If P = (1, 1),Q = (3, 2) and R is a point on | vedclass

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(A) Let the coordinates of $R$ be $(x, 0)$.Given $P = (1, 1)$ and $Q = (3, 2)$.The distance $PR + RQ = \sqrt{(x - 1)^2 + (0 - 1)^2} + \sqrt{(x - 3)^2

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$\left( \frac{5}{3}, 0 \right)$

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(A) Let the coordinates of $R$ be $(x, 0)$.Given $P = (1, 1)$ and $Q = (3, 2)$.The distance $PR + RQ = \sqrt{(x - 1)^2 + (0 - 1)^2} + \sqrt{(x - 3)^2 + (0 - 2)^2}$.$PR + RQ = \sqrt{x^2 - 2x + 2} + \sqrt{x^2 - 6x + 13}$.For the minimum value of $PR + RQ$,we set the derivative with respect to $x$ to zero:$\frac{d}{dx}(\sqrt{x^2 - 2x + 2} + \sqrt{x^2 - 6x + 13}) = 0$.$\frac{x - 1}{\sqrt{x^2 - 2x + 2}} + \frac{x - 3}{\sqrt{x^2 - 6x + 13}} = 0$.$\frac{x - 1}{\sqrt{x^2 - 2x + 2}} = -\frac{x - 3}{\sqrt{x^2 - 6x + 13}}$.Squaring both sides:$\frac{(x - 1)^2}{x^2 - 2x + 2} = \frac{(x - 3)^2}{x^2 - 6x + 13}$.$(x^2 - 2x + 1)(x^2 - 6x + 13) = (x^2 - 6x + 9)(x^2 - 2x + 2)$.Expanding both sides and simplifying leads to $3x^2 - 2x - 5 = 0$.$(3x - 5)(x + 1) = 0$,which gives $x = \frac{5}{3}$ or $x = -1$.Since $R$ lies between the projections of $P$ and $Q$ on the $x$-axis,$x$ must be in the interval $(1, 3)$.Thus,$x = \frac{5}{3}$.Therefore,the point $R$ is $\left( \frac{5}{3}, 0 \right)$.

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

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