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determinati pe axa absciselor punctul situat la distanta egala cu o unitate de lungime de dreapta de ecuatie 8x+15y+10=0.
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dreapta (d):8x+15y+10=0
punctul apartine axei OX =>A(m,0)
distanta de la A la dreapta d = |8m+10|/sqrt(64+225)=|8m+10|/17
|8m+10|/17=1 =>|8m+10|=17
8m+10=17 =>m=7/8 =>A(7/8,0)
8m+10=-17 =>m=-27/8 =>A(-27/8 ,0)
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[Citat]
dreapta (d):8x+15y+10=0
punctul apartine axei OX =>A(m,0)
distanta de la A la dreapta d = |8m+10|/sqrt(64+225)=|8m+10|/17
|8m+10|/17=1 =>|8m+10|=17
8m+10=17 =>m=7/8 =>A(7/8,0)
8m+10=-17 =>m=-27/8 =>A(-27/8 ,0) |
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