Questions: Find all points having an x-coordinate of 1 whose distance from the point (-3,-6) is 5.

Solution Steps

We need to find all points with an \( x \)-coordinate of \( 1 \) that are \( 5 \) units away from the point \( (-3, -6) \).

Using the distance formula, we have: \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = d \] Substituting \( x_1 = -3 \), \( y_1 = -6 \), \( x_2 = 1 \), and \( d = 5 \): \[ \sqrt{(1 - (-3))^2 + (y - (-6))^2} = 5 \]

This simplifies to: \[ \sqrt{(1 + 3)^2 + (y + 6)^2} = 5 \] which further simplifies to: \[ \sqrt{16 + (y + 6)^2} = 5 \]

Squaring both sides gives: \[ 16 + (y + 6)^2 = 25 \]

Rearranging the equation results in: \[ (y + 6)^2 = 9 \]

Taking the square root of both sides yields: \[ y + 6 = \pm 3 \] This leads to two equations:

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

Solving these equations gives:

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

Thus, the points that satisfy the conditions are: \[ (1, -3) \quad \text{and} \quad (1, -9) \]

Final Answer