Algebra/Algebra 1 honors help
hi my class is working on graphing and finding the equations for parallel and perpendicular lines and i am a little bit confused
the questions are asking me to write an equation of the line that passes through the given point and is parallel to the graph of the given equation
1. (-4, -1); y=2x+14
i understand how to find the equation is by using point slope form: y+1=2(x+4)
but heres what confuses me: the + 14 part; do i just add that there and make ii y+1=2(x+4) +14?
where does it go? what am i supposed to do with it?
Hi Olivia, I will be happy to help you.
Because parallel lines go in the same direction, they have the same slope.
The slope-intercept form of a linear equation is
y = mx + b
The slope is m, the coefficient of the x term (or simply the number in front of x)
With y = 2x + 14, 2 is your slope. Any line parallel to this will also have a slope of 2.
The point-slope form of a linear equation is
y - y1 = m(x - x1)
Where m is your slope, in your case m = 2, x1 and y1 are the ordinates of a point, in your case x1 is -4 and y1 is -1, and x and y are just your variables.
To find the equation of your parallel line plug in the slope and point's coordinates into slope-intercept form and solve
y - (-1) = 2(x - (-4))
Simplify to get
y + 1 = 2(x + 4)
y + 1 = 2x + 8
Solving will get
y = 2x +7
This is the equation of the parallel line. It has the same slope, 2, as the original line. Plus, you can check that it goes through (-4, -1) by plugging in those values for x and y.
As to your question about the +14, that is just the y-intercept of the original line. When finding parallel and perpendicular lines, all you are interested in from the original equation is the slope.
I hope this helps. Please let me know if you have any additional questions.