1. **Problem statement:** Find the coordinates of point X on line segment AB such that X is $\frac{1}{5}$ of the distance from A to B. 2. **Formula:** To find a point dividing a segment in ratio $t$ (from A to B), use the section formula: $$X = A + t(B - A)$$ where $A = (x_1, y_1)$, $B = (x_2, y_2)$, and $t$ is the fraction along the segment. 3. **Given:** $A = (-5, -5)$, $B = (2, 9)$, $t = \frac{1}{5}$ 4. **Calculate the difference vector:** $$B - A = (2 - (-5), 9 - (-5)) = (7, 14)$$ 5. **Multiply by $t$:** $$\frac{1}{5} \times (7, 14) = \left(\frac{7}{5}, \frac{14}{5}\right)$$ 6. **Add to point A:** $$X = (-5, -5) + \left(\frac{7}{5}, \frac{14}{5}\right) = \left(-5 + \frac{7}{5}, -5 + \frac{14}{5}\right)$$ 7. **Simplify:** $$-5 = \frac{-25}{5}$$ $$X = \left(\frac{-25}{5} + \frac{7}{5}, \frac{-25}{5} + \frac{14}{5}\right) = \left(\frac{-18}{5}, \frac{-11}{5}\right)$$ 8. **Final answer:** $$X = \left(-\frac{18}{5}, -\frac{11}{5}\right)$$ This means point X is at coordinates $\left(-3.6, -2.2\right)$ on the line segment AB, exactly $\frac{1}{5}$ of the way from A to B.