How do you graph the line through point (2,6) and slope m=2/3?

1 Answer
Jun 22, 2018

Answer:

See a solution process below:

Explanation:

First, we can plot the point given in the problem:

graph{((x-2)^2+(y-6)^2-0.1)=0 [-20, 20, -10, 10]}

Slope is #"rise"/"run"# or the change in the #y# values over the change in the #x# values:

The slope in the problem is #2/3#/3#

Therefore,

The change in the #"rise"# or #y# value is #color(red)(+2)#

The change in the #"run"# or #x# value is #color(blue)(+3)#

We can find another point by adding these to the values from the point in the problem:

#(2, 6) -> (2color(blue)(+ 3), 6color(red)(+2)) -> (5, 8)#

We can now plot this point:

graph{((x-2)^2+(y-6)^2-0.1)((x-5)^2+(y-8)^2-0.1)=0 [-20, 20, -10, 10]}

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

graph{(y-2/3x - 4.6)((x-2)^2+(y-6)^2-0.1)((x-5)^2+(y-8)^2-0.1)=0 [-20, 20, -10, 10]}