Question

# 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

224

likes
1121 views

## Answer to a math question 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

Frederik
4.6
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

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

Frequently asked questions $FAQs$
What is the limit of $3x^2 + 2x - 1$ / $2x^2 + 5x$ as x approaches infinity?
+
What is the square root of the number 25?
+
Question: Find the limit as x approaches 3 of $2x^2 - 7x + 4$/$x - 3$.
+