One line. One slope.

SparkClass

Move right triangles along a line. Compare their runs and signed rises, prove similarity with AA, then derive the two forms of a linear equation.

CCSS 8.EE.B.6
AA + Scale
x, y: −8 … 8
Slope triangles on the graph of y equals mx plus b Two right triangles have hypotenuses on the same line. Their horizontal runs and signed vertical rises demonstrate a constant slope.
AA

Two angles are enough. Right angles and congruent acute angles establish similarity.

AA similar
Current Line · Live Equationy = 1x + 1
rise / run = m = 1

Load A Preset · Challenges Always Use Nonzero Slope

Proof & ProportionAA Similarity
Corresponding Measurements

Live Measurements2 Triangles

P and Q are the hypotenuse endpoints on the line. R is the right-angle construction point. Rise is signed; side lengths are positive.

From Constant Slope To An EquationUse A General Point (x, y)
Line Through O(0, 0) · Dashed Reference

Derive y = mx

On the parallel line through the origin, use O(0, 0) and a general point (x, y). The same slope ratio gives:

m = (y − 0) / (x − 0) = y / x
y = mx

This line: y = 1x. It has the same m as the main line, but its intercept is 0.

The division step assumes x ≠ 0; the resulting equation also correctly gives O(0, 0).

Main Line · Intercept I(0, b)

Derive y = mx + b

Now use I(0, 1) and a general point (x, y) on the main line:

m = (y − b) / (x − 0)
y − b = mx
y = mx + b

Substituting today’s values: y = 1x + 1.

Changing b moves I and every triangle vertically; x-positions, runs, rises, and m stay fixed.

Teacher notes · making the proof precise

Use the sequence predict → measure → compare → prove → derive. The controls keep every displayed coordinate and formula tied to the same line equation.

1

Name the vertices accurately

Pᵢ and Qᵢ are the two hypotenuse endpoints and lie exactly on y = mx + b. Rᵢ is constructed at (xᵢ + runᵢ, yᵢ) to make horizontal and vertical legs. A nondegenerate right triangle cannot have all three vertices on one straight line; asking for that would collapse the figure.

2

Build the AA argument, not a visual guess

Each R angle is 90° because horizontal ⟂ vertical. Each P angle is formed by a rightward horizontal ray and the same line in the same left-to-right direction, so the acute angles are congruent for both positive and negative slopes. Conclude △P₁R₁Q₁ ∼ △P₂R₂Q₂ by AA.

3

Separate signed rise from side length

For slope, use signed rise = m · run; a downward change is negative. For similarity scale factors, use positive lengths: k₂ = run₂/run₁ = |rise₂|/|rise₁| = |P₂Q₂|/|P₁Q₁|. Triangle B is scaled from A by k₂ and then repositioned; the scale slider changes its run directly.

4

Do not force AA when m = 0

On a horizontal line, Qᵢ and Rᵢ coincide. The construction is a segment, not a triangle, so there are no two angles and no AA similarity claim. The numerical slope relationship still works: rise/run = 0/run = 0 = m, and the equation simplifies to y = b.

5

Use b and the challenges deliberately

Change only b and ask what stays invariant: runs, signed rises, ratios, and angle measures remain fixed while all y-coordinates shift. In either nonzero-slope challenge, ask students to predict all rises and the scale factor before reading the table. Add Triangle C to check whether the ratio survives a third, independently placed example.

CCSS 8.EE.B.6: Use similar triangles to explain why slope is the same between any two distinct points on a non-vertical line; derive y = mx for a line through the origin and y = mx + b for a line intercepting the vertical axis at b.