If `P=(1,1),Q=(3,2)` and `R` is a point on x-axis then the value of `PR+RQ` will be minimum at

Options

A. `(5/3,0)`
B. `(1/3,0)`
C. `(3,0)`
D. `(1,0)`

Answer

A

Explanation

Let the coordinates of `R` be `(x,0)`.
Given `P=(1,1), Q=(3,2)`
Now `PR+RQ=sqrt((x-1)^(2)+(0-1)^(2))`
`+sqrt((x-3)^(2)+(0-2)^(2))`
`=sqrt(x^(2)-2x+2)+sqrt(x^(2)-6x+13)`
For minimum value of `PR+RQ`, put
`d/(dx)(PR+RQ)=0`
`impliesd/(dx)(sqrt(x^(2)-2x+2))+d/(dx)(sqrt(x^(2)-6x+13))=0`
`implies ((x-1))/(sqrt(x^(2)-2x+2))=-((x-3))/(sqrt(x^(2)-6x+13))`
On squaring both sides we get
`((x-1)^(2))/((x^(2)-2x+2))=((x-3)^(2))/(x^(2)-6x+13)`
`implies3x^(2)-2x-5=0`
`implies(3x-5)(x+1)=0`
`impliesx=5/3,-1`
Also, `1 lt x lt 3`
`:.R=(5//3,0)`