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

To find the value of \( y \) such that the distance between the points \( P(2, -3) \) and \( Q(10, y) \) is 10 units, we use the distance formula:

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

Here, \( P(2, -3) = (x_1, y_1) \) and \( Q(10, y) = (x_2, y_2) \). We know the distance \( d \) is 10 units, so we substitute the values:

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

Simplifying:

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

Now, square both sides to eliminate the square root:

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

Subtract 64 from both sides:

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

Next, take the square root of both sides:

\[
y + 3 = \pm 6
\]

Now, solve for \( y \) in both cases:

1. \( y + 3 = 6 \) ⟹ \( y = 6 - 3 = 3 \)
2. \( y + 3 = -6 \) ⟹ \( y = -6 - 3 = -9 \)

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

Would you like a detailed explanation of these steps or have any questions?

Here are 5 related questions for practice:

1. Find the distance between \( P(2, -3) \) and \( Q(10, 5) \).
2. What is the midpoint between \( P(2, -3) \) and \( Q(10, 3) \)?
3. If the distance between \( P(2, -3) \) and \( Q(10, y) \) is 5 units, what are the possible values of \( y \)?
4. What is the equation of the circle centered at \( P(2, -3) \) with a radius of 10 units?
5. How do you derive the distance formula from the Pythagorean theorem?

**Tip:** Whenever you square both sides of an equation, don't forget to consider both the positive and negative 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 points P(2,-3) and Q(10,y) is 10 units. This problem involves applying the distance formula and solving a quadratic equation, making it suitable for students in Grades 8-10.

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