Put a line and a hyperbola in one plane: they meet where a quadratic is solved, one rides above the other, and the points cut tidy triangles.
The two function families of this strand — the straight line of Stage 21 and the hyperbola of this stage — meet on the same coordinate plane, and their meeting is rich. Where do they cross? Setting k∕x equal to ax + b and clearing the fraction turns the question into a quadratic equation, so the algebra of Stage 11 comes back to work. Once you have the crossings you can ask which graph is higher on each stretch, measure the triangle a point cuts off with the axes, and even work backward from a known crossing to pin down both graphs. Here the line becomes a tool for reading the curve.
A point that lies on both graphs must satisfy both equations. So the crossings of y = k∕x and y = ax + b are exactly the simultaneous solutions of the system { y = k∕x, y = ax + b } — the pairs (x, y) that work in each line at once.
Geometrically there is nothing to compute yet: just look. Wherever the amber line cuts through the blue curve, that crossing point sits on both graphs, so its coordinates solve the system. Our running example, y = 6∕x and y = −x + 5, crosses at the two green dots below.
A point shared by two graphs is a point shared by their two equations: the crossings of a line and a hyperbola are the simultaneous solutions of the system. A shared point = a shared solution.
To find the crossings exactly, set the two right-hand sides equal — at a shared point the two y-values agree:
ax + b = kx
The k∕x is awkward, so clear the fraction: multiply every term by x. We are allowed to, because x ≠ 0 on the hyperbola, so we are never multiplying by zero. That gives
ax² + bx − k = 0
— a plain quadratic. Solve it by factoring or with the quadratic formula, then read each y straight off the line y = ax + b.
Take y = 6∕x and y = −x + 5. Set them equal and clear the fraction:
−x + 5 = 6x ⇒ −x² + 5x = 6 ⇒ x² − 5x + 6 = 0 ⇒ (x − 2)(x − 3) = 0 ⇒ x = 2 or 3.
Now the y's, from the line y = −x + 5: at x = 2, y = 3; at x = 3, y = 2. The crossings are (2, 3) and (3, 2). Check on the curve: 2·3 = 6 ✓ and 3·2 = 6 ✓.
How many crossings? Exactly the number of roots of ax² + bx − k = 0: a quadratic can give two, one, or no real solutions, so a line can cut a hyperbola in two, one, or zero points. Try all three in the widget.
When you multiply by x, keep the term order straight: ax + b = k∕x becomes ax² + bx − k = 0 (the k moves across with a minus sign). A sign slip here turns the whole quadratic — and every crossing — wrong.
Often you don't want the crossings themselves but a comparison: where is k∕x bigger than ax + b, and where is it smaller? The graph answers at a glance — over any stretch of x, whichever graph is drawn higher has the larger value there. And the crossings are the boundaries: that is exactly where the two graphs are equal, so that is where they trade places.
Read it straight off the picture. For y = 6∕x against y = −x + 5 on the right branch: between x = 2 and x = 3 the line is on top; for 0 < x < 2 and x > 3 the curve is on top. No more algebra needed — the crossings did the sorting. The meet widget above reports which graph is higher between its crossings.
Higher graph = larger value. The crossings split the x-axis into stretches that alternate between "curve on top" and "line on top" — find the crossings, then read off which graph wins on each piece.
Here is a small surprise the hyperbola hands you. Take any point P(x, k∕x) on the curve and drop a perpendicular straight down to the x-axis, landing at A(x, 0). Join O, A, and P into a right triangle. Its base is the run along the axis, |x|; its height is the drop, |k∕x|. So
area = ½ · base · height = ½ · |x| · |k∕x| = ½|k|
— the x cancels! The area is the same wherever P sits on the curve. (Recall the whole rectangle under P had area |k|; the triangle is exactly half of it.) For y = 6∕x every such triangle has area ½·6 = 3, no matter where you slide P.
For any point P on y = k∕x, the right triangle O–A–P (down to the x-axis and back to the origin) always has area ½|k| — a constant, exactly half the |k| rectangle.
Every problem so far ran forward: given the graphs, find the crossings. Run it backward and the same equations become a way to recover unknown coefficients. A stated crossing, or any condition the graphs must obey, turns into equations for a, b, or k — usually just enough to solve.
