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A line passes through \( (2,2) \) and is perpendicular in the line \( 3 x+y=3 \) its \( y \)-intercepts is
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The correct answer is:
\( \frac{4}{3} \)
Given line, $3 x+y=3$
$\Rightarrow y=-3 x+3$
So, slope of this line is $m_{1}=-3$
Then, slope of line perpendicular to this line is $m_{2}=\frac{1}{3}$
General equation line passing through point $\left(x_{1}, y_{1}\right)$ is
$\left(y-y_{1}\right)=m\left(x-x_{1}\right)$
Here $m=\frac{1}{3}$ and $\left(x_{1}, y_{1}\right)=(2,2)$. So,
$y-2=\frac{1}{3}(x-2)$
$\Rightarrow y-2=\frac{x}{3}-\frac{2}{3}$
$\Rightarrow y=\frac{x}{3}+2-\frac{2}{3}$
$\Rightarrow y=\frac{x}{3}+\frac{4}{3}$
So, $y$-axis intercept is, $\frac{4}{3}$
$\Rightarrow y=-3 x+3$
So, slope of this line is $m_{1}=-3$
Then, slope of line perpendicular to this line is $m_{2}=\frac{1}{3}$
General equation line passing through point $\left(x_{1}, y_{1}\right)$ is
$\left(y-y_{1}\right)=m\left(x-x_{1}\right)$
Here $m=\frac{1}{3}$ and $\left(x_{1}, y_{1}\right)=(2,2)$. So,
$y-2=\frac{1}{3}(x-2)$
$\Rightarrow y-2=\frac{x}{3}-\frac{2}{3}$
$\Rightarrow y=\frac{x}{3}+2-\frac{2}{3}$
$\Rightarrow y=\frac{x}{3}+\frac{4}{3}$
So, $y$-axis intercept is, $\frac{4}{3}$
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