The graph of a linear equation contains the points (3.11) and (-2,1). Which point also lies on the graph?

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Answer 1 · 2018-05-12T20:59:28

(0, 5) [y-intercept], or any point on the graph below

First, find the slope with two points by using this equation:

#(Y_2 - Y_1)/(X_2 - X_1)# = #m#, the slope

Label your ordered pairs.

(3, 11) #(X_1, Y_1)#
(-2, 1) #(X_2, Y_2)#

Plug in your variables.

#(1 - 11)/(-2 - 3)# = #m#

Simplify.

#(-10)/(-5)# = #m#

Because two negatives divide to make a positive, your answer will be:

#2# = #m#

Part Two

Now, use point-slope formula to figure out what your equation in y = mx + b form is:

#y - y_1 = m(x - x_1)#

Plug in your variables.

#y - 11 = 2(x - 3)#

Distribute and simplify.

#y - 11 = 2x - 6#

Solve for each variable. To solve for the y = mx + b equation, add 11 to both sides to negate -11.

#y = 2x + 5#

Now, plot this on a graph:

graph{2x + 5 [-10, 10, -5, 5]}