How do you graph the lines using slope-intercept form #y = -2/3x + 1#?

1 Answer
smendyka
Nov 27, 2017

See a solution process below:

Explanation:

This equation is in slope intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.

#y = color(red)(-2/3)x + color(blue)(1)#

Therefore, the y-intercept is: #color(blue)(b = 1)# or #(0, color(blue)(1))#

We can plot this point on the grid as:

graph{(x^2 + (y-1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}

The slope is: #color(red)(m = -2/3)#

Slope is also: #"rise"/"run"#

In this case, the #"rise"# is #-2# so we need to go down 2 positions on the #y# value. And, the run is #3# so we need to go right 3 positions on the #x# value.

This second point is: #(0 + 3, 1 - 2) => (3, -1)#

We can now plot this point:

graph{(x^2 + (y - 1)^2 - 0.025)((x - 3)^2 + (y + 1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}

Now, we can draw a line through the two points giving:

graph{(y + (2/3)x - 1)(x^2 + (y - 1)^2 - 0.025)((x - 3)^2 + (y + 1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}

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