02 // Visual Math & Finance Solversm = (y₂ − y₁) / (x₂ − x₁), y = m·x + b
Visual Slope-Intercept ($y = mx + b$) & Rise-over-Run Triangle Grapher
A non-vertical line through $(x_1, y_1)$ and $(x_2, y_2)$ has slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta$ and y-intercept $b = y_1 - mx_1$, uniquely defining the slope-intercept equation $y = mx + b$.
m = (y₂ − y₁) / (x₂ − x₁), y = m·x + bInputs & Parameters (4)
-10±0.510
-10±0.510
-10±0.510
-10±0.510
Slope m = Δy / Δx
0.7
Δy = 7, Δx = 10 (θ = 35°)
Line Equation
y = 0.7x + 0.8
Y-int = (0, 0.8)
Euclidean Distance (d)
12.207 u
Manhattan |Δx|+|Δy| = 17
Midpoint M(x̄, ȳ)
(1, 1.5)
Normal slope m⊥ = -1.429
20-Row Comparison Table
Compare across:
| # | Point A — X₁ | Euclidean Distance (d) | Slope m | Midpoint M |
|---|---|---|---|---|
| 01 | -10 | 17.464 u | 0.438 | (-2, 1.5) |
| 02 | -9 | 16.553 u | 0.467 | (-1.5, 1.5) |
| 03 | -8 | 15.652 u | 0.5 | (-1, 1.5) |
| 04 | -7 | 14.765 u | 0.538 | (-0.5, 1.5) |
| 05 | -6 | 13.892 u | 0.583 | (0, 1.5) |
| 06 | -4.5 | 12.619 u | 0.667 | (0.8, 1.5) |
| 07 | -3.5 | 11.8 u | 0.737 | (1.3, 1.5) |
| 08 | -2.5 | 11.011 u | 0.824 | (1.8, 1.5) |
| 09 | -1.5 | 10.259 u | 0.933 | (2.3, 1.5) |
| 10 | -0.5 | 9.552 u | 1.077 | (2.8, 1.5) |
| 11 | 0.5 | 8.902 u | 1.273 | (3.3, 1.5) |
| 12 | 1.5 | 8.322 u | 1.556 | (3.8, 1.5) |
| 13 | 2.5 | 7.826 u | 2 | (4.3, 1.5) |
| 14 | 3.5 | 7.433 u | 2.8 | (4.8, 1.5) |
| 15 | 4.5 | 7.159 u | 4.667 | (5.3, 1.5) |
| 16 | 6 | 7 u | ∞ | (6, 1.5) |
| 17 | 7 | 7.071 u | -7 | (6.5, 1.5) |
| 18 | 8 | 7.28 u | -3.5 | (7, 1.5) |
| 19 | 9 | 7.616 u | -2.333 | (7.5, 1.5) |
| 20 | 10 | 8.062 u | -1.75 | (8, 1.5) |
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Key Analytical Takeaways
- 01The dashed right triangle between $A$ and $B$ isolates horizontal run $\Delta x = x_2 - x_1$ and vertical rise $\Delta y = y_2 - y_1$.
- 02When $x_1 = x_2$, $\Delta x = 0$ and the line is vertical ($x = x_1$) with undefined $(\infty)$ slope.
- 03Any line perpendicular to $y = mx + b$ has negative-reciprocal slope $m_\perp = -1/m$.
Frequently Asked Questions
What is the difference between zero slope and undefined slope?+
A horizontal line (y₁ = y₂) has Δy = 0, so its slope is 0. A vertical line (x₁ = x₂) has Δx = 0, causing division by zero and an undefined slope.
How do you find the x-intercept from y = mx + b?+
Set y = 0 and solve 0 = mx + b, which gives x_intercept = −b / m.
How are the slopes of parallel and perpendicular lines related?+
Non-vertical parallel lines share the exact same slope (m₁ = m₂), while perpendicular lines have negative-reciprocal slopes satisfying m₁ × m₂ = −1.