Question

A(–8, 0), B(0, 16) and C(0, 0) are the vertices 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.

Answer 1

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

The co-ordinates of point P are

`((3 xx 0 + 5 xx (-8))/(3 + 5),(3 xx 16 + 5 xx 0)/(3 + 5))`

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

= (–5, 6)

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

The co-ordinates of point Q are

`((3 xx 0 + 5 xx (-8))/(3 + 5),(3 xx 0 + 5 xx 0)/(3 + 5))`

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

= (–5, 0)

Using distance formula,

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

= `sqrt(0 + 36)`

= 6

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

= `sqrt(0 + 16^2)`

= 16

Now, PQ = `3/8` BC

= `3/8 xx 16`

= 6

Hence, proved