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.
the sum of the angles of any triangle is equal to 180°. So, we do 180-50= 130. Then we divide 30 by two because there are two remaining angles so 130/2=65. x=65
We have y - 2 = m(x + 5), where m = (-1 - 2)/(10 + 5) = -3/15 = -1/5; Then y - 2 = (-1/5)(x + 5); 5y - 10 = (-1)(x + 5); 5y - 10 = -x - 5; x + 5y - 5 = 0 is the equation of the line.
