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प्रश्न
A line segment of length 10 units has one end at A (-4 , 3). If the ordinate of te othyer end B is 9 , find the abscissa of this end.
उत्तर
A (-4 , 3) ,Let the other point B (x , 9)
Given , AB = 10 units
`sqrt ((-4-"x")^2 + (3 - 9)^2)` = 10
squaring both sides ,
⇒ 16 + x2 + 8x + 36 = 100
⇒ x2 + 8x - 48 = 0
⇒ x2 + 12x - 4x - 48 = 0
⇒ x (x + 12) - 4(x + 12) = 0
⇒ (x - 4) (x + 12) = 0
⇒ x = 4 or -12
The abscissa of other end is 4 or -12.
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