Q81 / 170 · Manipal (MET) 2021 · Mathematics
The point on the line $y=x$ equidistant from $(4,0)$ and $(5,1)$ is
- A$(2,2)$
- B$(3,3)$
- Correct answerC$\left(\frac{5}{2}, \frac{5}{2}\right)$
- D$\left(\frac{1}{2}, \frac{1}{2}\right)$
Solution
Let the required point is $(a, a)$.
$\therefore(4-a)^2+a^2=(5-a)^2+(1-a)^2$
$\begin{array}{ll}\Rightarrow & 16+a^2-8 a+a^2= 25+a^2 -10 a+1+a^2-2a \\ \Rightarrow & 16-25-1+a^2+a^2-a^2-a^2-8 a+10 a+2 a=0 \\ \Rightarrow & 4 a=10 \\ \Rightarrow & \quad a=\frac{10}{4}=\frac{5}{2}\end{array}$ $\therefore$ Required point is $\left(\frac{5}{2}, \frac{5}{2}\right)$.
$\begin{array}{ll}\Rightarrow & 16+a^2-8 a+a^2= 25+a^2 -10 a+1+a^2-2a \\ \Rightarrow & 16-25-1+a^2+a^2-a^2-a^2-8 a+10 a+2 a=0 \\ \Rightarrow & 4 a=10 \\ \Rightarrow & \quad a=\frac{10}{4}=\frac{5}{2}\end{array}$ $\therefore$ Required point is $\left(\frac{5}{2}, \frac{5}{2}\right)$.
Hence, the correct answer is option: 'C'.