If Q is the point with coordinates (b,1), and its distance from the origin is half its distance to the point (1,3), determine the value b

The distance from the origin (0,0) to the point Q(b,1) is:

\sqrt{b^2 + 1^2} = \sqrt{b^2 + 1}

The distance from Q to the point (1,3) is:

\sqrt{(b-1)^2 + (1-3)^2} = \sqrt{(b-1)^2 + 4}

Given that the distance from the origin to Q is half its distance to (1,3), we set up the equation:

\sqrt{b^2 + 1} = \frac{1}{2} \sqrt{(b-1)^2 + 4}

Square both sides to eliminate the square roots:

b^2 + 1 = \frac{1}{4} ((b-1)^2 + 4)

Multiply both sides by 4 to clear the fraction:

4(b^2 + 1) = (b-1)^2 + 4

Expand and simplify the equation:

4b^2 + 4 = b^2 - 2b + 1 + 4

4b^2 + 4 = b^2 - 2b + 5

4b^2 - b^2 + 4 + 2b = 5

3b^2 + 2b + 4 = 5

3b^2 + 2b - 1 = 0

Solve the quadratic equation using the quadratic formula:

b = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}

b = \frac{-2 \pm \sqrt{4 + 12}}{6}

b = \frac{-2 \pm \sqrt{16}}{6}

b = \frac{-2 \pm 4}{6}

So, we have two solutions:

b = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}

b = \frac{-2 - 4}{6} = \frac{-6}{6} = -1

Therefore, the values of b can be:

\frac{1}{3} \text{ or } -1