Question

Find all possible values of y for which the distance between the points A(2, –3) and B(10, y) is 10 units.

Answer 1

The given points are A(2,-3) and B(10, y)

`∴AB = sqrt((2-10)^2 +(-3-y)^2)`

`=sqrt((-8)^2 + (-3-y)^2)`

`=sqrt(64+9+y^2+6y)`

∵ AB= 10

`∴sqrt(64+9+y^2+6y =10)`

`⇒ 73+y^2+6y =100`                (Squaring both sides)

`⇒y^2+6y-27 = 0`

`⇒y^2+9y -3y-27=0`

`⇒ y(y+9)-3(y+9)=0`

`⇒(y+9)(y-3)=0`

⇒y+9=0 or y-3=0

⇒ y=-9 or y=3

Hence, the possible values of y are -9 and 3.