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Find the equation of the line midway between the parallel lines `9x+6y-7=0` and 3x+2y+6=0` |
Answer» Converting each of the given equation to the form `y=mx+C`, We get, `9x+6y-7=0 Rightarrow y=(-3)/(2)x+(7)/(6)....(i)` `3x+2y+6=0 Rightarrow y=(-3)/(2)x-3....(ii)` Clearly, the slope of each one of the given line is `(-3)/(2)` Let the given lines by `y=mx+C_(1) and y=mx+C_(2)`. Then, `m=(-3)/(2), C_(1)=(7)/(6) and C_(2)=-3` Let the required line. Then, L is parallel to each one of (i) and (ii) and equidistant from each one of them. `therefore "slope of L"=(-3)/(2)` Let the equation of L be y`=(-3)/(2)x+C......(iii)` Then, distance between (i) and (iii) must be equal to the distance between (ii) and (iii). `therefore (|C_(1)-C|)/(sqrt(1+m^(2)))=(|C_(2)-C|)/(sqrt(1+m^(2)))=|C_(1)-C|=|C_(2)-C|` `Rightarrow |(7)/(6)-C|=|-3-C| Rightarrow |(7)/(6)-C|=|3+c|` `Rightarrow (7)/(6)-C=3+C Rightarrow 2C=|(-11)/(6)Rightarrow C=(-11)/(2)` `therefore "equation of L is y"=(-3)/(2)x-(11)/(12),i.e. 18+12y+11=0` Hence, the line `18x+12y+11=0` is midway between the parallel line `9x+6y-7=0 and 3x+2y+6=0`. |
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