Example

Finding the Equation of a Line Through (3,1)(-3, -1) and (2,2)(2, -2)

To find the equation of a line containing the points (3,1)(-3, -1) and (2,2)(2, -2), apply the four-step two-points procedure and write the result in slope-intercept form. Step 11 — Find the slope. Apply the slope formula: m=2(1)2(3)=15=15m = \frac{-2 - (-1)}{2 - (-3)} = \frac{-1}{5} = -\frac{1}{5} Step 22 — Choose one point. Use (2,2)(2, -2), so x1=2x_1 = 2 and y1=2y_1 = -2. Step 33 — Substitute into the point-slope form yy1=m(xx1)y - y_1 = m(x - x_1): y(2)=15(x2)y - (-2) = -\frac{1}{5}(x - 2) Simplify the left side and distribute 15-\frac{1}{5} on the right: y+2=15x+25y + 2 = -\frac{1}{5}x + \frac{2}{5} Step 44 — Write in slope-intercept form by subtracting 22 from both sides: y=15x+252y = -\frac{1}{5}x + \frac{2}{5} - 2 y=15x85y = -\frac{1}{5}x - \frac{8}{5} The equation of the line is y=15x85y = -\frac{1}{5}x - \frac{8}{5}. This example illustrates how subtracting a negative coordinate (y(2)y - (-2)) becomes addition, and how a fractional slope produces fractional terms throughout the calculation.

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Updated 2026-05-03

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