Depth · one concept from first principles · step 2 of 4

A line through the origin sweeps between endpoints

The graph of y=ax is a line through the origin whose steepness is a. As a changes, the line turns about O. It touches segment AB exactly when it lies between the line through A and the line through B.

RatiosNumber systemExpressions

Your answer

How to solve

Strategy Visualize Spatial Relationships — Picture the line spinning about O. The steepest position that still hits the segment passes through A, the flattest passes through B, and every position in between crosses AB.

1STEP 1

Steepest line: through A

Put A into y=ax.

4 = 2a → a = 2

2STEP 2

Flattest line: through B

Put B into y=ax.

2 = 7a → a = 2/7

3STEP 3

Everything in between

Turning the line from OB to OA sweeps across every point of AB, and any steeper or flatter line misses the segment.

2/7 ≤ a ≤ 2

The whole segment sits in the wedge between the lines OB and OA.

7.EE.B.4Visualize Spatial Relationships

The line $y = ax$ meets segment AB exactly when $\frac{2}{7} \le a \le 2$ .

Why?

A line through O keeps the same ratio $\frac{y}{x}$ at every point, so $y = ax$ passes through a point of AB exactly when $a$ is that point's ratio $\frac{y}{x}$ .

🧱Proportional quantities scale togetherWhen two quantities keep a fixed ratio, doubling one doubles the other, so a ratio you know can stand in for one you do not.

Why?

That ratio is $2$ at A and $\frac{2}{7}$ at B.

Why?

The line through A must make $y = ax$ true at $(2, 4)$ , so $4 = 2a$ ; through B it must make $2 = 7a$ true.

🧱A solution makes the equation trueA solution is a number that, put in for the unknown, makes both sides come out equal; that is all being a solution means.

Why?

Dividing undoes the multiplying, so $a = 4 \div 2 = 2$ and $a = 2 \div 7 = \frac{2}{7}$ .

🧱Inverse operationsSubtraction undoes addition and division undoes multiplication — they reverse each other.

Why?

A line turning about O from OB up to OA, without passing vertical, runs through its slopes in order, so the lines through O that hit AB are exactly those with slopes from $\frac{2}{7}$ to $2$ , while a steeper or flatter line misses.

Why?

For a point to the right of O the run is positive, so multiplying by it keeps the order, and the point lies above $y = ax$ exactly when its own ratio $\frac{y}{x}$ is bigger than $a$ .

🧱Scaling an inequalityMultiplying both sides by the same positive number stretches or shrinks the gaps on the number line but keeps the order; multiplying by the same negative number also flips the line around zero, so the inequality turns over.

Why?

So for $\frac{2}{7}\le a\le2$ the end A is on or above the line and B is on or below it, while a steeper line has both ends below it and a flatter line has both ends above it.

🧱Transitivity of orderIf the first is smaller than the second and the second smaller than the third, the first is smaller than the third, so pairwise comparisons chain into one full order.

Why?

Along AB the gap between its height and the line's height changes by the same amount at every step, so it moves steadily between its two end values and reaches zero exactly when those values are not both of the same strict sign.

🧱A constant step reaches any termWhen a list climbs by the same amount every time, any term is the starting value plus that step repeated, so you never have to write out the ones between.

Answer

2/7 ≤ a ≤ 2

Take a=1 (inside the range): y=x meets AB where x = 4 - (2/5)(x-2), at x = 24/7, which is between 2 and 7. Take a=3 (outside): at x=2 the line is already at 6 ＞ 4, above A, so it misses.

Takeaway

A line through the origin is a door on a hinge: swing it to each end of the target, and the answer is everything in between.

Steepest line: through A
Flattest line: through B
Everything in between

Where next?

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▶ Practice — 12 problems