Let A be the points (a,2a-1)and B be point (2a+4,3a+9) where "a" is a constant , given that the distance AB is √(260) .find the possible values of "a".note a is an integer .
Answers
Therefore the possible values of a are 4 and - 18.
Given : Let A be the point (a, 2a - 1) and be the point (2a + 4, 3a + 9) where a is a constant. given that the distance AB is √(260).
To find : The possible values of a. [ a is an integer ]
solution : from distance formula,
distance between two points (x₁, y₁) and (x₂, y₂) = √{(x₂ - x₁)² + (y₂ - y₁)²}
so, distance between A and B = √{(2a + 4 - a)² + (3a + 9 - 2a + 1)²}
⇒√(260) = √{(a + 4)² + (a + 10)²}
⇒260 = (a + 4)² + (a + 10)²
⇒260 = a² + 8a + 16 + a² + 20a + 100
⇒260 = 2a² + 28a + 116
⇒2a² + 28a - 144 = 0
⇒a² + 14a - 72 = 0
⇒a² + 18a - 4a - 72 = 0
⇒a(a + 18) - 4(a + 18) = 0
⇒(a - 4)(a + 18) = 0
⇒a = 4 , - 18
Therefore the possible values of a are 4 and - 18.
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