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Question Number 75300 by vishalbhardwaj last updated on 09/Dec/19
A line is such that its segment between  the lines 5x−y+4 = 0 and 3x+4y−4 = 0  is bisected at the point (1,5). Obtain its  equation ?
$$\mathrm{A}\:\mathrm{line}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:\mathrm{its}\:\mathrm{segment}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{lines}\:\mathrm{5x}−\mathrm{y}+\mathrm{4}\:=\:\mathrm{0}\:\mathrm{and}\:\mathrm{3x}+\mathrm{4y}−\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{bisected}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{5}\right).\:\mathrm{Obtain}\:\mathrm{its} \\ $$$$\mathrm{equation}\:? \\ $$
Answered by mr W last updated on 09/Dec/19
say the eqn. is  ((y−5)/(x−1))=m  ⇒y=mx+5−m  5x−mx−5+m+4=0  ⇒x=((1−m)/(5−m))  3x+4mx+20−4m−4=0  ⇒x=((4m−16)/(3+4m))  (1/2)(((1−m)/(5−m))+((4m−16)/(3+4m)))=1  ⇒m=((107)/3)  ((y−5)/(x−1))=((107)/3)  ⇒107x−3y−92=0
$${say}\:{the}\:{eqn}.\:{is} \\ $$$$\frac{{y}−\mathrm{5}}{{x}−\mathrm{1}}={m} \\ $$$$\Rightarrow{y}={mx}+\mathrm{5}−{m} \\ $$$$\mathrm{5}{x}−{mx}−\mathrm{5}+{m}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}−{m}}{\mathrm{5}−{m}} \\ $$$$\mathrm{3}{x}+\mathrm{4}{mx}+\mathrm{20}−\mathrm{4}{m}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{4}{m}−\mathrm{16}}{\mathrm{3}+\mathrm{4}{m}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−{m}}{\mathrm{5}−{m}}+\frac{\mathrm{4}{m}−\mathrm{16}}{\mathrm{3}+\mathrm{4}{m}}\right)=\mathrm{1} \\ $$$$\Rightarrow{m}=\frac{\mathrm{107}}{\mathrm{3}} \\ $$$$\frac{{y}−\mathrm{5}}{{x}−\mathrm{1}}=\frac{\mathrm{107}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{107}{x}−\mathrm{3}{y}−\mathrm{92}=\mathrm{0} \\ $$

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