find-the-coordinate-of-the-line-prese-nted-with-line-2x-3y-7-0-which-is-equadistance-from-points-4-8-and-7-1-

Question Number 80869 by MASANJAJ last updated on 07/Feb/20

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

n t e d w i t h l i n e 2 x - 3 y + 7 = 0 . w h i c h i s

e q u a d i s t a n c e f r o m p o i n t s (- 4, 8) a n d

(7, 1)

Commented by ajfour last updated on 07/Feb/20

{(x + 4)}^{2} + {(\frac{2 x + 7}{3} - 8)}^{2}

= {(x - 7)}^{2} + {(\frac{2 x + 7}{3} - 1)}^{2}

\Rightarrow 11 (2 x - 3) = 7 (\frac{4 x + 14}{3} - 9)

66 x - 99 = 28 x - 91

\Rightarrow 38 x = 8 \Rightarrow x = \frac{4}{19}

y = \frac{\frac{8}{19} + 7}{3} = \frac{141}{57} = \frac{47}{19}

p o i n t i s (\frac{4}{19}, \frac{47}{19}) .

Commented by mr W last updated on 07/Feb/20

c o r r e c t s i r!

Answered by mr W last updated on 07/Feb/20

A (- 4, 8), B (7, 1)

C = m i d p o i n t o f A B

x_{c} = \frac{- 4 + 7}{2} = \frac{3}{2}

y_{c} = \frac{8 + 1}{2} = \frac{9}{2}

i n c l i n a t i o n o f A B = \frac{1 - 8}{7 - (- 4)} = - \frac{7}{11}

b i s e c t o r o f A B :

y = \frac{9}{2} + \frac{1}{- (- \frac{7}{11})} (x - \frac{3}{2})

\Rightarrow y = \frac{9}{2} + \frac{11}{7} (x - \frac{3}{2}) = \frac{11}{7} x + \frac{15}{7}

{\begin{cases} 2 x - 3 y + 7 = 0 \\ y = \frac{11}{7} x + \frac{15}{7} \end{cases}

\Rightarrow x = \frac{4}{19}

\Rightarrow y = \frac{47}{19}

\Rightarrow t h e p o i n t o n t h e l i n e i s (\frac{4}{19}, \frac{47}{19}) .

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