Subjects algebra

Perpendicular Line 26C5E6

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1. **State the problem:** Find the equation of a line passing through the point $(-4,2)$ and perpendicular to the line given by $3x + 4y - 7 = 0$. 2. **Identify the slope of the given line:** Rewrite the given line in slope-intercept form $y = mx + b$. $$3x + 4y - 7 = 0 \implies 4y = -3x + 7 \implies y = -\frac{3}{4}x + \frac{7}{4}$$ So, the slope of the given line is $m = -\frac{3}{4}$. 3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another is the negative reciprocal of the original slope. $$m_{\perp} = -\frac{1}{m} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}$$ 4. **Use point-slope form to find the equation of the perpendicular line:** The point-slope form is: $$y - y_1 = m(x - x_1)$$ Substitute $m = \frac{4}{3}$ and point $(-4, 2)$: $$y - 2 = \frac{4}{3}(x - (-4)) = \frac{4}{3}(x + 4)$$ 5. **Simplify the equation:** $$y - 2 = \frac{4}{3}x + \frac{16}{3}$$ Add 2 to both sides: $$y = \frac{4}{3}x + \frac{16}{3} + 2$$ Convert 2 to fraction with denominator 3: $$2 = \frac{6}{3}$$ So, $$y = \frac{4}{3}x + \frac{16}{3} + \frac{6}{3} = \frac{4}{3}x + \frac{22}{3}$$ **Final answer:** $$\boxed{y = \frac{4}{3}x + \frac{22}{3}}$$