Question

find the values of y for which the distance between the points P(2,-3) and Q(10,Y) is 10 units.

Answer 1

Hey there,

It is given that the distance between (2, - 3) and (10, y) is 10.

[tex]Therefore, \sqrt{(2 - 10)^2 + (-3 - y)^2 = 10 \sqrt{(-8)^2 + (3 + y)^2 = 10 [/tex]

 64 + (y + 3)2 = 100

(y +3)2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = -6
Therefore, y = 3 or -9 

Hope this helps!

Answer 2

distance of PQ=√(10-2)²+(y+3)²
                         √(8)²+(y+3)²=10
SOBS
64+(y+3)²=100
(y+3)²=100-64
(y+3)²=36
y+3=√36
y+3=6
y=3