Question

find the value of 'a' when the distance between the points (3,a) and (4,1) is
$\sqrt{10}$





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Answer 1

Answer:

√(4-3)^2+(1-a)^2=√10

√1+1+a^2-2a=√10

2+a^2-2a=10

a^2+2-10-2a=0

a^2-2a-8=0

a^2-4a+2a-8=0

a(a-4) 2(a-4)

(a+2)(a-4)=0

a=-2;4

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