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]}