Question 217087
base  eqn,,,x+3y =6,,,,,change  to  slope  intercept  form,,,,3y=-x+6,,,y=(-1/3)x +2,,,,,with  m1 = (-1/3)  and  y intercept  = +2
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We  know  a  perpendicular  has  negative  reciprocal  slope,,,(m2=-1/m1),,,,therefore  the  slope  of  the  perpendicular  to  the  base  eqn  is,,,,,,,,m2= +3
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For  the  perpendicular  line  eqn,,,,start  with  y=mx+b,,,,and  y=3x +b
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To  find  b,,,,,subst  ( -3,5)  into  eqn
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(5)= (+3)(-3) +b
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5=-9=b
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14=b,,,,,,,,subst  back  into  eqn
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y=3x + 14,,,,,,answer  to  perpendicular  line in  slope  int form
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-14 = 3x-y,,,,,or 3x-y = -14,,,,in  standard  form
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check,  m2= -1/m1,,,,,,,3=-1/(-1/3)= +3,,,,ok
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(-3,5),,,,check  fit,,,,,(5) = 3(-3) +14,,,,,,5=-9+14,,,,,,ok