Question 150744
Hi, Hope I can help,
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what is the equation in the standard form of the line through (4,1) and (-2,3)?
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First, we need to find the slope of the line.
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We can find the slope by using the two points ( the slope is equal to {{{ (y2 - y1)/(x2 - x1) }}} or {{{ (y1 - y2)/ (x1 - x2) }}}
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We will use the first equation
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(4,1)( x1 , y1 )  and (-2,3) ( x2 , y2 )
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{{{ (y2 - y1)/(x2 - x1) }}}
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{{{ (3 - 1) / ((-2) - 4 ) }}}
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{{{ 2/ (-6) }}}
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It will reduce to {{{ (-1)/3 }}}
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 - {{{ 1/3 }}} is the slope
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The slope intercept form of a line = {{{ y = mx + b }}} m = slope, b = unknown
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We can replace "m" with - {{{ 1/3 }}}
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It becomes {{{ y = (-1/3)x + b }}}
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We can find "b" by replacing "x" and "y" with one of our points
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(4,1)(x,y) and (-2,3) (x,y)
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We will use the first point, (4,1) (x,y)
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{{{ y = (-1/3)x + b }}}
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{{{ (1) = (-1/3)(4) + b }}}
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{{{ 1 = (-4/3) + b }}}
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We will move {{{ (-4/3) }}} to the left
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{{{ 1 + (4/3) = (-4/3) + (4/3) + b }}}
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{{{ (7/3) =  b }}}
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{{{ b = (7/3) }}}
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We can replace "b" with {{{ 7/3 }}} in our equation
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{{{ y = (-1/3)x + b }}}
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{{{ y = (-1/3)x + (7/3) }}}
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That is the slope intercept form of the equation, The standard form is {{{ Ax + By = C }}}
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To get it to the standard form, we need to get rid of the fractions
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{{{ y = (-1/3)x + (7/3) }}}
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We will multiply everything by "3"
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{{{ (3)y = (3)(-1/3)x + (3)(7/3) }}}
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{{{ 3y = (-1)x + 7 }}}
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{{{ 3y = (-x) + 7 }}}
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We will move (-x) to the left side
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{{{ 3y + x = (-x) + x + 7 }}}
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{{{ 3y + x = 7 }}}
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{{{ x + 3y = 7 }}}
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{{{ x + 3y = 7 }}} is the standard form,
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{{{ Ax + By = C }}}
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{{{ (1)x + (3)y = (7) }}}
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{{{ x + 3y = 7 }}}
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We can check by replacing "x" and "y" with the two points
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(4,1) and (-2,3)
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We will use the first point, (4,1) (x,y)
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{{{ x + 3y = 7 }}}
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{{{ 4 + 3(1) = 7 }}}
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{{{ 4 + 3 = 7 }}}
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{{{ 7 = 7 }}} True
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We will use the second point, (-2,3) (x,y)
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{{{ x + 3y = 7 }}}
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{{{ (-2) + 3(3) = 7 }}}
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{{{ (-2) + 9 = 7 }}}
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{{{ 7 = 7 }}} True
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Some other points to {{{ x + 3y = 7 }}} = (1,2), (7,0), and (-8,5)
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{{{ x + 3y = 7 }}} is the standard equation for the line
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Hope I helped, Levi