Find x so the distance between (x,2) and (1, 5) is square root of l3.

The formula for the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

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

Given the two points \((x, 2)\) and \((1, 5)\), and the distance is \(\sqrt{13}\), we can set up the equation as:

\[
\sqrt{(1 - x)^2 + (5 - 2)^2} = \sqrt{13}
\]

Simplify the equation:

\[
\sqrt{(1 - x)^2 + 3^2} = \sqrt{13}
\]
\[
\sqrt{(1 - x)^2 + 9} = \sqrt{13}
\]

Square both sides to eliminate the square roots:

\[
(1 - x)^2 + 9 = 13
\]

Subtract 9 from both sides:

\[
(1 - x)^2 = 4
\]

Now, take the square root of both sides:

\[
1 - x = \pm 2
\]

Solve for \(x\):

1. \(1 - x = 2 \Rightarrow x = -1\)
2. \(1 - x = -2 \Rightarrow x = 3\)

So, \(x = -1\) or \(x = 3\).

Would you like more details or have any questions?

Here are some related questions for further practice:
1. Find the distance between \((2, 3)\) and \((5, 7)\).
2. If the distance between \((x, 1)\) and \((3, 4)\) is 5, what is \(x\)?
3. What is the midpoint between the points \((x, 2)\) and \((1, 5)\)?
4. How would the distance change if one of the coordinates were \((1, 7)\)?
5. If \(y = 2x + 1\), find the intersection point with the line \(x = 3\).

**Tip**: For finding the distance between two points, it's always helpful to remember the Pythagorean Theorem, as it underlies the distance formula.

Find x so the distance between (x,2) and (1, 5) is square root of 13.

Discover how to find the value of x such that the distance between the points (x, 2) and (1, 5) equals the square root of 13. This solution involves applying the distance formula and solving a quadratic equation. Suitable for students in Grades 8-10.

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