1. The problem asks to find points on each of the given lines to capture them. 2. The lines are: - $y = 3x + 6$ - $y = 3(x - 5)$ which simplifies to $y = 3x - 15$ - $y = -\frac{1}{2}x - 15$ - $-x + y = 12$ which can be rewritten as $y = x + 12$ 3. To find points on each line, choose convenient $x$ values and calculate corresponding $y$ values. 4. For $y = 3x + 6$: - Let $x=0$, then $y=3(0)+6=6$, point is $(0,6)$ - Let $x=1$, then $y=3(1)+6=9$, point is $(1,9)$ 5. For $y = 3x - 15$: - Let $x=5$, then $y=3(5)-15=0$, point is $(5,0)$ - Let $x=6$, then $y=3(6)-15=3$, point is $(6,3)$ 6. For $y = -\frac{1}{2}x - 15$: - Let $x=0$, then $y=-\frac{1}{2}(0)-15=-15$, point is $(0,-15)$ - Let $x=2$, then $y=-\frac{1}{2}(2)-15=-1-15=-16$, point is $(2,-16)$ 7. For $y = x + 12$: - Let $x=0$, then $y=0+12=12$, point is $(0,12)$ - Let $x=1$, then $y=1+12=13$, point is $(1,13)$ 8. Since only three zaps are allowed, select one point from three different lines to capture them. 9. Zap 1: $(0,6)$ from $y=3x+6$ 10. Zap 2: $(5,0)$ from $y=3x-15$ 11. Zap 3: $(0,-15)$ from $y=-\frac{1}{2}x-15$ 12. The fourth line $y=x+12$ is not captured due to zap limit. Final answer: Zap 1: $(0,6)$ Zap 2: $(5,0)$ Zap 3: $(0,-15)$