Find-the-equation-of-the-locus-of-a-moving-point-P-such-that-its-distance-from-the-point-A-4-3-and-B-5-1-is-in-the-ratio-3-1-

Question Number 167185 by pete last updated on 09/Mar/22

Find the equation of the locus of a moving point P,

such that its distance from the point

A (4, 3) and B (5, 1) is in the ratio 3 : 1

Commented by mr W last updated on 09/Mar/22

t h e l o c u s o f a m o v i n g p o i n t, w h o s e

d i s t a n c e s t o t w o f i x e d p o i n t s a r e i n

a c o n s t a n t r a t i o, i s a c i r c l e .

Commented by otchereabdullai@gmail.com last updated on 21/Mar/22

Nice!

Answered by Rasheed.Sindhi last updated on 09/Mar/22

P = (x, y)

∣ AP ∣:∣ BP ∣= 3 : 1

∣ AP ∣= 3 ∣ BP ∣

\sqrt{{(4 - x)}^{2} + {(3 - y)}^{2}} = 3 \sqrt{{(5 - x)}^{2} + {(1 - y)}^{2}}

x^{2} - 8 x + 16 + y^{2} - 6 y + 9 = 9 (x^{2} - 10 x + 25 + y^{2} - 2 y + 1)

8 x^{2} + 8 y^{2} - 82 x - 12 y + 209 = 0

Answered by som(math1967) last updated on 09/Mar/22

l e t P (h, k)

\frac{P A}{P B} = \frac{3}{1}

\frac{\sqrt{{(h - 4)}^{2} + {(k - 3)}^{2}}}{\sqrt{{(h - 5)}^{2} + {(k - 1)}^{2}}} = \frac{3}{1}

\frac{h^{2} - 8 h + 16 + k^{2} - 6 k + 9}{h^{2} - 10 h + 25 + k^{2} - 2 k + 1} = \frac{9}{1}

9 h^{2} - 90 h + 225 + 9 k^{2} - 18 k + 9

= h^{2} - 8 h + k^{2} - 6 k + 25

8 h^{2} + 8 k^{2} - 82 h - 12 k + 209 = 0

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

8 x^{2} + 8 y^{2} - 82 x - 12 y + 209 = 0

Commented by pete last updated on 09/Mar/22

Thank you very much sir . But sir, why is the locus

of the point not a straight line but a circle ?

That is my problem .

Commented by som(math1967) last updated on 09/Mar/22

A a n d B i s f i x e d p t b u t P i s m o v a b l e p t

s u c h P A : P B = 3 : 1 s o p a t h o f P i s n o t

s t . l i n e

Commented by MJS_new last updated on 09/Mar/22

on a straight line (= the x - axis) to given

points A & B, find P with

∣ A P ∣:∣ B P ∣= 3 : 1

this is the same problem as this :

split A B with ratio 3 : 1

which has exactly 2 solutions (internal and

external split point) P_{i}, P_{e}

these are on the line itself but if you allow

“ split {points}^{″} in 2 dimensions, how could

they all form a straight line ?

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