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Give answer! - Co-ordinate geometry - JEE Main-6

Let P=(-1,0),Q=(0,0)\; and\; R=(3,3\sqrt{3}) be three points. The equation of the bisector of the angle PQR  is

  • Option 1)

    \frac{\sqrt{3}}{2}x+y=0\;

  • Option 2)

    \; x+\sqrt{3y}=0\;

  • Option 3)

    \; \sqrt{3}x+y=0\;

  • Option 4)

    \; x+\frac{\sqrt{3}}{2}y=0

 
Answers (1)
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As learnt in

Slope of a line -

m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

- wherein

Slope of line joining A(x1,y1) and  B(x2,y2) .

 

 P(-1,0)\:\:\:\:Q(0,0)\:\:\:\:and\:\:\:\:R(3,3\sqrt 3)

\tan (\angle RQX)=\frac{3 \sqrt 3}{3}=\sqrt 3\:\:\:\:\Rightarrow \angle RQX=\frac{\pi}{3}

Slope of line TQ is \tan 120^{o}=-\sqrt3

Equation is  y=mx

y=-\sqrt 3x\:\:\:\:\Rightarrow \sqrt3x+y=0 


Option 1)

\frac{\sqrt{3}}{2}x+y=0\;

This option is incorrect.

Option 2)

\; x+\sqrt{3y}=0\;

This option is incorrect.

Option 3)

\; \sqrt{3}x+y=0\;

This option is correct.

Option 4)

\; x+\frac{\sqrt{3}}{2}y=0

This option is incorrect.

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