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प्रश्न
Find all possible values of y for which distance between the points is 10 units.
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उत्तर
The given points are A(2,-3) and B(10, y)
`∴AB = sqrt((2-10)^2 +(-3-y)^2)`
`=sqrt((-8)^2 + (-3-y)^2)`
`=sqrt(64+9+y^2+6y)`
∵ AB= 10
`∴sqrt(64+9+y^2+6y =10)`
`⇒ 73+y^2+6y =100` (Squaring both sides)
`⇒y^2+6y-27 = 0`
`⇒y^2+9y -3y-27=0`
`⇒ y(y+9)-3(y+9)=0`
`⇒(y+9)(y-3)=0`
⇒y+9=0 or y-3=0
⇒ y=-9 or y=3
Hence, the possible values of y are -9 and 3.
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संबंधित प्रश्न
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Case Study -2
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