How do you graph #x - 2/3y = 1 # by plotting points?

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Answer 1 · 2017-09-04T16:23:20

See a solution process below:

First, solve for two points which solve the equation and plot these points:

First Point:

For #y = 0#

#x - (2/3 * 0) = 1#

#x - 0 = 1#

#x = 1# or #(1, 0)#

Second Point:

For #y = 3#

#x - (2/3 * 3) = 1#

#x - 2 = 1#

#x - 2 + color(red)(2) = 1 + color(red)(2)#

#x - 0 = 3#

#x = 3# or #(3, 3)#

We can next graph the two points on the coordinate plane:

graph{((x-1)^2+y^2-0.035)((x-3)^2+(y-3)^2-0.035)=0 [-10, 10, -5, 5]}

Now, we can draw a straight line through the two points to graph the line:

graph{(x-2/3y-1)((x-1)^2+y^2-0.035)((x-3)^2+(y-3)^2-0.035)=0 [-10, 10, -5, 5]}