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  • Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Q : 21     Find equation of the line which is equidistant from parallel lines  9x+6y-7=0   and  3x+2y+6=0.

Answers (1)

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Let's take the point p(a,b) which is equidistance from  parallel lines  9x+6y-7=0   and  3x+2y+6=0
Therefore,
d_1= \left | \frac{9.a+6.b-7}{\sqrt{9^2+6^2}} \right |                                     d_2= \left | \frac{3.a+2.b+6}{\sqrt{3^2+2^2}} \right |
d_1= \left | \frac{9a+6b-7}{\sqrt{117}} \right |                                       d_2= \left | \frac{3a+2b+6}{\sqrt{13}} \right |
It is that   d_1=d_2
Therefore,
\left | \frac{9a+6b-7}{3\sqrt{13}} \right |= \left | \frac{3a+2b+6}{\sqrt{13}} \right |
(9a+6b-7)=\pm 3(3a+2b+6)
Now, case (i)
(9a+6b-7)= 3(3a+2b+6)
25=0
Therefore, this case is not possible 

Case (ii)
(9a+6b-7)= -3(3a+2b+6)
18a+12b+11=0

Therefore, the required equation of the line is  18a+12b+11=0

 

Posted by

Gautam harsolia

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