Question 201322
For  this problem,  lets  first  find  the  slope  of  the  original  equation,  using  y=mx +b form,  and  then  define  the  slope  of  the  second  eqn  as, m2=-1/m1,perpendicular.  Finally,  we  can  define  the  eqn  of  the  second  line  from  the  slope  and  the  given  point.
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-8x +5y =77
5y = 8x +77
y = (8/5)x + 77/5
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or  from  form  ,,,,y= m x +b,,,m = (8/5)
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m2=  - 1/m1 = (-5/8)
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Second  eqn  is  ,  y= mx +b,,,,y = (-5/8) x +b
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As  it  contains  the  point  ( -4,3),,,,substitute
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(3) = (-5/8) (-4) + b
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3 = 20/8 +b
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24/8 - 20/8 = 4/8 =1/2 = b
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subst  m,,and  b,,in  form  eqn,, y = (-5/8)x + (1/2)
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mult  thru  by  8
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8y = -5x + 4
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checking,  subst ( -4,3),,,,8*3=(-5)(-4) +4 = 24 ,,,,ok