Question 1135207
To begin with, relax  :-)
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slope intercept form is
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y = mx +b, where m is the slope and b is the y axis intercept
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point slope form is
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(y-y1) = m(x-x1), m is the slope and the line passes through the point (x1,y1)
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standard form is
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ax +by +c = 0, where a, b, c are constants
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two point form is
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(x-x1)/(x2-x1) = (y-y1)/(y2-y1), where the two points are (x1,y1) and (x2,y2)
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1) 4x - 2y = 8 
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lets write the point slope form of this line
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-2y = 8 -4x
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divide both sides of the = by -2
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2) y = 2x +8
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a) the line parallel to equation 2 has the same slope
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3) y = 2x +b
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we are given the point (4,2) which the parallel line passes through
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substitute x = 4 and y = 2 in equation 3
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2 = 2(4) + b
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2 = 8 +b
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b = -6
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slope intercept form is
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y = 2x -6
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point slope form is
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(y-2) = 2(x-4)
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standard form is
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2x -y -6 = 0
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two point form is
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use the slope intercept form to find the y axis intercept point
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set x = 0, then
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y = 2(0) -6 = -6
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therefore the second point is (0, -6)
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two points we are using are (4,2) and (0,-6)
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(x-4)/(0-4) = (y-2)/(-6-2)
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b) the slope of the perpendicular line to equation 2 is the negative
reciprocal of equation 2's slope
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therefore, the slope is -1/2
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y = -x/2 +b
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2 = -4/2 +b
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b = 4
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slope intercept form is 
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y = -x/2 +4
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point slope form is
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(y-2) = -(x-4)/2
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standard form is
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x/2 +y -4 = 0
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the points to use are (4,2) and (0,4)
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two point form is 
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(x-4)/(0-4) = (y-2)/(4-2)
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c)  I will use the slope intercept form for the three lines
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4) y = 2x +8
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points are (1,10),(2,12),(0,8),(-4,0)
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5) y = 2x -6
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points are (4,2),(2,-2),(0,-6),(3,0)
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6) y = -x/2 +4
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points are (4,2),(2,5),(0,4),(8,0)
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here is the graph, red line is equation 4, green line is equation 5, and blue line is equation 6
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{{{ graph( 300, 200, -15, 15, -15, 15, 2x+8, 2x-6, -x/2 +4) }}}
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