A line passes through (2,2) and is perpendicu | vedclass

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(C) The slope of the given line $3x+y=3$ is $m_1 = -3$. Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(C) The slope of the given line $3x+y=3$ is $m_1 = -3$. Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$. Thus,$m_2 = \frac{-1}{-3} = \frac{1}{3}$. The equation of the line passing through $(2,2)$ with slope $m_2 = \frac{1}{3}$ is given by $y - 2 = \frac{1}{3}(x - 2)$. Multiplying by $3$,we get $3y - 6 = x - 2$,which simplifies to $x - 3y = -4$. To find the $y$-intercept,we set $x = 0$: $0 - 3y = -4 \Rightarrow y = \frac{4}{3}$. Therefore,the $y$-intercept is $\frac{4}{3}$.

$A$ line passes through $(2,2)$ and is perpendicular to the line $3x+y=3$. Its $y$-intercept is

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