Sum

A(-8, 0), B(0, 16) and C(0, 0) are the verticals of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5

Show that : PQ =3/8 BC

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#### Solution

Given that, point P lies on AB such that AP: PB = 3: 5.

The co-ordinates of point P are

`((3xx0+5xx(-8))/(3+5),(3xx16+5xx0)/(3+5))`

`=(-40/8,48/8)`

`=(-5,6)`

Also, given that, point Q lies on AB such that AQ: QC = 3: 5.

The co-ordinates of point Q are

`((3xx0+5xx(-8))/(3+5),(3xx0+5xx0)/(3+5))`

`=(-40/8,0/8)`

`=(-5,0)`

Using distance formula,

`PQ=sqrt((-5+5)^2+(0-6)^2)=sqrt(0+36)=6`

Hence, proved

`BC=sqrt((0-0)^2+(0-16)^2)=sqrt(0+16^2)=16`

Now `3/8BC=3/8xx16=6=PQ

Concept: Distance Formula

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