Question

If\[P{\text{ }} = {\text{ }}\left( {1,{\text{ }}1} \right),{\text{ }}Q{\text{ }} = {\text{ }}(3,{\text{ }}2)\] and $ R $ is a point on \[x{\text{ }} - {\text{ }}axis\] then the value of \[PR{\text{ }} + {\text{ }}RQ\]will be minimum at
\[1)\] \[\left( {\dfrac{5}{3},{\text{ }}0} \right)\]
\[2)\] \[\left( {\dfrac{1}{3},{\text{ }}0} \right)\]
\[3)\] \[\left( {3,{\text{ }}0} \right)\]
\[4)\] \[\left( {1,{\text{ }}0} \right)\]

Answer 1

Hint: We need to find the point $ R $ such that \[PR{\text{ }} + {\text{ }}RQ\] has the minimum value . We solve this by using the concept of coordinate system and applications of derivatives . Firstly we equate a relation in terms of a variable $ x $ by using the distance formula then by using the concept of increasing and decreasing functions and computing it to zero we find the point $ R $ .

Complete step-by-step answer:
Given :
Let point $ R $ be \[\left( {x{\text{ }},{\text{ }}0} \right)\][ as the point $ R $ lies on \[x - axis\]the value of y - coordinate is zero ]
Point \[P{\text{ }} = {\text{ }}\left( {1{\text{ }},{\text{ }}1} \right)\]and point \[Q{\text{ }} = {\text{ }}\left( {3{\text{ }},{\text{ }}2} \right)\]
Now , using distance formula we get the value of $ PR $ and $ RQ $
We know ,
Distance formula $ = \sqrt {[{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}]} $
Using distance formula ,
 $ PR = \sqrt {[{{(x - 1)}^2} + {{(0 - 1)}^2}]} $
 $ PR = \sqrt {[{{(x - 1)}^2} + 1]} $
Expanding the term , using $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
 $ PR = \sqrt {[{x^2} + 1 - 2x + 1]} $
 $ PR = \sqrt {[{x^2} - 2x + 2]} $
Similarly , using distance formula we find \[RQ\]
 $ RQ = \sqrt {[{{(3 - x)}^2} + {{(2 - 0)}^2}]} $
 $ RQ = \sqrt {[{{(3 - x)}^2} + 4]} $
Expanding the term , using $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
 $ RQ = \sqrt {[{x^2} + 9 - 6x + 4]} $
 $ RQ = \sqrt {[{x^2} - 6x + 13]} $

Let \[y{\text{ }} = {\text{ }}PR{\text{ }} + {\text{ }}RQ\]
 $ y = \sqrt {[{x^2} - 2x + 2]} + \sqrt {[{x^2} - 6x + 13]} $
For minimum value , we differentiate\[\;y\] with respect to
 $ \dfrac{{dy}}{{dx}} = \dfrac{1}{2} \times \dfrac{1}{{\sqrt {[{x^2} - 2x + 2]} }} \times [2x - 2] + \dfrac{1}{2} \times \dfrac{1}{{\sqrt {[{x^2} - 6x + 13]} }} \times [2x - 6] $
Put , \[\dfrac{{dy}}{{dx}}{\text{ }} = {\text{ }}0\]
 $ \dfrac{1}{2} \times \dfrac{1}{{\sqrt {[{x^2} - 2x + 2]} }} \times [2x - 2] + \dfrac{1}{2} \times \dfrac{1}{{\sqrt {[{x^2} - 6x + 13]} }} \times [2x - 6] = 0 $
On simplifying , we get
 $ \dfrac{1}{2} \times \dfrac{1}{{\sqrt {[{x^2} - 2x + 2]} }} \times [2x - 2] = \dfrac{{ - 1}}{2} \times \dfrac{1}{{\sqrt {[{x^2} - 6x + 13]} }} \times [2x - 6] $
Cancelling the terms and squaring both sides , we get
 $ \dfrac{{{{(x - 1)}^2}}}{{[{x^2} - 2x + 2]}} = \dfrac{{{{(x - 3)}^2}}}{{[{x^2} - 6x + 13]}} $
Cross multiplying terms , we get
 $ {(x - 1)^2} \times [{x^2} - 6x + 13] = {(x - 3)^2} \times [{x^2} - 2x + 2] $
Expanding the terms using $ {(a + b)^2} = {a^2} + {b^2} + 2ab $ , we get
 $ [{x^2} - 2x + 1] \times [{x^2} - 6x + 13] = [{x^2} - 6x + 9] \times [{x^2} - 2x + 2] $
Expanding the terms , we get
\[{x^4} - 2{x^3} + {x^2} - 6{x^3} + 12{x^2} - 6x + 13{x^2} - 26x + 13 = {x^4} - 6{x^3} + 9{x^2} - 2{x^3} + 12{x^2} - 18x + 2{x^2} - 12x + 18\]
After further simplifying the terms and after cancelling the terms , we get
 $ 3{x^2} - 2x - 5 = 0 $
Using splitting of roots ,
\[3{x^2} - 5x + 3x - 5 = 0\]
The roots of the equation becomes as
\[\left( {3x - 5} \right) \times \left( {x + 1} \right) = 0\]
So,
\[x = \dfrac{5}{3}\] or \[x = - 1\]
Hence , the value of point $ R $ is \[\left( {\dfrac{5}{3},0} \right)\]
Thus , the correct option is \[\left( 1 \right)\]
So, the correct answer is “Option 1”.

Note: We get two values of $ x $ . Neglecting \[x = - 1\]is a negative value for $ \dfrac{{dy}}{{dx}} $ which makes the values of \[PR + RQ\]maximum . But in the question we had to find the value of point R for which\[PR + RQ\] was minimum , which is satisfied by \[\left( {\dfrac{5}{3},0} \right)\] as it gives a positive value for $ \dfrac{{dy}}{{dx}} $ .
If the value of the double derivative of a function is negative at a point then it is the point of minima and vice versa .