Find the equation of a line that is perpendicular to the lin

Find the equation of a line that is perpendicular to the line

y equals one half x plus 5y=12x+5

and contains the point left parenthesis negative 3 comma 0 right parenthesis .(3,0).

5y = 12x + 5

writing the equation of line in slope intercept form

isolating y we get

y = 12/5 x + 1

hence, slope of this line = 12/5

slope of perpendicular lines are negative reciprocal of each other

so, slope pf perpendicular line would be -5/12

and passes throught (-3,0)

plugging the values in the equation y = mx+b

0 = -5/12(-3) + b

b = - 5/4

hence, equation of perpendicular line is

y = -5/12 x - 5/4

12y = - 5x - 15