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 .