We are asked to find points on the x-axis that are 5 units away from the point $A(5, -3)$.

Step 1: Generalize the form of points on the x-axis

Any point on the x-axis can be written as $P(x, 0)$, since the y-coordinate of any point on the x-axis is 0.

Step 2: Apply the distance formula

The distance $d$ between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by the formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Here, $P(x, 0)$ is the point on the x-axis, and $A(5, -3)$ is the given point. The distance between these points is 5 units. So, we can set up the equation as follows: $\sqrt{(x - 5)^2 + (0 - (-3))^2} = 5$ This simplifies to: $\sqrt{(x - 5)^2 + 3^2} = 5$ $\sqrt{(x - 5)^2 + 9} = 5$

Step 3: Solve the equation

Square both sides to eliminate the square root: $(x - 5)^2 + 9 = 25$ $(x - 5)^2 = 16$ Take the square root of both sides: $x - 5 = \pm 4$ This gives two possible solutions:

1. $x - 5 = 4$ ⟹ $x = 9$
2. $x - 5 = -4$ ⟹ $x = 1$

Step 4: Conclusion

The points on the x-axis that are 5 units away from $A(5, -3)$ are $(9, 0)$ and $(1, 0)$.

Would you like more details or have any questions?

Here are 5 relative questions:

1. What is the distance formula, and how is it derived?
2. How can we generalize finding points on a line at a specific distance from another point?
3. What are other geometric methods to approach this problem?
4. How do you apply the Pythagorean Theorem in such problems?
5. How can this concept be extended to find points in 3D space?

Tip: Always check if the points satisfy the distance condition by substituting back into the distance formula.