how do I know what formulas to use when trying to find a slope for linear equation

Question

^{I a having a hard time with intermediate algebra, I just don't get how to use the formulas to find the slopes to a lot of problems and it is getting really frustrating,  I need to know step by step what formulas to use and how to use them, so I can get this graphing stuff down packed, please help.}

Answer 1

Tamyra,

 

I can think of think of several types of problems that you might be given where you're asked to find the slope of a line.

 

1st, you might be given an equation of the type:

 

2·y - 6·x = 12  or maybe 5·y + 3·x =14, or ( x - 3 ) = (2·y + 5), or something similar.

 

The point-slope equation for a line is : y = m·x + b, where m is the slope and b is a number that represents the value of y when x = 0.

 

We need to put our examples into this form in order to determine the slope.  For our examples:

 

(1) 2y - 6x = 12 ⇒ 2y = 6x + 12 ⇒ y = 3x =6; we can now identify m = 3

 

(2) 5y + 3x = 14 ⇒ 5y = -3x + 14 ⇒ y = -(3/5)x + 14/5;  m = -(3/5)

 

(3) x - 3 = 2y + 5 ⇒ x = 2y + 8 ⇒ 2y = x - 8 ⇒ y = ½x - y;  m = ½

 

A second type of problem might be where you given the values of the x and y intercepts of a line and asked to find the slope.  This might be something like:

 

x = 3, y = -5, or maybe x = -2, y = -7

 

In cases like these we need to first express each point as an ordered pair (x,y).  It can be helpful when your learning to find slopes to print these points on a graph and connect them with a straight line.  It helps you to "see" the slope.  The slope is going to be the "change in y" divided by the "change in x"

 

(1) x = 3, y = -5 becomes (3, 0) and (0, -5)

 

For these two point we subtract the value of the first y - coordinate from the second, then do the same for x and finally divide (y₂ - y₁) by (x₂ - x₁) to get the slope.

 

This would be for (1):  for y: (-5 - 0) = -5, and x:  (0 - 3) = -3.  Then the slope is: (-5)/(-3) = 5/3

 

(2) x = -2, y = -7 becomes (-2,0) and (0,-7), so y = (-7-0) = -7 and x = (0-(-2)) = 2, 

      Slope = (-7)/(2) = -7/2

 

A third type of situation you might be given is where two points on the line are specified; say you are asked to find the slope from line passing through the points.

 

(2,4) and (7, 14)

 

Handle just like you did the x,y intercept form.

 

for y:  (14-4) = 10, and for x:  (7-2) = 5, then Slope = 10/5 = 2

 

Hope this helps.

 

Answer 2

Tamyra, what are the multiple formulae you are being asked to choose between?

 

I hit the web for a very brief moment to answer this question because, as far as I'm aware, there is only one equation for slope, though it can be worded differently and applied differently.

 

The simplest, perhaps, is merely to take two points on the line (which remember, will be described as (x,y)).  Let's say . . .

 

(7,13)

(10,31)

 

You just subtract the two y coordinates:

 

31-13=18

 

And then the two x-coordinates:

 

10-7=3

 

And then divide the first by the second:

 

18/3=6

 

And bam!  You have the slope!  This is generally written as such:

 

(y₂-y₁)/(x₂-x₁)

 

However, mathematicians often use the Greek letter delta, which just looks like a triangle (Δ) to mean "the change in".  Since what we're measuring in slope is how the two coordinates change, we can write the same formula as:

 

Δy/Δx

 

Means the same thing.  As does the third way of saying this, which is:

 

Rise/Run

 

 

The "rise" of a line just means how much it goes up.  You can find it by choosing two points on the line and drawing a little right triangle connecting them.  The side of that triangle which is vertical is the rise.  This is functionally the same as y₂-y₁ or Δy.  The "run" is how far forward it goes; this will be the horizontal line of that right triangle, which is again functionally the same as Δx or x₂-x₁.

 

Does that make sense?

 

 

Answer 3

Click on the link below to see a graphic explaining how to find the slope given two points on a line:

 

http://www.wyzant.com/resources/files/284889/computing_the_slope_of_a_line