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Question Number 121418 by faysal last updated on 08/Nov/20

a c h a n g e a b l e s t r a i g h t l i n e g o i n g t h e p o i n t (a, b)

c r e a t e s a t r i a n g l e w i t h a x e s . w h a t i s

t h e e q u a t i o n o f l o c u s o f c e n t r o i d o f

t h e t r i a n g l e

Answered by TANMAY PANACEA last updated on 08/Nov/20

l e t t h e e q n b e

\frac{x}{p} + \frac{y}{q} = 1 \to s t l i n e m e e t x a x i s a t (p, 0) y a x i s

a t (0, q)

c e n t r o i d = (\frac{p + 0 + 0}{3}, \frac{0 + q + 0}{3}) = (\frac{p}{3}, \frac{q}{3})

s a y α = \frac{p}{3} β = \frac{q}{3}

n o w s t l i n e \frac{x}{p} + \frac{y}{q} = 1 p a s s e s t h r o u g h (a, b) s o

\frac{a}{p} + \frac{b}{q} = 1

\frac{a}{3 α} + \frac{b}{3 β} = 1

s o t h e l o c u s i s \frac{a}{3 x} + \frac{b}{3 y} = 1

\frac{a}{x} + \frac{b}{y} = 3

Commented by peter frank last updated on 08/Nov/20

thank you

Commented by TANMAY PANACEA last updated on 08/Nov/20

m o s t w e l c o m e