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Finding the Equation of a Line Through (−4,−3)(-4, -3) and (1,−5)(1, -5)

To find the equation of a line containing the points (−4,−3)(-4, -3) and (1,−5)(1, -5), apply the two-points procedure and write the result in slope-intercept form. Step 11 — Find the slope. Apply the slope formula: m=−5−(−3)1−(−4)m = \frac{-5 - (-3)}{1 - (-4)} Subtracting a negative is equivalent to addition: m=−5+31+4=−25=−25m = \frac{-5 + 3}{1 + 4} = \frac{-2}{5} = -\frac{2}{5} Step 22 — Choose one point. Use (1,−5)(1, -5), so x1=1x_1 = 1 and y1=−5y_1 = -5. Step 33 — Substitute into the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1): y−(−5)=−25(x−1)y - (-5) = -\frac{2}{5}(x - 1) Simplify the left side and distribute the slope on the right: y+5=−25x+25y + 5 = -\frac{2}{5}x + \frac{2}{5} Step 44 — Write in slope-intercept form by subtracting 55 from both sides: y=−25x+25−5y = -\frac{2}{5}x + \frac{2}{5} - 5 y=−25x−235y = -\frac{2}{5}x - \frac{23}{5} The equation of the line is y=−25x−235y = -\frac{2}{5}x - \frac{23}{5}.

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Updated 2026-05-03

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References

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OpenStax

Intermediate Algebra @ OpenStax

Ch.3 Graphs and Functions - Intermediate Algebra @ OpenStax

Algebra

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