write the slope- intercept form of the equation of the line parallel to - x+5y=25, passing through the point (-5,-5). simplify

1 answer:

5 0

To find the equation of a line that is parallel to your original equation and goes through a certain point on a graph, here's what you need to know:
First you need to find the slope of your original equation.
To do that, you need to convert it to slope intercept form (y = mx+b).
Add the x over, and then divide everything by 5 to get the y by itself.
Here's what that would look like (without the small steps that I mentioned):
-x + 5y = 25
5y = x + 25
y = 1/5x + 5

That's the original equation rewritten in slope intercept form.
The m represents the slope, so this equation's slope is 1/5.

Because you are given a point, and now you have a slope, the best and easiest route is using point slope form.
I've seen different versions of the equation base but I prefer y - y(sub1) = m(x - x(sub1))
But since I can't use subscripts in this, I'll use the one with h and k. The h is the x value of the point, and the k is the y value.
(h,k)
Then just substitute the values in and solve for y.
y - k = m(x - h)
y + 5 = 1/5(x + 5)
y + 5 = 1/5x + 1
y = 1/5x - 4

Your final answer is
y = 1/5x - 4
You can double check by using a graph. If the slopes are the same, the lines should be parallel.

I hope that helps. If anything didn't make sense, feel free to ask me.

You might be interested in

use the guide to construct a two column proof proving △aec≅△deb, given that ca is parallel to db and e is the midpoint of ad. gi

MrMuchimi

By using the AAA congruence property of Triangles, △aec≅△deb is proved

It is given two triangles, Δaec and Δdeb, where e is the common point known as the midpoint of ad

Also, it is given that,

ca ║db

We need to prove that, △aec≅△deb

Then we'll use AAA congruence property of Triangles to prove the situation

As ca ║db

then, ∠cae = ∠ebd (Alternate angles)

∠ace = ∠edb (Alternate angles)