jonathanvaldes98
  • jonathanvaldes98
Write the equation of the line in slope-intercept form that passes through the point (3, -2) and has a slope of 2.
Mathematics
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jamiebookeater
  • jamiebookeater
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jonathanvaldes98
  • jonathanvaldes98
@Eric__Cartman @amistre64 #help
anonymous
  • anonymous
Write an equation of the Line 1 passing through (3,-4) that is parallel to the Line II 5x-3y-9=0. Does the line I pass through the point (6,7)? Does the line II pass through the point (6,7)? . First, we have to put Line II in slope-intercept form (y = Ax+b)("A" is the slope) . 5x-3y-9=0 . We will move (-3y) to the right side . 5x - 3y + 3y - 9 = 0 + 3y . 5x - 9 = 3y . 3y = 5x - 9 . We will now divide each side by "3" . +3y%2F3+=+%285x%2F3%29+-+%289%2F3%29+ . y = %285%2F3%29+x - 3 . (y = %285%2F3%29+x - 3) is the slope-intercept form of Line II . Since Line I is parallel to Line II, it has the same slope +%285%2F3%29+ . Equation to Line I . y = %285%2F3%29+x + b . We can solve "b" by replacing "x" and "y" with (3,-4)(x,y)(Since the line contains that point . +%28-4%29+=+%285%2F3%29%283%29+%2B+b+ . We will multiply the right side . +%28-4%29+=+5+%2B+b+ . We will move "5" to the left side . +%28-4%29+-+5+=+5+-+5+%2B+b+ . +%28-9%29+=+b+ . b = (-9), we can replace "b" in our equation . y = %285%2F3%29+x + b . y = %285%2F3%29+x + (-9) . y = %285%2F3%29+x - 9 . (y = %285%2F3%29+x - 9) is the slope-intercept equation to Line I . You can check by replacing "x" and "y" with (3, -4)(x,y) . +%28-4%29+=+%285%2F3%29%283%29+-+9+ . +%28-4%29+=+5+-+9+ . +%28-4%29+=+%28-4%29+ . (y = %285%2F3%29+x - 9) is the correct equation to Line I . y = %285%2F3%29+x - 9( to get a different form of equation, you get rid of the fraction)(We will multiply each side by "3") . +y%283%29+=+%285%2F3%29%28x%29%283%29+-+9%283%29+ . 3y = 5x - 27 . We will move "3y" to the right side . 3y - 3y = 5x - 3y - 27 0 = 5x - 3y - 27 (5x - 3y - 27 = 0) is another form,( you can move (-27) over, and it will become +5x+-+3y+=+27+ ( this is the standard form) . To see if (6,7) is a point of either of the two lines, you will replace "x" and "y" with (6,7)(x,y) . First equation, . +5x+-+3y+=+27+ . +5%286%29+-+3%287%29+=+27+ . +30+-+21+=+27+ . +9+=+27+ This is not true, so point (6,7) is not a point on the first line. . Second line . 5x-3y-9=0 ( standard form = (5x - 3y = 9)(moved (-9) to the right side) . 5x - 3y = 9 . We can replace "x" and "y" with (6,7)(x,y) . 5(6) - 3(7) = 9 . 30 - 21 = 9 . (9 = 9) This is true, (6,7) is a point on Line II
jonathanvaldes98
  • jonathanvaldes98
O.O

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anonymous
  • anonymous
><
anonymous
  • anonymous
o.o
anonymous
  • anonymous
O.O
jonathanvaldes98
  • jonathanvaldes98
i have to have them in that form but jeezus lol y = 2x - 8 y = 2x - 4 y = 2x + 7 y = 2x + 5

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