1. **Problem:** Find the coordinates of point P which is on the x-axis and at a distance of $\sqrt{10}$ units from the line $3x - y + 2 = 0$. 2. **Formula:** The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ 3. **Step 1:** Since point P is on the x-axis, its coordinates are $P(p, 0)$. 4. **Step 2:** Substitute $A=3$, $B=-1$, $C=2$, $x_0=p$, $y_0=0$, and $d=\sqrt{10}$ into the distance formula: $$\sqrt{10} = \frac{|3p - 0 + 2|}{\sqrt{3^2 + (-1)^2}} = \frac{|3p + 2|}{\sqrt{9 + 1}} = \frac{|3p + 2|}{\sqrt{10}}$$ 5. **Step 3:** Multiply both sides by $\sqrt{10}$: $$|3p + 2| = 10$$ 6. **Step 4:** Solve the absolute value equation: $$3p + 2 = 10 \quad \text{or} \quad 3p + 2 = -10$$ 7. **Step 5:** Solve each linear equation: - For $3p + 2 = 10$: $$3p = 8 \implies p = \frac{8}{3}$$ - For $3p + 2 = -10$: $$3p = -12 \implies p = -4$$ 8. **Answer:** The coordinates of point P are: $$P\left(\frac{8}{3}, 0\right) \quad \text{and} \quad P(-4, 0)$$