Find the equation of a line that meets the given condition

Find the equation of a line that meets the given condition : perpendicular to 6x-2y=10,passes through (6,-3)

Solution

6x-2y=10
put the equation in the form y=mx+b, where m is the slope and b is the y-intercept
6x-2y=10
-2y=-6x+10
y=3x+(-5) or y=3x-5
the slope is 3 and the y-intercept is -5
if two lines are perpendicular to each other their slopes are negative reciprocals of each other
the slope of the perpendicular line is -1/3
y=(-1/3)x+b
substitute 6 for x and -3 for y to find b
-3=(-1/3)6 +b
-3=-2+b
-1=b
therefore the equation of the line is y=(-1/3)x-1