Which of the following lines is perpendicular to $y = - 3 x + 2$ and has the same $y$ -intercept as $y = 3 x - 2$ ?

y = - \frac{1}{3} x

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y = - \frac{1}{3} x + 2

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y = - \frac{1}{3} x - 2

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y = \frac{1}{3} x - 2

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y = \frac{1}{3} x + 2

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Solution

The correct option is D $y = \frac{1}{3} x - 2$
The equation of given line $y = - 3 x + 2$ is $- 3$
Let the equation of line perpendicular to given line be $m$
Therefore we have $m \times (- 3) = - 1$
$\Rightarrow$ $m = \frac{1}{3}$
The $y$ - intercept of given line $y = 3 x - 2$ is $- 2$
Therefore the equation of required line with slope $\frac{1}{3}$ and $y$ -intercept $- 2$ is
$y = \frac{1}{3} x - 2$

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Similar questions

y = 3 x + 1

y = \frac{1}{4} x + 2

Equation of the line passing through (1, 2) and parallel to the line

$y = 3 x - 1$ is

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}, \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}, 3 x + y \neq 0, 3 x - y \neq 0

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}, \frac{1}{2 (3 x - y)} - \frac{1}{2 (3 x - y)} = - \frac{1}{8}

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}

\frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}

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