A) \[p=q~\]
B) \[3p=4q\]
C) \[4p=3q\]
D) \[p+q=5\]
Correct Answer: C
Solution :
[c] \[P=\sqrt{{{(4-4)}^{2}}+{{(3-0)}^{2}}}=3\] \[q=\sqrt{{{(4-0)}^{2}}+{{(3-3)}^{2}}}=4\] Now, \[4p=4\times 3=12\] \[3q=3\times 4=12\] \[\therefore \,\,\,\,\,\,\,\,4p=3q\]You need to login to perform this action.
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