Question

If P(2,P) is the midpoint of the line segment joining the points A(6,-5) and B(-2,11) find the valuve of P.

Answer 1

GIVEN points are P(2,p), A(6,-5) , B(-2,11)
P is the mid point of A & B
for distance AP : x1= 6 ,y1= -5 , x2 = 2, y2= p
AP =√(x1 - x2)² + (y1 - y2)²
[ By Distance formula]
AP =√[(6 -2)]² + (-5 - p)²
AP = √(4)² + [(-5)² +p² +2×5p]
AP = √ 16+ [25 +p² +10p]
AP = √ 16 + 25 +p² +10p
AP = √41 +p² +10p
For distance BP: x1 = -2 ,y1= 11, x2= 2, y2 = p
BP=√(-2 -2)² + (11 - p)²
= √(-4)² + [(11)² +p² -2×11p]
= √ 16 + [121 +p² - 22p]
= √ 16 +121 +p² - 22p
BP = √137 +p² -22p
ATQ,
AP = BP
√41 +p² +10p = √137 +p² -22p
[On squaring both sides]
41 -137 +p² -p² +10p +22p = 0
-96 +32p = 0
32p = 96
p = 96/32
p = 3
Hence, the value of p is 3.

HOPE THIS WILL HELP YOU....

Answer 2

We know that

P( x, y) = P(2,p)

X=2. , y=p

By section formula

y = my2+ny1/m+n

p=my2+ny1/m+n. (y=p)

p=11+(-5)/1+1

p=3

Hope it help!!!!!