*find the equation?

Question

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

I am so lost:(

Answer 1

First, you would change x - 5y = 4 into standard form, which is y = mx + b, with m being the slope of the line and b being the y-intercept (where the line passes through the y-axis).

After you do that, find what "m" is. Flip the fraction and make it negative to find the slope of the perpendicular line. The coordinates (-3, 5) are part of this line, with -3 being the x value and 5 being the y value. Fill them in to your perpendicular line equation, and you will get the value of b. Replace x and y into the equation and you have your answer.

So, your solution would be like this:

x-5y = 4
5y = x - 4
y = 1/5x - 4
m (or the slope) = 1/5

Flip it over and make negative is slope of the perpendicular line, so it would be -5

New equation: y = -5x + b
Fill in -3 and 5: 5 = -3 times -5 + b
5 = 15 + b
b = -10

y = -5x -10 is your answer

Answer 2

when ax + by = c, any line ax + by = d is parallel, since -a/b is the slope; likewise bx - ay = d will be perpendicular, so your line starts out 5x + y = something, and you plug in the (-3,5):

5(-3) + (5) = -10, so
5x + y = -10 is the line you want.