Question

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

• A. $(\frac{5}{3},0)$
• B. $(\frac{1}{3},0)$
• C. $(3,0)$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

• Consider the reflection of point
  

PP in the x-axis,P′.P′.

• Then, clearly
  

P′≡(1,−1)P′≡(1,−1).

• Further, because
  

PP was reflected across the x-axis to get P′P′, the x-axis is the perpendicular bisector of PP′.PP′.

• Hence, for any point
  

RR lying on it,

RP=RP′RP=RP′

Adding RQRQ to both sides,

RP+RQ=RP′+RQRP+RQ=RP′+RQ

The sum (RP′(RP′+RQ)+RQ) is minimised when RR lies on the line joining P′P′ and Q.Q.

Now, the equation of the line joining P′(1,−1)P′(1,−1) and Q(3,2)Q(3,2) is,

y+1=((2+1)/(3-1))(x-1)

y=-1+(3/2)x-(3/2)

y=(3/2)x-(5/2)

y=32x−The point

The Point RR lying on x-axis, has y-coordinate zero, so let it have coordinates (a,0).(a,0).

Since, RR lies on P′QP′Q , it should satisfy the equation of P′Q.P′Q.

Thus,

0=(3/2)a-(5/2)

a=(5/3)

Therefore Point= R(5/3,0)