Question

Find the possible value of a when the distance between (3, a) and (4, 1) is √10 units

Answer 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

Answer 2

√{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