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Re: In the figure above, the point on segment PQ that is twice as far from [#permalink]
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Roosh18 wrote:
Any proper solution to this question?


Hi ,

The points can be found out by the section formula as below
P = (mx2+nx1/m+n , my2+ny1/m+n)
P= [(2*3 + 1*0 / 2+1), 2*2+1*-1 /2+1]
P= 2.1,1
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Re: In the figure above, the point on segment PQ that is twice as far from [#permalink]
Bunuel wrote:

In the figure above, the point on segment PQ that is twice as far from P as from Q is

(A) (3,1)
(B) (2,1)
(C) (2,–1)
(D) (1.5, 0.5)
(E) (1,0)

Attachment:
2017-12-12_2124.png


The distance between P & Q is sqrt((3-0)^2+(2+1)^2), which is 3sqrt(2).

The question asked for the point on PQ segment which is twice as far from P as from Q so if the total length of the segment is 3sqrt(2), the distance of the point from P should be 2sqrt(2) and from Q should be sqrt(2).

So I am getting no answer. Can anybody please explain where I am going wrong?
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Re: In the figure above, the point on segment PQ that is twice as far from [#permalink]
raoshahb wrote:
Bunuel wrote:

In the figure above, the point on segment PQ that is twice as far from P as from Q is

(A) (3,1)
(B) (2,1)
(C) (2,–1)
(D) (1.5, 0.5)
(E) (1,0)

Attachment:
2017-12-12_2124.png


The distance between P & Q is sqrt((3-0)^2+(2+1)^2), which is 3sqrt(2).

The question asked for the point on PQ segment which is twice as far from P as from Q so if the total length of the segment is 3sqrt(2), the distance of the point from P should be 2sqrt(2) and from Q should be sqrt(2).

So I am getting no answer. Can anybody please explain where I am going wrong?


I think using the midpoint formula might help. However in this case, you weight point P by 2 and point Q by 1. And so the divisor will be 3.

Posted from my mobile device
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Re: In the figure above, the point on segment PQ that is twice as far from [#permalink]
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