Find the possible value of a when the distance between (3, a) and (4, 1) is √10 units
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Answered by
1
let first point is A(3,a)and second one is B(4,1)
then
AB^2=(x1-x2)^2+(y1-y2)^2
10=(3-4)^2+(a-1)^2
10=1+a^2-2a+1
a^2-2a-8=0
solving this
we get two value of a
first is a=4or
second is a=-2
then
AB^2=(x1-x2)^2+(y1-y2)^2
10=(3-4)^2+(a-1)^2
10=1+a^2-2a+1
a^2-2a-8=0
solving this
we get two value of a
first is a=4or
second is a=-2
Answered by
0
√{3-4}^2+(a-1)^2=√10
(-1)^2+a^2+1+2a=10
1+a^2+1+2a=10
a^2+2a-8=0
Splitting the middle term
We get,
a=2,-4
Neglecting the negative term we get ,
a=2
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