1.

Find a point on the line through A(5, -4)and B(-3, 2),that is, twice as far from A as from B.

Answer»

SOLUTION :Let `P(x, y)` be a point on AB such that
`""PA=2PB`

`rArr""(PA)/(PB)=(2)/(1)""rArr" "PA:PB=2:1`
So, by using SECTION formula for internal division,
`""x=(mx_(2)+nx_(1))/(m+n), y= (my_(2)+ny_(1))/(m+n)`
`rArr""x=(2(-3)+1(5))/(2+1)rArr" "x=-(1)/(3)`
and `""y=(2(2)+1(-4))/(2+1)rArr""y=0`
`rArr" "` Required point `-=(-(1)/(3), 0)`
`"""BUT"""`
This is not the END of this question.
Think : Is it not POSSIBLE that `P(x,y)` divides AB externally in the ratio 2 : 1.

So, by using section formula for external division,
`therefore""x=(2(-3)-1(5))/(2-1)rArr" "x=-(11)/(1)rArr" "x=-11`
and `""y=(2(2)-1(-4))/(2-1)rArr" "y=(8)/(1)rArr" "y=8`
So, co-ordinates of `p` are (-11, 8) also.
Hence, required POINTS are `(-(1)/(3),0) and (-11, 8)`.


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