Find all points on the \( x \)-axis that are 12 units from the point \( (4,-6) \)

To find all points on the \( x \)-axis that are 12 units away from the point \( (4, -6) \), we need to apply the distance formula. The general distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] In this case, our reference point is \( (4, -6) \), and any point on the \( x \)-axis can be denoted as \( (x, 0) \). We set the distance equal to 12: \[ \sqrt{(x - 4)^2 + (0 + 6)^2} = 12 \] Squaring both sides: \[ (x - 4)^2 + 6^2 = 12^2 \] This simplifies to: \[ (x - 4)^2 + 36 = 144 \] Subtract 36 from both sides: \[ (x - 4)^2 = 108 \] Now we take the square root of both sides: \[ x - 4 = \pm \sqrt{108} \] Calculating \( \sqrt{108} \): \[ \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \] So we have: \[ x - 4 = 6\sqrt{3} \quad \text{or} \quad x - 4 = -6\sqrt{3} \] This gives us: \[ x = 4 + 6\sqrt{3} \quad \text{and} \quad x = 4 - 6\sqrt{3} \] Therefore, the points on the \( x \)-axis that are 12 units from the point \( (4, -6) \) are: \[ (4 + 6\sqrt{3}, 0) \quad \text{and} \quad (4 - 6\sqrt{3}, 0) \]