Show that the pair of the lines $3 x - y = 0$ and $x + 3 y - 3 = 0$ is perpendicular.

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Solution

Step 1: Converting the given equations in standard form of a line.
The given lines are $3 x - y = 0$ and $x + 3 y - 3 = 0$ .
Let, $m_{1,} m_{2}$ be the slopes of lines $3 x - y = 0$ and $x + 3 y - 3 = 0$ respectively.
Converting the given line in the form $y = m x + c$
$\Rightarrow y = 3 x$ and $y = - \frac{x}{3} + 1$
Step 2: Showing that lines are perpendicular:
On comparing the above equations with $y = m x + c$
$m_{1} = 3$ and $m_{2} = \frac{- 1}{3}$
For any two lines are perpendicular, the product of their slopes must be $- 1$
Now, $m_{1} \times m_{2} = 3 \times \frac{- 1}{3}$
$\Rightarrow m_{1} \times m_{2} = - 1$
Therefore , the given lines $3 x - y = 0$ and $x + 3 y - 3 = 0$ are perpendicular.

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