A) \[\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{6}}\]
B) \[\frac{7}{\sqrt{6}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{10}}\]
C) \[\frac{7}{\sqrt{5}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{15}}\]
D) \[\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}}\]
Correct Answer: D
Solution :
\[{{L}_{12}}\equiv x-3y+1=0\] \[{{L}_{23}}\equiv 2x+y-12=0\] \[{{L}_{13}}\equiv 3x-2y-4=0\] Therefore, the required distances are \[{{D}_{3}}=\left| \frac{4-3\times 4+1}{\sqrt{10}} \right|=\frac{7}{\sqrt{10}}\] \[{{D}_{1}}=\left| \frac{4+1-12}{\sqrt{5}} \right|=\frac{7}{\sqrt{5}}\] \[{{D}_{2}}=\left| \frac{3\times 5-2\times 2-4}{\sqrt{9+4}} \right|=\frac{7}{\sqrt{13}}\].
You need to login to perform this action.
You will be redirected in
3 sec
