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The equation of the lines which passes through the point (3,-2) and are inclined at 60^{\circ} to the line \sqrt{3} x+y=1

Option: 1

y+2=0, \quad \sqrt{3} x-y-2-3 \sqrt{3}=0


Option: 2

x-2=0, \sqrt{3} x-y+2+3 \sqrt{3}=0


Option: 3

\sqrt{3} x-y-2-3 \sqrt{3}=0


Option: 4

\mathrm{None \, \, of \, \, these}


Answers (1)

best_answer

The equation of lines passing through (3,-2) is (y+2)=m(x-3)       .....(i)

The slope of the given line is -\sqrt{3}

So, \tan 60^{\circ}= \pm \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} . On solving, we get m=0 \text { or } \sqrt{3}

Putting the values of m in (i), the required equation is y+2=0 and  \sqrt{3} x-y-2-3 \sqrt{3}=0

Posted by

shivangi.shekhar

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