More algebra help please.?

Question

Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x – 5y = 4 and passing through (-3, 5).

The only thing I know so far is, the format for a standard equation is Ax + By = C. Do I turn the points into an equation and solve the system? (I am so lost).

Answer 1

The slope of the line x - 5y = 4 is m = 1/5. Write the equation in slope-intercept form and you will see it. A perpendicular line has the opposite inverse as its slope (-1/m) in this case -5.
Use the point slope formula y - 5 = -5(x + 3)
y - 5 = -5x - 15
y = -5x - 10
5x + y = -10

Answer 2

typing
x-5y -4=0
thus 5y=x-4
the slope is one
thus the perpindicular equations is the same
making it 5= 1(-3)+b
b=8
thus the equation
is y= x + 8

Answer 3

a line perp. to x-5y=4 is -1/x - 5(y)=4 and substitute the -3 for x and 5 for y and evaluate it . There you go

Answer 4

first put this equation in y=mx+b form
y= -1/5x + 4/5
to find the line perpendicular to this u need the opposite reciprocal slope which would be 5
then use the point slope formula (y-y1= m (x-x1)
so it would be 5-y= 5 (x-3)
y= -5x +20
that is ur perpendicular line
now put that in Ax+By=C which is 5x+y=20

Answer 5

x-5y=4
-5y=4-x
y=x/5-4/5

Gradient=1/5

For a line to be perpendicular to this line, its gradient has to be -5

This line is of the form y=-5x+c

To find c, substitute the coordinates given for x & y

5=-5*-3+c
5=15+c
c=-10

Answer y=-5x-10

Answer 6

Well! Three different answers and this case it is only possible for one to be correct. One of them has done it correctly but I'll leave you to decide which.