Find a point on the y-axis that is equidistant from the points 
(8, −8)
 and 
(3, 3).

To find a point on the y-axis that is equidistant from the points \((8, -8)\) and \((3, 3)\), we need to determine the coordinates of the point on the y-axis, which will have the form \((0, y)\).

### Step 1: Use the distance formula
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the distance formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

### Step 2: Set up two distance equations
Let the point on the y-axis be \((0, y)\).

- The distance between \((0, y)\) and \((8, -8)\) is:
\[
d_1 = \sqrt{(0 - 8)^2 + (y - (-8))^2} = \sqrt{8^2 + (y + 8)^2} = \sqrt{64 + (y + 8)^2}
\]

- The distance between \((0, y)\) and \((3, 3)\) is:
\[
d_2 = \sqrt{(0 - 3)^2 + (y - 3)^2} = \sqrt{3^2 + (y - 3)^2} = \sqrt{9 + (y - 3)^2}
\]

Since the point on the y-axis is equidistant from both \((8, -8)\) and \((3, 3)\), we set \(d_1 = d_2\).

### Step 3: Set up the equation
\[
\sqrt{64 + (y + 8)^2} = \sqrt{9 + (y - 3)^2}
\]

### Step 4: Square both sides
Squaring both sides of the equation to eliminate the square roots:

\[
64 + (y + 8)^2 = 9 + (y - 3)^2
\]

### Step 5: Expand both sides
Expand the squares on both sides:

\[
64 + (y^2 + 16y + 64) = 9 + (y^2 - 6y + 9)
\]

Simplifying:

\[
128 + y^2 + 16y = y^2 - 6y + 18
\]

### Step 6: Eliminate \(y^2\) and simplify
Since \(y^2\) appears on both sides, we can subtract it from both sides:

\[
128 + 16y = -6y + 18
\]

Now, combine like terms:

\[
128 + 16y + 6y = 18
\]
\[
128 + 22y = 18
\]

### Step 7: Solve for \(y\)
Subtract 128 from both sides:

\[
22y = 18 - 128
\]
\[
22y = -110
\]

Now, divide by 22:

\[
y = \frac{-110}{22} = -5
\]

### Step 8: Conclusion
The point on the y-axis that is equidistant from \((8, -8)\) and \((3, 3)\) is \((0, -5)\).

---

Would you like further details on this solution? Here are five related questions to explore:

1. How can you use the midpoint formula to check if the point \((0, -5)\) lies on the perpendicular bisector of the two given points?
2. What is the significance of symmetry when finding points equidistant from two given points?
3. How would the problem change if the point was restricted to the x-axis instead of the y-axis?
4. Can you find the equation of the circle that passes through the points \((8, -8)\) and \((3, 3)\) with center \((0, -5)\)?
5. How does this method generalize for finding points equidistant from two points in three-dimensional space?

**Tip:** When solving distance problems, always use the symmetry of the coordinate axes to simplify the computation process.

Find a point on the y-axis that is equidistant from the points (8, −8) and (3, 3).

Learn how to find a point on the y-axis equidistant from the points (8, -8) and (3, 3) using the distance formula and algebraic techniques. This problem explores key concepts in coordinate geometry and algebra. Suitable for students in Grades 9-11.

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