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  • How to solve this problem- - Co-ordinate geometry - JEE Main-2

If the equation of the locus of point equidistant from the points (a_{1},b_{1})\; and\; (a_{2},b_{2}) is (a_{1}-a_{2})x+(b_{1}-b_{2})y+c=0,\; then\; c=

  • Option 1)

    (a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2})

  • Option 2)

    \frac{1}{2}(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2})

  • Option 3)

    \sqrt{(a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2})}

  • Option 4)

    \frac{1}{2}(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})

 

Answers (3)

As we learnt in 

Distance formula -

The distance between the point A\left ( x_{1},y_{1} \right )\: and \: B\left ( x_{2},y_{2} \right )

is \sqrt{\left ( x_{1} -x_{2}\right )^{2}+\left ( y_{1} -y_{2}\right )^{2}}

- wherein

 

 

According to distance formula, PA=PB

i.e. PA2=PB2

(h-a_{1})^{2}+(k-b_{1})^{2}=(h-a_{2})^{2}+(k-b_{2})^{2}

2(a_{1}-a_{2})h+2(b_{1}-b_{2})k+(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})

On comparing, C=\frac{1}{2}(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})


Option 1)

(a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2})

This option is incorrect 

Option 2)

\frac{1}{2}(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2})

This option is incorrect 

Option 3)

\sqrt{(a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2})}

This option is incorrect 

Option 4)

\frac{1}{2}(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})

This option is correct 

Posted by

Vakul

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JEE Main high-scoring chapters and topics

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option 4 is correct

Posted by

priyanka Bundela

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option 4

Posted by

priyanka Bundela

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