"The line y = ax + 1 meets y = 6∕x at the point (2, 3). Find a, and the other crossing."
Use that (2, 3) lies on both graphs. On the curve: 2·3 = 6 ✓ — so k = 6 is consistent. On the line: 3 = a·2 + 1 ⇒ 2a = 2 ⇒ a = 1. So the line is y = x + 1.
For the other crossing, solve the system x + 1 = 6∕x ⇒ x² + x − 6 = 0 ⇒ (x − 2)(x + 3) = 0 ⇒ x = 2 or x = −3. The new root x = −3 gives y = −3 + 1 = −2, the point (−3, −2) on the far branch.
The pattern is always the same: a known point sits on both graphs, so substitute it into both equations and solve for what's missing. Then, if you like, finish the system to find any second crossing.
| The question | The move |
|---|---|
| Where do they meet? | solve ax² + bx − k = 0, then y from the line |
| Which graph is higher? | compare heights; crossings are the boundaries |
| Triangle O–A–P from a point? | area = ½|k|, the same for every P |
| Find a, b, or k? | use the known point(s) in both equations |
Drop a line y = ax + b into the same plane as a hyperbola y = k∕x and four ideas fall out, all from one move — clearing the fraction:
Find where y = 4∕x meets the line y = x.
Set x = 4∕x and clear the fraction: x² = 4, so x = ±2. The crossings are (2, 2) and (−2, −2) — one on each branch.
Solve the system y = 6∕x, y = x + 1 for its crossing points.
x + 1 = 6∕x ⇒ x² + x − 6 = 0 ⇒ (x − 2)(x + 3) = 0 ⇒ x = 2 or −3. The points are (2, 3) and (−3, −2).
For y = 6∕x and y = −x + 5, on the right branch which graph is higher between x = 2 and x = 3?
The line. The crossings are at x = 2 and x = 3; between them the line rides above the curve (test x = 2.5: line gives 2.5, curve gives 2.4).
A point on y = 10∕x is joined to the x-axis and the origin to make a right triangle. What is its area?
The O–A–P triangle always has area ½|k| = ½·10 = 5 — the same wherever the point sits on the curve.
The line y = ax meets y = 8∕x at (2, 4). Find a.
The point is on the line, so 4 = a·2 ⇒ a = 2; the line is y = 2x. (Check: 2·4 = 8 ✓ puts (2, 4) on the curve too.)
Does the line y = −x − 1 meet y = 6∕x? Explain.
−x − 1 = 6∕x ⇒ x² + x + 6 = 0. Its discriminant is 1 − 24 = −23 < 0, so there are no real roots — the line misses the curve entirely. No crossing.
Six questions to lock it in. Tap the answer you think is right.
This lesson fuses the two function families of the strand. The single most important move is clearing the fraction: a point lies on both y = k∕x and y = ax + b exactly when ax + b = k∕x, and multiplying through by x (legal because x ≠ 0 on the curve) turns the encounter into the quadratic ax² + bx − k = 0. From there everything is Stage-11 algebra: factor or use the quadratic formula, and the count of real roots — two, one, or none — is the count of crossings. The "which graph is higher" question is then a reading task: heights compare directly, and the crossings are the only places the two graphs can swap order.
The misconceptions to watch are concrete. Students forget to clear the fraction (and lose the x ≠ 0 condition that justifies it); they slip a sign when moving k across, writing ax² + bx + k = 0 instead of the correct ax² + bx − k = 0; and they assume a line must cross a hyperbola twice, when a tangent meets it once and a "missing" line not at all (a negative discriminant). The constant-area triangle O–A–P (area ½|k|, half the |k| rectangle from 22.3) is a favorite because the answer is independent of where the point sits — encourage students to prove the x cancels rather than measure one case.
Common Core. This lesson supports A-REI.C.7 and A-REI.D.11 (solve a system of a linear and a non-linear equation; the intersection is the solution), A-REI.B.4 (solve the resulting quadratic), and A-REI.D.10 / F-IF.C.7 (compare two functions read from their graphs). It builds on the inverse-proportion graph of 22.3 and the linear-function tools of Stage 21, and closes the algebraic heart of the stage before 22.5 turns inverse proportion on the real world.