Find the equation of the straight line perpendicular to 3x+5y+6=0 that passes through (3,4)

First we need to write the equation in the form y = mx + c 3x + 6= 5y (3x+6) / 5 = y -3/5x + 6/5 = ynow we know that the gradient is -3/5 therefore the perpendicular gradient is 5/3.so our line equation is 5/3x + c =y, we now need to find the value of our constant, c , using the coordinate given so x=3 and y=4 5/3(3) + c = 4 5 + c = 4 c = -1therefore we now know our line equation is 5/3x - 1 = y