|March 4th, 2015, 12:17 AM||#1|
Joined: Mar 2015
Hi I'm having a really hard time on these question and honestly don't know how to solve them.
Find the equation (in the form y=mx+c) of the line
a) passing through (1,5) with slope 6
c) passing through (3,-1 1over3) and having slope -1
e) vertical and passing through (6,-3)
Equations of linear functions
1. Determine whether each of the following lines is increasing or decreasing and determine whether the given point is above, below or on the line
c) y=4-x (1,0)
e) x+3y-8=0 (14,-2)
Find the equation (in the form y=mx+c) of the straight line passing through the points
a) (1,3) and (4,1)
c) (3,5) and (3,-5)
e) (-2,6) and (1,0)
I know it's a lot but I would really appreciate it if someone could show the working for all if not at least one of these questions.
Last edited by skipjack; March 4th, 2015 at 01:38 AM.
|March 4th, 2015, 01:46 AM||#3|
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city
Math Focus: Calculus/ODEs
Posting a long list of problems especially with no work shown is a huge turnoff to many folks who post help. So, to maximize your chances of getting prompt help, it is best to post one (or two at the most) problems along with what you have tried. Just constructive advice, my aim here is not to chide.
Now, let's look at the first 2 problems. What you need here is the point-slope formula:
Now, we can rewrite this in the required slope-intercept form:
where you are given the point $\left(x_1,y_1\right)$, and slope $m$.
So, can you take the given points and slopes and plug them into (2), and then simplify by combining like terms?
|March 4th, 2015, 01:54 AM||#4|
Joined: Dec 2006
The line through (p, q) with slope m has equation y = mx + (q - mp).
a). y = 6x + (-1)
c). y = -x + 1$\frac23$
e). x = 6 (A vertical line doesn't have a finite slope.)
The line though the points (m, p) and (n, q) (where m and n are not equal) has slope (q - p)/(n - m).
Given two points, find the slope then the equation of the line follows by the method used for (a) and (c) above.
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