How do you write an equation in standard form given a line that passes through (4,-3) with m=2?

1 Answer
Jun 4, 2015

The equation in standard form is #2x-y=5#.

The general equation for the standard form of a line is #"A"x+"B"y="C"#, where #"A"# and #"B"# are integers, and #"A"!=0#, and #"B"!=0#.
http://www.mathwarehouse.com/algebra/linear_equation/standard-form-equation-of-a-line.php

Start with the equation to find the point-slope form of a line: #y-y_1=m(x-x_1)#, where #m# is the slope, and #x_1# and #y_1# are the known point.

#m=2#
#x_1=4#
#y_1=-3#

Solution:

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

Substitute known values into the equation.

#(y-(-3))=2(x-4)# =

#y+3=2(x-4)# = Point-slope form.

Continue to the Standard Form.

Distribute the #2#.

#y+3=2x-8#

Subtract #y# from both sides.

#3=2x-y-8#

Add #8# to both sides.

#11=2x-y#

Flip the equation.

#2x-y=11# = Standard Form

graph{2x-y=11 [-32.48, 32.43, -16.23, 16.24]}