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Question

P(x, y), A(3, 4) and B (5, -2) are the vertices of triangle PAB such that |PA|=|PB| and area of ΔPAB=10 sq. units, then PA=k5 units. Find the value of k.

A
2
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B
1
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C
5
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D
3
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Solution

The correct option is A 2

Given, |PA|=|PB|
PA2=PB2(x3)2+(y4)2=(x5)2+(y+2)2x26x+9+y28y+16=x210x+25+y2+4y+44x12y=4x3y=1(i)
area of ΔPAB=12|x(4+2)+3(2y)+5(y4)|10=12|6x63y+5y20|20=6x+2y266x+2y26=20 or 6x+2y26=206x+2y=46 or 6x+2y=63x+y=23(2)

or 3x+y=3(3)
Solving (1) and (3), we get
x=1,y=0
Solving (1) and (2), we get
x=7,y=2
Coordinates of P are (7, 2) and (1, 0)
Length of PA=(73)2+(24)2=16+4=2525=k5k=2
Also, Length of PA=(13)2+(04)2=4+16=2525=k5k=2

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