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

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

$${find}\:{the}\:{coordinate}\:{of}\:{the}\:{line}\:{prese} \\ $$$${nted}\:{with}\:{line}\:\mathrm{2}{x}−\mathrm{3}{y}+\mathrm{7}=\mathrm{0}.{which}\:{is} \\ $$$${equadistance}\:{from}\:{points}\:\left(−\mathrm{4},\mathrm{8}\right)\:{and}\: \\ $$$$\left(\mathrm{7},\mathrm{1}\right) \\ $$

Commented by ajfour last updated on 07/Feb/20

(x+4)^2 +(((2x+7)/3)−8)^2 = (x−7)^2 +(((2x+7)/3)−1)^2 ⇒ 11(2x−3)=7(((4x+14)/3)−9) 66x−99=28x−91 ⇒ 38x=8 ⇒ x=(4/(19)) y=(((8/(19))+7)/3) = ((141)/(57))=((47)/(19)) point is ((4/(19)), ((47)/(19))) .

$$\left({x}+\mathrm{4}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{x}+\mathrm{7}}{\mathrm{3}}−\mathrm{8}\right)^{\mathrm{2}} \\ $$$$\:\:=\:\left({x}−\mathrm{7}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{x}+\mathrm{7}}{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{11}\left(\mathrm{2}{x}−\mathrm{3}\right)=\mathrm{7}\left(\frac{\mathrm{4}{x}+\mathrm{14}}{\mathrm{3}}−\mathrm{9}\right) \\ $$$$\mathrm{66}{x}−\mathrm{99}=\mathrm{28}{x}−\mathrm{91} \\ $$$$\Rightarrow\:\:\:\mathrm{38}{x}=\mathrm{8}\:\:\:\Rightarrow\:\:{x}=\frac{\mathrm{4}}{\mathrm{19}} \\ $$$${y}=\frac{\frac{\mathrm{8}}{\mathrm{19}}+\mathrm{7}}{\mathrm{3}}\:=\:\frac{\mathrm{141}}{\mathrm{57}}=\frac{\mathrm{47}}{\mathrm{19}} \\ $$$${point}\:{is}\:\:\left(\frac{\mathrm{4}}{\mathrm{19}},\:\frac{\mathrm{47}}{\mathrm{19}}\right)\:\:. \\ $$

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

correct sir!

$${correct}\:{sir}! \\ $$

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))).

$${A}\left(−\mathrm{4},\mathrm{8}\right),\:{B}\left(\mathrm{7},\mathrm{1}\right) \\ $$$${C}={midpoint}\:{of}\:{AB} \\ $$$${x}_{{c}} =\frac{−\mathrm{4}+\mathrm{7}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${y}_{{c}} =\frac{\mathrm{8}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${inclination}\:{of}\:{AB}=\frac{\mathrm{1}−\mathrm{8}}{\mathrm{7}−\left(−\mathrm{4}\right)}=−\frac{\mathrm{7}}{\mathrm{11}} \\ $$$${bisector}\:{of}\:{AB}: \\ $$$${y}=\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{1}}{−\left(−\frac{\mathrm{7}}{\mathrm{11}}\right)}\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{y}=\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{11}}{\mathrm{7}}\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{11}}{\mathrm{7}}{x}+\frac{\mathrm{15}}{\mathrm{7}} \\ $$$$ \\ $$$$\begin{cases}{\mathrm{2}{x}−\mathrm{3}{y}+\mathrm{7}=\mathrm{0}}\\{{y}=\frac{\mathrm{11}}{\mathrm{7}}{x}+\frac{\mathrm{15}}{\mathrm{7}}}\end{cases} \\ $$$$\Rightarrow{x}=\frac{\mathrm{4}}{\mathrm{19}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{47}}{\mathrm{19}} \\ $$$$ \\ $$$$\Rightarrow{the}\:{point}\:{on}\:{the}\:{line}\:{is}\:\left(\frac{\mathrm{4}}{\mathrm{19}},\frac{\mathrm{47}}{\mathrm{19}}\right). \\ $$

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