Find all points having an​ x-coordinate of 4 whose distance from the point left parenthesis negative 2 comma negative 1 right parenthesis is 10.

To find all points with an \(x\)-coordinate of 4 whose distance from the point \((-2, -1)\) is 10, we can use the distance formula:

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

where \((x_1, y_1)\) is the point \((-2, -1)\), and \((x_2, y_2)\) is the point \((4, y)\). We are given that \(d = 10\), so:

\[
10 = \sqrt{(4 - (-2))^2 + (y - (-1))^2}
\]

Simplifying:

\[
10 = \sqrt{(4 + 2)^2 + (y + 1)^2}
\]
\[
10 = \sqrt{6^2 + (y + 1)^2}
\]
\[
10 = \sqrt{36 + (y + 1)^2}
\]

Square both sides to eliminate the square root:

\[
100 = 36 + (y + 1)^2
\]

Subtract 36 from both sides:

\[
64 = (y + 1)^2
\]

Now, take the square root of both sides:

\[
y + 1 = \pm 8
\]

This gives two cases:

1. \(y + 1 = 8 \Rightarrow y = 7\)
2. \(y + 1 = -8 \Rightarrow y = -9\)

Thus, the two points are \((4, 7)\) and \((4, -9)\).

### Solution:
The points are \((4, 7)\) and \((4, -9)\).

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How is the distance formula derived from the Pythagorean theorem?
2. What would the distance be for points with an \(x\)-coordinate of \(4\) and distances different from 10?
3. How do you solve similar problems if the given point is in three-dimensional space?
4. Can you find the midpoint between the points \((4, 7)\) and \((4, -9)\)?
5. What happens if the given distance is less than the direct distance between the points?

**Tip:** Always check your final answers by plugging them back into the original equation to confirm accuracy.

Find all points having an​ x-coordinate of 4 whose distance from the point (-2, -1) is 10.

Learn how to find points with an x-coordinate of 4 whose distance from the point (-2, -1) is 10. This problem uses the distance formula and involves solving a quadratic equation. Suitable for students in Grades 9-11.

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