A 900 N weight needs to be hung
(at a distance of NxL from the man's end)
on a uniform, 100 N horizontal pole (of length L)
so that a boy at one end supports one-third as
much as the man at the other end. Find N.
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A(5,1)≡(x
1
,y
1
)
B(−1,5)≡(x
2
,y
2
)
P≡(x,y)
PA=PB ....(given)
∴ By distance formula
(x−x
1
)
2
+(y−y
1
)
2
we have,
⇒(x−5)
2
+(y−1)
2
=(x+1)
2
+(y−5)2
⇒x
2
−10x+25+y
2
−2y+1=x
2
+2x+1+y
2
−10y+25
∴−2y+10y=2x+10x
∴8y=12y
2y=3x. [henceproved]
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