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  • (i)Do the following equations represent a pair of coincident linesJustify your answer. 3x pus 1by7=3 7x + 3y = 7

Do the following equations represent a pair of coincident lines? Justify your answer.

(i)3x+\frac{1}{7}y= 3; 7x + 3y = 7
(ii) –2x – 3y = 1 ; 6y + 4x = –2
(iii)  \frac{x}{2}+y+\frac{2}{5}= 0;4x+8y+\frac{5}{6}= 0

 

 

Answers (1)

(i)Solution:
Equation are 3x + \frac{1}{7}y = 3
7x + 3y = 7
In equation
3x + \frac{1}{7}y – 3 = 0
a1 = 3 ; b1 = \frac{1}{7} ; c1 = –3
In equation
7x + 3y – 7 = 0
a2 = 7 ; b2 = 3; c2 = –7

\frac{a_{1}}{a_{2}}= \frac{3}{7};\frac{b_{1}}{b_{2}}= \frac{1}{21};\frac{c_{1}}{c_{2}}= \frac{-3}{-7}= \frac{3}{7}

For coincident lines \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}} but here \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}

Hence the given pair of equations does not represent a pair of coincident lines. 

(ii)Solution:

Equation are   –2x – 3y = 1
4x + 6y = –2
In equation
–2x – 3y  – 1 = 0
a1 = –2 ; b1 = –3 ; c1 = –1
In equation
4x + 6y + 2 = 0
a2 = 4 ; b2 = 6; c2 = 2
\frac{a_{1}}{a_{2}}= \frac{-2}{4}= \frac{-1}{2};\frac{b_{1}}{b_{2}}= \frac{-3}{6}= \frac{-1}{2},\frac{c_{1}}{c_{2}}= \frac{-1}{2}

For coincident lines \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}} also here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
Hence the given pair of equation represents a pair of coincident lines.

Solution: (iii)
Equation are     \frac{x}{2}+y+\frac{2}{5}= 0           

4x + 8y + \frac{5}{16}= 0
In equation

\frac{x}{2}+y+\frac{2}{5}= 0
a1 = 1/2 ; b1 = 1 ; c1 = \frac{2}{5}
In equation
4x + 8y +  \frac{5}{16}=  0
a2 = 4 ; b2 = 8; c2 =  \frac{5}{16}

\frac{a_{1}}{a_{2}}= \frac{1}{8};\frac{b_{1}}{b_{2}}= \frac{1}{8};\frac{c_{1}}{c_{2}}\Rightarrow \frac{2}{5}\times \frac{16}{5}\Rightarrow \frac{32}{25}

For coincident lines \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}but here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

Hence the given pair of equations does not represent a pair of coincident lines.

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infoexpert27

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