Two lines are given to be parallel. The equation of one of the lines is 4x + 3y = 14. Find the equation of the second line.

Answer : 

We know if we have two parallel lines a1x + b1y + c1  =  0  And a2x + b2y + c2  = 0  , Then the value of all cofficient are As : 

a1a2 = b1b2    c1c2

Here we have a line 4x + 3y = 14   , So  a1  =  4  ,  b1  =  3 and  c1  = - 14  
let our line that is parallel to given line is a2x + b2y + c2 = 0 , So 
We get 

4a2 = 3b2    - 14c2

So,

a2 : b2  =  4k : 3k                                 ( Where k is a ratio constant that could be any integer ) 

So,
We can form any equation that has the co efficients in the ratios 4k : 3k : n  (where n is any integer and n 14k) 

So,
Our parallel equation could be  

8x  + 6y  =  12                                                  ( Ans )


 

  • 24

since lines are parallel,

therefore a1/a2=b1/b2

so the othe equation can be 8x + 6y = 10

  • -7

If the two lines are parallel, then a1/a2==b1/b2=/=c1/c2

Then, 4/a2==3/b2=/=14/c2

Therefore to satisfy this situation a2:b2=4k:3k(wherek is any integer)

Therefore we can form any equation that has the co efficients in the ratios 4k:3k:n(where n is any integer and n=/=14k)

  • 7
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