| The distance between the lines \[4x+3y=11\] and \[8x+6y=15\] is |
A) 4
B) \[\frac{7}{10}\]
C) \[\frac{5}{7}\]
D) 26
Correct Answer: B
Solution :
| \[4x+3y=11\] ...(i) |
| \[8x+6y=15\,\,\Rightarrow \,\,4x+3y=\frac{15}{2}\] ... (ii) |
| Since, Eqs. (i) and (ii) are equations of parallel lines in the form of \[ax+by+{{c}_{1}}=0\]and \[ax+by+{{\text{c}}_{2}}=0\] |
| Hence, distance between them d |
| \[=\,\,\frac{\left| {{c}_{2}}-{{c}_{1}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\,\,=\,\,\frac{\left| \frac{15}{2}-11 \right|}{\sqrt{{{(4)}^{2}}+{{(3)}^{2}}}}\,\,=\,\,\frac{3.5}{5}=\frac{7}{10}\] |
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