Find the equation of the line passing through | vedclass

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(A) Given the line equation: $y = 3x - 1$. Comparing this with the slope-intercept form $y = mx + c$,the slope $m_1 = 3$. Since the required line is p

Vedclass · Accepted Answer

$x + 3y - 7 = 0$

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(A) Given the line equation: $y = 3x - 1$. Comparing this with the slope-intercept form $y = mx + c$,the slope $m_1 = 3$. Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \cdot m_2 = -1$. Therefore,$3 \cdot m_2 = -1$,which gives $m_2 = -\frac{1}{3}$. The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $(y - y_1) = m(x - x_1)$. Substituting $(x_1, y_1) = (1, 2)$ and $m = -\frac{1}{3}$: $(y - 2) = -\frac{1}{3}(x - 1)$. Multiplying both sides by $3$: $3(y - 2) = -(x - 1) \Rightarrow 3y - 6 = -x + 1$. Rearranging the terms: $x + 3y - 7 = 0$.

Find the equation of the line passing through the point $(1, 2)$ and perpendicular to the line $y = 3x - 1$.

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