Question-168609

Question Number 168609 by infinityaction last updated on 14/Apr/22

Commented by infinityaction last updated on 14/Apr/22

c o o r d i n a t e o f p o i n t (a, b)

Answered by mr W last updated on 14/Apr/22

r = \sqrt{{(12 - 6)}^{2} + {(- 6 - 4)}^{2}} = 2 \sqrt{34}

\sin α = \frac{4 - (- 6)}{2 \sqrt{34}} = \frac{5}{\sqrt{34}}

\cos α = \frac{6 - 12}{2 \sqrt{34}} = - \frac{3}{\sqrt{34}}

a = 12 + 2 \sqrt{34} \cos (α + \frac{π}{3})

= 12 + 2 \sqrt{34} (- \frac{3}{\sqrt{34}} \times \frac{1}{2} - \frac{5}{\sqrt{34}} \times \frac{\sqrt{3}}{2})

= 9 - 5 \sqrt{3}

b = - 6 + 2 \sqrt{34} \sin (α + \frac{π}{3})

= - 6 + 2 \sqrt{34} (\frac{5}{\sqrt{34}} \times \frac{1}{2} - \frac{3}{\sqrt{34}} \times \frac{\sqrt{3}}{2})

= - 1 - 3 \sqrt{3}

Commented by infinityaction last updated on 14/Apr/22

w h e r e i s a α i n t h i s d i a g r a m

Commented by mr W last updated on 14/Apr/22

Commented by infinityaction last updated on 14/Apr/22

t h a n k y o u s i r

Answered by cortano1 last updated on 14/Apr/22

l e t ℓ i s l i n e p a s s e s t h r o u g t p o i n t

B (\frac{6 + 12}{2}, \frac{4 - 6}{2}) = B (9, - 1)

w i t h g r a d i e n t = - \frac{1}{(\frac{4 + 6}{6 - 12})} = \frac{3}{5}

t h e n ℓ \equiv 3 x - 5 y = 32

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

{(x - 12)}^{2} + {(y + 6)}^{2} = 136

t h e n t h e l i n e ℓ c u t t h e c i r c l e

\Rightarrow {(\frac{32 + 5 y}{3} - 12)}^{2} + {(y + 6)}^{2} = 136

\Rightarrow {(\frac{5 y - 4}{3})}^{2} + {(y + 6)}^{2} = 136

\Rightarrow {\begin{cases} y_{1} = - 3 \sqrt{3} - 1 = b \\ x_{1} = - 5 \sqrt{3} + 9 = a \end{cases}

∴ P (- 5 \sqrt{3} + 9, - 3 \sqrt{3} - 1)

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