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
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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....
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....
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