A) \[\frac{35}{\sqrt{34}}\]
B) \[\frac{1}{3\sqrt{34}}\]
C) \[\frac{35}{3\sqrt{34}}\]
D) \[\frac{35}{2\sqrt{34}}\]
Correct Answer: C
Solution :
Given lines are \[5x+3y-7=0\]and\[15x+9y+14=0\] The distance from origin to the lines are \[{{d}_{1}}=\frac{0+7-7}{\sqrt{{{5}^{2}}+{{3}^{2}}}}=\frac{-7}{\sqrt{34}}\] and \[{{d}_{2}}=\frac{0+0+14}{\sqrt{225+81}}=\frac{14}{\sqrt{306}}=\frac{14}{3\sqrt{34}}\] Since, the distance is on opposite sign, it means that the given lines are on opposite side of the origin, therefore the distance between them is \[{{d}_{1}}+{{d}_{2}}=\frac{7}{\sqrt{34}}+\frac{14}{3\sqrt{34}}=\frac{35}{3\sqrt{34}}\] Alternative Solution: Given lines are \[5x+3y-7=0\] and \[15x+9y+14=0\] or \[5x+3y+\frac{14}{3}=0\] Since, the given lines are parallel, therefore the distance between them is \[d=\frac{|{{c}_{1}}-{{c}_{2}}|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\frac{|-7-\frac{14}{3}|}{\sqrt{{{5}^{2}}+{{3}^{2}}}}\] \[=\frac{35}{3\sqrt{34}}\] Note: Let\[ax+by+{{c}_{1}}=0\]and\[ax+by+{{c}_{2}}=0\] are two parallel lines, then the distance between them is \[d=\frac{{{c}_{1}}-{{c}_{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
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