Q61 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics

If $M$ is a point on the line $y=x$ and points $P(0,1), Q(2,0)$ are such that $P M+Q M$ is minimum, the coordinates of $M$ are

  • A
    $(0,0)$
  • B
    $\left(\frac{13}{17}, \frac{13}{17}\right)$
  • Correct answer
    C
    $\left(\frac{2}{3}, \frac{2}{3}\right)$
  • D
    $\left(\frac{31}{7}, \frac{31}{7}\right)$

Solution

Given, $$ \begin{aligned} & P=(0,1) \\ & Q=(2,0) \end{aligned} $$ Let $\quad M=(a, a) \quad[\because M$ lies on the line $y=x]$ In order to Make $P M+Q M$ is minimum $\Rightarrow P, M, Q$ Must be collinear $\therefore$ Slope of $P M=$ Slope of $Q M$ $$ \begin{aligned} & \frac{a-1}{a-0}=\frac{a-0}{a-2} \\ &(a-1)(a-2)=a^2 \\ & a^2-3 a+2=a^2 \\ &-3 a=-2 \\ & a=\frac{2}{3} \\ & \therefore M \text { is }\left(\frac{2}{3}, \frac{2}{3}\right) \end{aligned} $$

Hence, the correct answer is option: 'C'.

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