Question 832619
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Hi, Max--

THE PROBLEM:
How do you algebraically find the equation of a line whose slope is -2/3 and contains the point (-18,14)?

A SOLUTION:

Use the point-slope form. Here m is the slope, while {{{x[1]}}} and {{{y[1]}}} are the x- and y-values of a 
point on the line

{{{y-y[1]=m(x-x[1])}}}

Substitute -2/3 for m, -18 for {{{x[1]}}} and 14 for {{{y[1]}}}.
{{{y-14=(-2/3)(x-(-18))}}}

Simplify.
{{{y-14=(-2/3)(x+18)}}}

You can rearrange the terms to get this equation in slope-intercept form (y=mx+b) or in general form if 
you like, but this is a perfectly valid equation for the line with a slope of -2/3 that passed through the 
point(-18, 14).


Hope this helps! Feel free to email if you have any questions about the solution.

Good luck with your math,

Mrs. F
math.in.the.vortex@gmail.com
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