Question

If p x,6 is the midpoint of the line segment joining A 6,5 and B 4, x find the value of x

Answer 1

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

Answer 2

Answer:

X=5.

Step by step explanation is in the attachment given ⬇⬇⬇

Hope it helps you✌✌