a point over the line #y=x# is #(x,y) -> (y,x)#
1) reflect #P(5,3)# over the line #y=x#
#=> P(5,3) -> P'(3,5)#
so long as #R# moves along the line #y=x, color(red)(RP'=RP)#
The rule for a reflection of a point over the x-axis is #(x,y) -> (x,-y)#
2) reflect #P(5,3)# over the x-axis,
#=> P(5,3) -> P''(5,-3)#
so long as #Q# moves along the x-axis, #color(red)(P''Q=PQ)#
See Fig 2.
When #P'',Q,R, and P'# lie on a straight line,
minimum value of #PQ+QR+RP (=P''Q+QR+RP')# can be obtained.
Let coordinates of #Q# be #(x,0)#
slope of #P'P''= (-3-5)/(5-3)=-4#
#=># slope of #P'Q=(0-5)/(x-3)=-4#
#=> x=(17/4)#

Hence, coordinates of #Q=(17/4,0)#

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