1Cademy - Finding the Equation of a Line Perpendicular to $$y = \frac{1}{2}x

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Finding an Equation of a Line Perpendicular to a Given Line

Example

Finding the Equation of a Line Perpendicular to $y = \frac{1}{2}x - 3$ Through (6, 4)

To successfully assemble the formal equation for a line striking perpendicularly against $y = \frac{1}{2}x - 3$ exactly located at the point (6, 4), follow the procedural protocol designed to wrap the solution conclusively into slope-intercept form. Step 1 — Extract the initial slope parameter. Since the provided baseline equation $y = \frac{1}{2}x - 3$ arrives formatted in standard slope-intercept orientation, the slope behaves transparently as $m = \frac{1}{2}$ . Step 2 — Determine the inverted perpendicular equivalent. The nature of perpendicular crossings ensures opposing negative reciprocals. Directly flipping the fraction $\frac{1}{2}$ upside down and affixing a rigid negative marker results in $-2$ . The applicable perpendicular slope activates as $m_{\perp} = -2$ . Step 3 — Catalog the required intercept point. The trajectory of the perpendicular offshoot remains absolutely bound to passing through the spatial coordinate $(x_1, y_1) = (6, 4)$ . Step 4 — Migrate collected details into point-slope construction. Position the curated components inside the equation block $y - y_1 = m_{\perp}(x - x_1)$ : $y - 4 = -2(x - 6)$ Propagate the $-2$ multiplier carefully against both inner parentheses participants: $y - 4 = -2x + 12$ Step 5 — Stabilize purely into slope-intercept limits. Inject a robust addition of $4$ against both flanking sides to perfectly detach $y$ : $y = -2x + 16$ The algebraic transcription verifying the finalized perpendicular path computes definitively as $y = -2x + 16$ .

Updated 2026-06-17

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OpenStax Intermediate Algebra

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