Question -
Answer -
Given:
The equation isperpendicular to x тАУ 3y + 5 = 0 and passes through (3,-2)
The equation of theline perpendicular to x┬атИТ┬а3y + 5 = 0 is
3x + y +┬а╬╗┬а=0,
Where,┬а╬╗┬аisa constant.
It passes through(3,┬атИТ┬а2).
Substitute the valuesin above equation, we get
3 (3) + (-2) +╬╗┬а= 0
9 тАУ 2 +┬а╬╗┬а=0
╬╗┬а= тАУ 7
Now, substitute thevalue of ╬╗┬а=┬атИТ┬а7 in 3x + y +┬а╬╗┬а= 0, we get
3x + y тАУ 7 = 0
тИ┤ The required line is3x + y тАУ 7 = 0.