find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 unites

We are given two points, \( P(2, -3) \) and \( Q(10, y) \), and the distance between them is 10 units. We need to find the values of \( y \) for which this is true.

We can use the distance formula:

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

Here:
- \( d = 10 \) (the given distance),
- \( x_1 = 2 \), \( y_1 = -3 \) (coordinates of point P),
- \( x_2 = 10 \), \( y_2 = y \) (coordinates of point Q).

Substituting these into the distance formula:

\[
10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}
\]

Simplify the expression:

\[
10 = \sqrt{8^2 + (y + 3)^2}
\]
\[
10 = \sqrt{64 + (y + 3)^2}
\]

Now square both sides to eliminate the square root:

\[
100 = 64 + (y + 3)^2
\]

Subtract 64 from both sides:

\[
36 = (y + 3)^2
\]

Now, take the square root of both sides:

\[
y + 3 = \pm 6
\]

This gives two cases:
1. \( y + 3 = 6 \), so \( y = 3 \),
2. \( y + 3 = -6 \), so \( y = -9 \).

Thus, the values of \( y \) are \( 3 \) and \( -9 \).

Would you like more details or have any questions?

Here are some related questions:
1. How is the distance formula derived?
2. What other methods can be used to find the distance between two points?
3. What if the distance was different? How would the solutions change?
4. How would the problem change if one of the points had unknown x-coordinates?
5. Can we generalize this solution for points in 3D space?

**Tip:** When solving distance problems, squaring both sides early helps avoid dealing with square roots.

Find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units.

Learn how to find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units. This problem involves applying the distance formula and solving a quadratic equation. Suitable for students in Grades 9-10.

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