How do you write the equation of the line that passes through #(-3, 3)# and #(2,-7)#?

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Answer 1 · 2017-08-04T05:02:44

#y=-2x-3#

The gradient/intercept form of a linear equation is:

#y=mx+c#

We need to solve for the two unknowns #m# and #c# to find the equation.

First, solve for the gradient, #m#:

#m=(y_2-y_1)/(x_2-x_1)=(-7-3)/(2-(-3))=-2#

Substitute this value into the gradient/intercept form:

#y=-2x+c#

Now substitute in any point on the line to solve for #c#:

#(-3,3)#:

#3=-2(-3)+crArr3=6+crArrc=-3#

#c# is the y-intercept.

Inserting #c# and #m# into the original equation gives the answer:

#:.y=mx+c=-2x-3#