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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.
Solution
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
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