find the coordinate of the line prese nted with line 2x−3y+7=0.which is equadistance from points (−4,8) and (7,1)


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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
	$A(−4,8), B(7,1) C=midpoint of AB x_c =((−4+7)/2)=(3/2) y_c =((8+1)/2)=(9/2) inclination of AB=((1−8)/(7−(−4)))=−(7/(11)) bisector of AB: y=(9/2)+(1/(−(−(7/(11)))))(x−(3/2)) ⇒y=(9/2)+((11)/7)(x−(3/2))=((11)/7)x+((15)/7) { ((2x−3y+7=0)),((y=((11)/7)x+((15)/7))) :} ⇒x=(4/(19)) ⇒y=((47)/(19)) ⇒the point on the line is ((4/(19)),((47)/(19))).$
	$A (- 4, 8), B (7, 1)$ $C = m i d p o i n t o f A B$ $x_{c} = \frac{- 4 + 7}{2} = \frac{3}{2}$ $y_{c} = \frac{8 + 1}{2} = \frac{9}{2}$ $i n c l i n a t i o n o f A B = \frac{1 - 8}{7 - (- 4)} = - \frac{7}{11}$ $b i s e c t o r o f A B :$ $y = \frac{9}{2} + \frac{1}{- (- \frac{7}{11})} (x - \frac{3}{2})$ $\Rightarrow y = \frac{9}{2} + \frac{11}{7} (x - \frac{3}{2}) = \frac{11}{7} x + \frac{15}{7}$ ${\begin{cases} 2 x - 3 y + 7 = 0 \\ y = \frac{11}{7} x + \frac{15}{7} \end{cases}$ $\Rightarrow x = \frac{4}{19}$ $\Rightarrow y = \frac{47}{19}$ $\Rightarrow t h e p o i n t o n t h e l i n e i s (\frac{4}{19}, \frac{47}{19}) .$
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