Question

If point $Q(0,1)$ is equidistant from points $P(5,-3)$ and $R(x,6)$, find the values of $x$.

Hint:

For "equidistant from two points", apply the distance formula, equate the expressions, and simplify by squaring both sides.

Answer 1

Step 1: Write the condition of equidistance.
\[ QP = QR \] Using the distance formula, \[ \sqrt{(0-5)^2 + (1-(-3))^2} = \sqrt{(0-x)^2 + (1-6)^2}. \]

Step 2: Simplify both sides.
Left side: $(0-5)^2 = 25$, $(1+3)^2 = 16$ \[ QP = \sqrt{25+16} = \sqrt{41}. \] Right side: $(0-x)^2 = x^2$, $(1-6)^2 = 25$ \[ QR = \sqrt{x^2 + 25}. \]

Step 3: Equating and solving.
\[ \sqrt{41} = \sqrt{x^2 + 25} \ \Rightarrow\ 41 = x^2 + 25 \ \Rightarrow\ x^2 = 16. \] \[ x = \pm 4. \] Conclusion:
The values of $x$ are $4$ and $-4$.