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  • Consider the point <img alt="\mathrm{\mathrm{A} \equiv(0,1)\: and \: \mathrm{B}(2,0). ' \mathrm{P} ' }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BA%7D%20%5Cequiv%

Consider the point \mathrm{\mathrm{A} \equiv(0,1)\: and \: \mathrm{B}(2,0). ' \mathrm{P} ' }be a point on the line \mathrm{4 x+3 y+9=0. } Coordinates of the point \mathrm{' \mathrm{P} '} such that \mathrm{|\mathrm{PA}-\mathrm{PB}|} is maximum, is -
 

Option: 1

\left(-\frac{12}{5}, \frac{17}{5}\right)


 


Option: 2

\left(-\frac{84}{5}, \frac{13}{5}\right)
 


Option: 3

\mathrm{\left(-\frac{6}{5}, \frac{17}{5}\right)}
 


Option: 4

\text{None of these}


Answers (1)

best_answer

We have |\mathrm{PA}-\mathrm{PB}| \leq \mathrm{AB}. Thus for \mathrm{|\mathrm{PA}-\mathrm{PB}|} to be maximum, point \mathrm{A, B \: and \: ' P ' }must be collinear equation of \mathrm{A B} is \mathrm{x+2 y=2}.

Solving it with given line, we get \mathrm{P \equiv\left(-\frac{84}{5}, \frac{13}{5}\right).}

Hence option 2 is correct.

 

Posted by

Rishabh

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