Question

find the equation of the line passing through the point (4,5) and perpendicular to line 7x-5y=420​

Answer 1

Answer:

5x+7y-55=0

Step-by-step explanation:

7x-5y=420 can also be written as y=7x/5-84

therefore, it's slope is 7/5 and the slope of its perpendicular is -5/7. The perpendicular passes through (4,5). therefore, using point-slope form it's eqn is y-5=-5(x-4)/7

7y-35=-5x+20

5x+7y==55

Answer 2

Step-by-step explanation:

by using slope point form

Y-Y1=m(X-X1)

Y-5=-5/7(X-4)

7Y-36=-5X+20

5X+7Y-35-20=0

:• 5X+7Y-55=0