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$\textsf {If the ordered pair satisfying the equations}$
$\mathsf {a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0 \: has \: 1 \: as \: its \: }$
$\textsf {first coordinate, then which of the following is correct ?}$

$\mathsf {a) \: \: {\dfrac {a_1 + b_1}{a_2 + b_2} = {\dfrac {c_1}{c_2}}}}$


$\mathsf {b) \: \: {\dfrac{b_1 + c_1}{b_2 + c_2} = {\dfrac {a_1}{b_2}}}}$


$\mathsf {c) \: \: {\dfrac {c_1 + a_1}{c_2 + a_2} = {\dfrac {b_1}{b_2}}}}$


$\textsf {d) All the above}$


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Answer 1

$\textsf{Given , the two equations are : }$

$\mathsf{a_1x+b_1y+c_1=0}$

$\mathsf{a_2x+b_2+c_2=0}$

$\underline{\underline{\huge\textsf{{{Cross Multiplication Method}}}}}$

$\textsf{By cross multiplication method , we know that :- }$

$\mathsf{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}}$

$\implies \mathsf{\frac{x}{b_1c_2-b_2c_1}=\frac{1}{a_1b_2-a_2b_1}}$

$\textsf{The solution of the 2 equation lies on (x,y)}\\\\\textsf{The first coordinate is 1  }\\\\\\\textsf{Hence, value of x is 1.}$

$\textsf{Put this value of x in the cross multiplication formula to get :}$

$\implies \:\:\:\mathsf{\frac{1}{b_1c_2-b_2c_1}=\frac{1}{a_1b_2-a_2b_1}}$

$\implies \:\:\:\mathsf{a_1b_2-a_2b_1=b_1c_2-b_2c_1}$

$\implies \:\:\:\mathsf{a_1b_2+b_2c_1=b_1c_2+b_1a_2}$

$\implies \:\:\:\mathsf{b_2(a_1+c_1)=b_1(a_2+c_2)}$

$\implies \:\:\:\mathsf{\frac{b_1}{b_2}=\frac{a_1+c_1}{a_2+c_2}}$

$\textsf{\huge{\underline{\underline{\underline{\underline{ANSWER}}}}}}$

$\bf{\underline{\textsf{OPTION\:\:C}}}$

$\textsf{Hope it helps !}$

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Answer 2

$\textsf{Given , the two equations are : }\\\\\\</p><p></p><p></p><p></p><p>\mathsf{a_1x+b_1y+c_1=0} \\\\\\</p><p></p><p></p><p></p><p>\mathsf{a_2x+b_2+c_2=0}\\\\\\</p><p></p><p></p><p>\underline{\underline{\huge\textsf{{{Cross Multiplication Method}}}}} \\\\\\</p><p></p><p> </p><p> </p><p> </p><p> </p><p></p><p></p><p>\textsf{By cross multiplication method , we know that :- }\\\\\\</p><p></p><p></p><p>\mathsf{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}} \\\\\\</p><p></p><p> </p><p></p><p></p><p>\implies \mathsf{\frac{x}{b_1c_2-b_2c_1}=\frac{1}{a_1b_2-a_2b_1}}\\\\\\</p><p> </p><p> </p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p>\textsf{Put this value of x in the cross multiplication formula to get :}\\\\\\</p><p></p><p></p><p>\implies \:\:\:\mathsf{\frac{1}{b_1c_2-b_2c_1}=\frac{1}{a_1b_2-a_2b_1}}\\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p>\implies \:\:\:\mathsf{a_1b_2-a_2b_1=b_1c_2-b_2c_1}\\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p>\implies \:\:\:\mathsf{a_1b_2+b_2c_1=b_1c_2+b_1a_2}\\\\\\</p><p> </p><p> </p><p></p><p></p><p>\implies \:\:\:\mathsf{b_2(a_1+c_1)=b_1(a_2+c_2)}\\\\\\</p><p></p><p></p><p>\implies \:\:\:\mathsf{\frac{b_1}{b_2}=\frac{a_1+c_1}{a_2+c_2}}\\\\\\</p><p> </p><p> </p><p> </p><p> </p><p></p><p></p><p></p><p></p><p>\textsf{\huge{\underline{\underline{\underline{\underline{ANSWER}}}}}} \\\\\\</p><p></p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p></p><p>\bf{\underline{\textsf{OPTION\:\:C}}} \\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p></p><p>\textsf{Hope it helps !}$

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