1. Problem 19: Show that points (-1, -1), (2, 8), and (5, 17) are collinear. 2. To prove collinearity, we check if the slope between each pair of points is the same. 3. Slope formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 4. Calculate slope between (-1, -1) and (2, 8): $$m_1 = \frac{8 - (-1)}{2 - (-1)} = \frac{9}{3} = 3$$ 5. Calculate slope between (2, 8) and (5, 17): $$m_2 = \frac{17 - 8}{5 - 2} = \frac{9}{3} = 3$$ 6. Since $m_1 = m_2 = 3$, the points are collinear. 7. Problem 20: Find the relation between $x$ and $y$ so that $(x, y)$ is 3 units from $(-1, 2)$. 8. Distance formula between $(x, y)$ and $(-1, 2)$ is $$\sqrt{(x + 1)^2 + (y - 2)^2} = 3$$ 9. Square both sides: $$ (x + 1)^2 + (y - 2)^2 = 9 $$ 10. This is the equation of a circle centered at $(-1, 2)$ with radius 3. 11. Problem 21: Find the relation between $x$ and $y$ so that $(x, y)$ is equidistant from the y-axis and the point $(4, 0)$. 12. Distance from $(x, y)$ to y-axis is $|x|$. 13. Distance from $(x, y)$ to $(4, 0)$ is $$\sqrt{(x - 4)^2 + y^2}$$ 14. Set distances equal: $$|x| = \sqrt{(x - 4)^2 + y^2}$$ 15. Square both sides: $$x^2 = (x - 4)^2 + y^2$$ 16. Expand and simplify: $$x^2 = x^2 - 8x + 16 + y^2$$ 17. Cancel $x^2$ on both sides: $$\cancel{x^2} = \cancel{x^2} - 8x + 16 + y^2$$ 18. Rearrange: $$0 = -8x + 16 + y^2$$ 19. Or equivalently: $$y^2 = 8x - 16$$ This is the relation between $x$ and $y$.