# 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 - Mathematics

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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

#### 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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#### APPEARS IN

Selina Concise Maths Class 10 ICSE
Chapter 13 Section and Mid-Point Formula
Exercise 13 (C) | Q 3 | Page 183