1. **Problem Statement:** Find the coordinates of the points $P$ and $Q$ on the line $x + 5y = 13$ which are at a distance of 2 units from the line $12x - 5y + 26 = 0$. 2. **Formula for distance from a point to a line:** 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. **Given:** - Line 1: $x + 5y = 13$ (points $P$ and $Q$ lie here) - Line 2: $12x - 5y + 26 = 0$ - Distance from points $P$ and $Q$ to Line 2 is 2 units. 4. **Express $x$ from Line 1:** $$x = 13 - 5y$$ 5. **Distance condition:** Substitute $(x, y) = (13 - 5y, y)$ into the distance formula for Line 2: $$2 = \frac{|12(13 - 5y) - 5y + 26|}{\sqrt{12^2 + (-5)^2}} = \frac{|156 - 60y - 5y + 26|}{\sqrt{144 + 25}} = \frac{|182 - 65y|}{13}$$ 6. **Solve for $y$:** $$2 = \frac{|182 - 65y|}{13} \implies |182 - 65y| = 26$$ This gives two cases: - Case 1: $182 - 65y = 26 \implies 65y = 182 - 26 = 156 \implies y = \frac{156}{65} = \frac{12}{5} = 2.4$ - Case 2: $182 - 65y = -26 \implies 65y = 182 + 26 = 208 \implies y = \frac{208}{65} = \frac{16}{5} = 3.2$ 7. **Find corresponding $x$ values:** - For $y = 2.4$: $$x = 13 - 5(2.4) = 13 - 12 = 1$$ - For $y = 3.2$: $$x = 13 - 5(3.2) = 13 - 16 = -3$$ 8. **Coordinates of points $P$ and $Q$ are:** $$P = (1, 2.4), \quad Q = (-3, 3.2)$$ **Final answer:** The points on the line $x + 5y = 13$ at a distance 2 units from the line $12x - 5y + 26 = 0$ are $(1, 2.4)$ and $(-3, 3.2)$.