A) 1 unit
B) \[1\frac{6}{25}\]units
C) 25 units
D) None of these
Correct Answer: A
Solution :
Given, lines are \[7x+24y+3=0\] and \[7x+24y+28=0\] Distance between lines, \[d=\left| \frac{{{c}_{1}}\tilde{\ }{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}} \right|\] \[=\left| \frac{28-3}{\sqrt{{{(7)}^{2}}+{{(24)}^{2}}}} \right|\] \[=\frac{25}{\sqrt{49+576}}=\frac{25}{\sqrt{625}}\] \[=\frac{25}{25}=1\,unit\]
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