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प्रश्न
Find all possible values of x for which the distance between the points
A(x,-1) and B(5,3) is 5 units.
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उत्तर
Given AB = 5 units
Therefore, (AB)2 = 25 units
`⇒(5-a)^2 +{3-(-1)}^2 = 25`
`⇒(5-a)^2+(3+1)^2 = 25`
`⇒(5-a)^2 + (4)^2 =25`
`⇒(5-a)^2 +16 =25`
`⇒(5-a)^2 =25-16`
`⇒(5-a)^2 = 9`
`⇒(5-a)=+-sqrt(9)`
`⇒ 5-a=+-3`
`⇒5-a =3 or 5-a=-3`
`⇒ a=2 or 8`
Therefore, a = 2 or 8.
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