1Cademy - Finding the Equation of a Line Perpendicular to $$y = 3x + 1$$ Through $$(4, 2)$$

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Finding an Equation of a Line Perpendicular to a Given Line

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Finding the Equation of a Line Perpendicular to $y = 3x + 1$ Through $(4, 2)$

To systematically unravel the exact equation of a line that travels perpendicular to $y = 3x + 1$ while crossing flawlessly through the specified point $(4, 2)$ , adopt the five-step algebraic mapping method to synthesize the final model in slope-intercept format. Step 1 — Identify the primary line's slope. Examining the reference equation $y = 3x + 1$ , it is actively structured in standard slope-intercept form ( $y = mx + b$ ), ensuring its current slope is $m = 3$ . Step 2 — Pinpoint the perpendicular slope. Truly perpendicular lines dictate slopes functioning as opposite reciprocals. Thus, the negative reciprocal counterpart of $3$ flips cleanly down to $-\frac{1}{3}$ , solidifying the active perpendicular slope as $m_{\perp} = -\frac{1}{3}$ . Step 3 — Note the targeted location. The newly formulated line must be routed unequivocally through the defined coordinate pair $(x_1, y_1) = (4, 2)$ . Step 4 — Substitute into the functional point-slope framework. Embed these specific variables directly into $y - y_1 = m_{\perp}(x - x_1)$ : $y - 2 = -\frac{1}{3}(x - 4)$ Properly route the fractional slope against the right-side terms: $y - 2 = -\frac{1}{3}x + \frac{4}{3}$ Step 5 — Condense completely into slope-intercept form. Separate the solitary $y$ variable by pushing $2$ uniformly across the equality sign. Transforming the whole integer $2$ neatly into the fraction $\frac{6}{3}$ mitigates calculation hazards: $y = -\frac{1}{3}x + \frac{4}{3} + \frac{6}{3}$ $y = -\frac{1}{3}x + \frac{10}{3}$ Summarily, the finalized equation characterizing the perpendicular offshoot is confidently recorded as $y = -\frac{1}{3}x + \frac{10}{3}$ .

Updated 2026-05-03

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OpenStax Intermediate Algebra

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