Question

If p(2,p) is the mid-point of the line-segment joining the points A(6.-5) and
B(-2, 11), find the value of p.​

Answer 1

Answer:

Step-by-step explanation:

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.