Questions: Find all points having an x-coordinate of 1 whose distance from the point (-3,-2) is 5. The point(s) is(are) . (Type an ordered pair. Use a comma to separate answers as needed.)

Solution Steps

To find all points with an $x$-coordinate of 1 that are 5 units away from the point $(-3, -2)$, we can use the distance formula. The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

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

Given \(x_1 = -3\), \(y_1 = -2\), \(x_2 = 1\), and \(d = 5\), we can solve for \(y_2\).

1. Substitute the known values into the distance formula.
2. Solve the resulting equation for \(y_2\).

We need to find all points \((1, y_2)\) that are 5 units away from the point \((-3, -2)\). Using the distance formula:

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

Substitute the known values \(x_1 = -3\), \(y_1 = -2\), \(x_2 = 1\), and \(d = 5\):

\[ \sqrt{(1 - (-3))^2 + (y_2 - (-2))^2} = 5 \]

Simplify the expression inside the square root:

\[ \sqrt{(1 + 3)^2 + (y_2 + 2)^2} = 5 \]

\[ \sqrt{4^2 + (y_2 + 2)^2} = 5 \]

\[ \sqrt{16 + (y_2 + 2)^2} = 5 \]

Square both sides to eliminate the square root:

\[ 16 + (y_2 + 2)^2 = 25 \]

Subtract 16 from both sides:

\[ (y_2 + 2)^2 = 9 \]

Take the square root of both sides:

\[ y_2 + 2 = \pm 3 \]

Solve for \(y_2\):

\[ y_2 + 2 = 3 \quad \text{or} \quad y_2 + 2 = -3 \]

\[ y_2 = 1 \quad \text{or} \quad y_2 = -5 \]

Final Answer