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 for distance from a point to a line:** The distance $d$ from a point $P(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by: $$ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} $$ 3. **Given:** - Point $P$ lies on the x-axis, so $P = (p, 0)$. - Distance $d = \sqrt{10}$. - Line equation: $3x - y + 2 = 0$ with $A=3$, $B=-1$, $C=2$. 4. **Apply 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. **Multiply both sides by $\sqrt{10}$:** $$ |3p + 2| = 10 $$ 6. **Solve the absolute value equation:** $$ 3p + 2 = 10 \quad \text{or} \quad 3p + 2 = -10 $$ 7. **Find $p$ for each case:** - For $3p + 2 = 10$: $$ 3p = 8 \implies p = \frac{8}{3} $$ - For $3p + 2 = -10$: $$ 3p = -12 \implies p = -4 $$ 8. **Coordinates of point P:** $$ P_1 = \left(\frac{8}{3}, 0\right), \quad P_2 = (-4, 0) $$ **Final answer:** The points on the x-axis at a distance $\sqrt{10}$ from the line $3x - y + 2 = 0$ are $\boxed{\left(\frac{8}{3}, 0\right)}$ and $\boxed{(-4, 0)}$.