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Help me please, - Co-ordinate geometry - JEE Main-5

Slope of a line passing through P(2,3) and intersecting the line , x+y=7 at a distance of 4 units from P, is :

  • Option 1)

      \frac{1-\sqrt{5}}{1+\sqrt{5}}               

  • Option 2)

    \frac{1-\sqrt{7}}{1+\sqrt{7}}

  • Option 3)

     \frac{7-\sqrt{1}}{7+\sqrt{1}}

  • Option 4)

     \frac{5-\sqrt{1}}{5+\sqrt{1}}

 
Answers (1)
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x+y=7

P\left ( 2,3 \right )

Let point Q  be (a,7-a)

distance b/w    PQ=\sqrt{\left ( a-2 \right )^{2}+\left ( 7-a-3 \right )^{2}}=4   

                        \left ( a-2 \right )^{2}+\left ( 4-1 \right )^{2}=16

                       a=3\pm \sqrt{7}

     When          a=3- \sqrt{7}

   slope of    PQ=M=\frac{1+\sqrt{7}}{1-\sqrt{7}}

            When  a=3+\sqrt{7}

   Slope of   PQ=\frac{1-\sqrt{7}}{1+\sqrt{7}}

  


Option 1)

  \frac{1-\sqrt{5}}{1+\sqrt{5}}               

Option 2)

\frac{1-\sqrt{7}}{1+\sqrt{7}}

Option 3)

 \frac{7-\sqrt{1}}{7+\sqrt{1}}

Option 4)

 \frac{5-\sqrt{1}}{5+\sqrt{1}}

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