1. **Stating the problem:** We have a table of Distance (Km) and Time values: | Distance (Km) | 10 | 15 | 20 | 25 | 30 | |--------------|----|----|----|----|----| | Time | 2 | 3 | 4 | 5 | 6 | We need to determine if the data shows direct variation. Also, identify which of the given lines are parallel or perpendicular: - $Y=5X$ - $Y=5(X-2)$ - $Y=5$ - $X=6$ - $Y=-4X+9$ - $Y=\frac{1}{4}X+8$ 2. **Direct Variation Check:** Direct variation means $Y=kX$ for some constant $k$. Calculate $\frac{Distance}{Time}$ for each pair: $$\frac{10}{2}=5, \quad \frac{15}{3}=5, \quad \frac{20}{4}=5, \quad \frac{25}{5}=5, \quad \frac{30}{6}=5$$ Since the ratio is constant ($k=5$), the data shows direct variation. 3. **Parallel and Perpendicular Lines:** - Lines are parallel if they have the same slope. - Lines are perpendicular if the product of their slopes is $-1$. Find slopes: - $Y=5X$ slope $m_1=5$ - $Y=5(X-2) = 5X - 10$ slope $m_2=5$ - $Y=5$ is a horizontal line slope $m_3=0$ - $X=6$ is a vertical line slope undefined - $Y=-4X+9$ slope $m_4=-4$ - $Y=\frac{1}{4}X+8$ slope $m_5=\frac{1}{4}$ Check parallel lines: - $Y=5X$ and $Y=5(X-2)$ both have slope 5, so they are parallel. Check perpendicular lines: - $Y=5$ (slope 0) and $X=6$ (vertical) are perpendicular. - $Y=-4X+9$ and $Y=\frac{1}{4}X+8$ slopes multiply to $-4 \times \frac{1}{4} = -1$, so they are perpendicular. **Final answers:** - The data shows direct variation with constant $k=5$. - Parallel lines: $Y=5X$ and $Y=5(X-2)$. - Perpendicular lines: $Y=5$ and $X=6$; $Y=-4X+9$ and $Y=\frac{1}{4}X+8$.