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
MT
Related Maths A Level answers
All answers ▸How many books and modules and what are they all about?
If I had an equation with both 'x' and 'y' present, how would I find the gradient?
Show, by first principles, that the differential of x^2 is 2x.
A block of temperature H=80ºC sits in a room of constant temperature T=20ºC at time t=0. At time t=12, the block has temperature H=50ºC. The rate of change of temperature of the block (dH/dt) is proportional to the temperature difference of the block ...