Question

find the value of Y for which the distance between the point as (3, -1) and b (11, y) is 10 units
friends please ans me

Answer 1

When in doubt, sketch it out. 
Show the two points on graph paper..............


guess that point B is in the upper left quadrant of graph. 


Connect A and B....

draw a vertical line through A and a horizontal line through B. 


You have a right triangle.........

hypotenuse = 10.......

one side = (3 - 11)......

other side = (y + 1) 

Use Pythagoras for equation of y..........

factorise to show

100 = (y+1)^2 + ( -8)^2

100 = y^2 + 2y + 1 + 64

35 = y^2 + 2y

y^2 + 2y - 35 = 0

y = 7 ,5


Check both values to prove each is OK. 


Answer ...B is either (11, 7) or (11,5)

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.