1. **State the problem:** Find the equations of the two lines that pass through the point $(1, 2)$ and are at a distance of $\sqrt{13}$ from the point $(6, 1)$. 2. **Set up the line equation:** Let the slope of the line be $m$. The line passing through $(1, 2)$ can be written as: $$y - 2 = m(x - 1)$$ Rearranged to standard form: $$mx - y + 2 - m = 0$$ 3. **Distance formula from a point to a line:** The perpendicular distance $d$ from point $(x_0, y_0)$ to line $Ax + By + C = 0$ is: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ 4. **Apply distance condition:** The distance from point $(6, 1)$ to the line is $\sqrt{13}$: $$\frac{|m(6) - 1 + 2 - m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{13}$$ Simplify numerator: $$|5m + 1| = \sqrt{13} \sqrt{m^2 + 1}$$ 5. **Square both sides:** $$(5m + 1)^2 = 13(m^2 + 1)$$ $$25m^2 + 10m + 1 = 13m^2 + 13$$ 6. **Bring all terms to one side:** $$25m^2 + 10m + 1 - 13m^2 - 13 = 0$$ $$12m^2 + 10m - 12 = 0$$ Divide entire equation by 2: $$\cancel{2}(6m^2 + 5m - 6)\cancel{2} = 0$$ 7. **Factor quadratic:** $$(3m - 2)(2m + 3) = 0$$ 8. **Solve for $m$:** $$m = \frac{2}{3} \quad \text{or} \quad m = -\frac{3}{2}$$ 9. **Write equations of the two lines:** - For $m = \frac{2}{3}$: $$y - 2 = \frac{2}{3}(x - 1)$$ Multiply both sides by 3: $$3y - 6 = 2x - 2$$ Rearranged: $$2x - 3y = -4$$ - For $m = -\frac{3}{2}$: $$y - 2 = -\frac{3}{2}(x - 1)$$ Multiply both sides by 2: $$2y - 4 = -3x + 3$$ Rearranged: $$3x + 2y = 7$$ **Final answer:** The two lines are: $$2x - 3y = -4$$ and $$3x + 2y = 7$$