Question

find the value of Y for which the distance between the points a ( 3 , - 1 ) and b ( 11 , y ) is 10 units.

Answer 1

$\sqrt{64 + (y + 1) {}^{2} }$
64+y2+1+2y=10
y2+2y=-55
y2+2y+55=0
now rationalize it

Answer 2

Answer:

It is given that the distance between the points AB = 10.  

Therefore, AB= √(11-3)2 + (y+1)2  = 10

                      = √82 + (y+1)2 =10

                     = √64+(y+1)2  =10

Squaring on both sides, we get,

 64 +(y+1)2 = 100

64+ y2 + 2y + 1 = 100

y2 + 2y +1 = 100-64=36

y2+2y - 35 = 0

Now, by splitting the middle term method, we get,

y2+ 7y - 5y - 35= 0

y(y+7) -5(y+7)

(y-5) ( y+7)

y = 5 or y = -7

Therefore, the value of 'y' can be either 5 or -7.