How do you write an equation in standard form given (2, 2) and (6, 3)?

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Answer 1 · 2018-02-24T01:12:58

#x-4y=-6#

I'm assuming that this is a linear equation...

The standard form of a linear equation is

#ax+by=c#

where #a# is non-negative and an integer.

You can also get the standard form by moving the #mx# to the left side in slope-intercept form.

#y=mx+b->-mx+y=b#

First, find the slope-intercept form by finding the slope:

#"slope"=(Δy)/(Δx) or (y_2-y_1)/(x_2-x_1)#

Plug in:

#(3-2)/(6-2)=1/4#

The slope is #1/4#.

Now you have #y=1/4x+b#

To find #b#, plug in any point. You are given #(2,2),(6,3)#.

#2=1/4*2+b=>2=2/4+b=>3/2=b# or

#3=1/4*6+b=>3=3/2+b=>3/2=b#

Either way, you'll get #3/2# as #b#.

Now, write out what you have:

#y=1/4x+3/2#

Convert to standard form:

#-1/4x+y=3/2#

#-4(-1/4x+y=3/2)#

#x-4y=-6#