Question 161909
Hi, Hope I can help,
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can somebody help me solve this problem. i need to find the equation of the line with these characteristics. I need to express my answer using either the general form or the slope-intercept form of the equation of the line.
x-intercept=2; containing the point (4,-5)
i forgot how to find the slope with only the, x-intercept = 2 given.
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The slope intercept form of the equation is {{{ y = mx + b }}}, where "m" is the slope, "b" is the y-intercept
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The x-intercept is the point where the line crosses the "x" axis
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x intercept is given as the point ( p, 0 ) where "p" is the "x intercept number", in our case, it would be ( 2,0 )
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Now we have another point (2,0)
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Now that we have two points we can find the slope of the line
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The two points are (2,0) and (4, -5)
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(2,0)(x1,y1) and (4, -5)(x2,y2)
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The slope equation is {{{ ( y2-y1)/(x2-x1) }}}, we can replace the variables with numbers
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Our slope = {{{ ( y2-y1)/(x2-x1) }}} = {{{ ( (-5)-(0))/((4)-(2)) }}} = {{{ -5/2 }}}
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Our slope = {{{ -5/2 }}}
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Our slope intercept form is {{{ y = mx + b }}}, "m" is our slope, so we can replace "m" with {{{ -5/2 }}}
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Our equation is now {{{ y = (-5/2)x + b }}}, to find "b", we will replace "x" and "y" with one of our two points ( doesn't matter which one ) ( we will use point (2,0), replace "x" with "2", "y" with "0")
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(2,0)(x,y) , {{{ y = (-5/2)x + b }}} = {{{ (0) = (-5/2)(2) + b }}} = {{{ 0 = (-5) + b }}}, we will move (-5) to the left side
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{{{ 0 = (-5) + b }}} = {{{ 0 + 5 = (-5) + 5 + b }}} = {{{  5 = b }}} = {{{ b = 5 }}}, we found "b", we can replace "b" with "5" in the original equation
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{{{ y = (-5/2)x + b }}} = {{{ y = (-5/2)x + (5) }}} = {{{ y = (-5/2)x + 5 }}}
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{{{ y = (-5/2)x + 5 }}} is the slope intercept form of the equation, for the general equation, we will multiply each side by "2"
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{{{ y = (-5/2)x + 5 }}} = {{{ (2)y = (2)((-5/2)x + 5) }}} = {{{ 2y = (2)(-5/2)x + (2)(5) }}} = {{{ 2y = (-5)x + 10 }}} = {{{ 2y = -5x + 10 }}}, we will move (-5x) to the left side
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{{{ 2y = -5x + 10 }}} = {{{ 2y + 5x = -5x + 5x + 10 }}} = {{{ 2y + 5x = 10 }}}, rearranging it becomes
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{{{ 2y + 5x = 10 }}} = {{{ 5x + 2y = 10 }}}
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{{{ 5x + 2y = 10 }}} is the general/standard form, we can check our equation by replacing "x" and "y" with both of our two points
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The two points are (2,0) and (4, -5)
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(2,0)(x,y) = {{{ 5x + 2y = 10 }}} = {{{ 5(2) + 2(0) = 10 }}} = {{{ 10 + 0 = 10 }}} = {{{ 10 = 10 }}} ( True )
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(4, -5)(x,y) = {{{ 5x + 2y = 10 }}} = {{{ 5(4) + 2(-5) = 10 }}} = {{{ 20 + (-10) = 10 }}} = {{{ 20 - 10 = 10 }}} = {{{ 10 = 10 }}} ( True )
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{{{ y = (-5/2)x + 5 }}} is the slope-intercept form of the equation
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{{{ 5x + 2y = 10 }}} is the general/standard form of the equation
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This is the line on a graph
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{{{ drawing (800,800,-20,20,-20,20,grid(1), graph (800,800,-20,20,-20,20,(-5/2)x + 5), blue (circle (4,-5,0.1)), blue (circle (2,0,0.1)), circle (4,-5,0.2), circle (2,0,0.2)) }}}
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Hope I helped, Levi