Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0.
Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0.

According to ques,:

The equation of the line perpendicular to:

    \[~x~-~3y\text{ }+\text{ }5\text{ }=\text{ }0\]

is

    \[3x\text{ }+\text{ }y\text{ }+~\lambda ~=\text{ }0,\]

Where, λ is a constant.

It passes through

    \[\left( 3,~-~2 \right).\]

Substitute the values in above equation, we get

    \[3\text{ }\left( 3 \right)\text{ }+\text{ }\left( -2 \right)\text{ }+\text{ }\lambda ~=\text{ }0\]

    \[9\text{ }\text{ }2\text{ }+~\lambda ~=\text{ }0\]

Or ,

    \[\lambda ~=\text{ }\text{ }7\]

Now, substituting the value of

    \[\lambda ~=~-~7\text{ }in\text{ }3x\text{ }+\text{ }y\text{ }+~\lambda ~=\text{ }0\]

, we have

    \[3x\text{ }+\text{ }y\text{ }\text{ }7\text{ }=\text{ }0\]

∴ The required line is

    \[3x\text{ }+\text{ }y\text{ }\text{ }7\text{ }=\text{ }0.\]