A line through the origin sweeps between endpoints · 12 practice problems

Take $a=\frac{3}{5}$ (inside the range): the line passes through the midpoint $\left(\frac{5}{2}, \frac{3}{2}\right)$ of AB, since $\frac{3}{5} \times \frac{5}{2} = \frac{3}{2}$. Take $a=3$ (outside): at $x=1$ the line is already at $3 > 2$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $2$ to $\frac{1}{4}$, taking every value in between.

Take $a=\frac{5}{6}$ (inside the range): the line passes through the midpoint $\left(3, \frac{5}{2}\right)$ of AB, since $\frac{5}{6} \times 3 = \frac{5}{2}$. Take $a=4$ (outside): at $x=1$ the line is already at $4 > 3$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $3$ to $\frac{2}{5}$, taking every value in between.

Take $a=\frac{7}{4}$ (inside the range): the line passes through the midpoint $\left(2, \frac{7}{2}\right)$ of AB, since $\frac{7}{4} \times 2 = \frac{7}{2}$. Take $a=6$ (outside): at $x=1$ the line is already at $6 > 5$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $5$ to $\frac{2}{3}$, taking every value in between.

Take $a=\frac{3}{4}$ (inside the range): the line passes through the midpoint $\left(4, 3\right)$ of AB, since $\frac{3}{4} \times 4 = 3$. Take $a=3$ (outside): at $x=2$ the line is already at $6 > 5$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{5}{2}$ to $\frac{1}{6}$, taking every value in between.

Take $a=1$ (inside the range): the line passes through the midpoint $\left(\frac{7}{2}, \frac{7}{2}\right)$ of AB, since $1 \times \frac{7}{2} = \frac{7}{2}$. Take $a=5$ (outside): at $x=1$ the line is already at $5 > 4$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $4$ to $\frac{1}{2}$, taking every value in between.

Take $a=\frac{2}{3}$ (inside the range): the line passes through the midpoint $\left(\frac{9}{2}, 3\right)$ of AB, since $\frac{2}{3} \times \frac{9}{2} = 3$. Take $a=3$ (outside): at $x=2$ the line is already at $6 > 4$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{4}{2}$ to $\frac{2}{7}$, taking every value in between.

Take $a=\frac{7}{10}$ (inside the range): the line passes through the midpoint $\left(5, \frac{7}{2}\right)$ of AB, since $\frac{7}{10} \times 5 = \frac{7}{2}$. Take $a=2$ (outside): at $x=3$ the line is already at $6 > 5$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{5}{3}$ to $\frac{2}{7}$, taking every value in between.

Take $a=\frac{10}{7}$ (inside the range): the line passes through the midpoint $\left(\frac{7}{2}, 5\right)$ of AB, since $\frac{10}{7} \times \frac{7}{2} = 5$. Take $a=4$ (outside): at $x=2$ the line is already at $8 > 7$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{7}{2}$ to $\frac{3}{5}$, taking every value in between.

Take $a=\frac{9}{11}$ (inside the range): the line passes through the midpoint $\left(\frac{11}{2}, \frac{9}{2}\right)$ of AB, since $\frac{9}{11} \times \frac{11}{2} = \frac{9}{2}$. Take $a=3$ (outside): at $x=3$ the line is already at $9 > 6$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{6}{3}$ to $\frac{3}{8}$, taking every value in between.

Take $a=\frac{10}{11}$ (inside the range): the line passes through the midpoint $\left(\frac{11}{2}, 5\right)$ of AB, since $\frac{10}{11} \times \frac{11}{2} = 5$. Take $a=4$ (outside): at $x=2$ the line is already at $8 > 6$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{6}{2}$ to $\frac{4}{9}$, taking every value in between.

Take $a=\frac{8}{13}$ (inside the range): the line passes through the midpoint $\left(\frac{13}{2}, 4\right)$ of AB, since $\frac{8}{13} \times \frac{13}{2} = 4$. Take $a=2$ (outside): at $x=4$ the line is already at $8 > 6$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{6}{4}$ to $\frac{2}{9}$, taking every value in between.

Take $a=\frac{11}{13}$ (inside the range): the line passes through the midpoint $\left(\frac{13}{2}, \frac{11}{2}\right)$ of AB, since $\frac{11}{13} \times \frac{13}{2} = \frac{11}{2}$. Take $a=3$ (outside): at $x=3$ the line is already at $9 > 8$, above A, so it misses.

Another way: Each point of AB has its own ratio $\frac{y}{x}$, and $a$ must equal one of them. Moving from A to B, $y$ falls and $x$ grows, so $\frac{y}{x}$ falls steadily from $\frac{8}{3}$ to $\frac{3}{10}$, taking every value in between.