
Let's solve ~
The equation of line perpendicular to required line is :

now, let's write it in alope intercept form to get its slope.


so, slope of this line is 3, and we know that product of slopes of two perpendicular lines is -1.
So, slope of required line is :


Now, we know slope of required line ( i.e - 1/3 ) and ir goes through point (6 , 3), let's write it's equation in point slope form ~



