Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The equations of the lines passing through the point 1, 0 and at a distance square root 3 divided 2 from the origin are A square root 3 x plus y minus square root 3 equal to 0 square root 3 x minus y minus square root 3 equal 0

The equations of the lines passing through the point (1, 0) and at a distance \frac{\sqrt{3}}{2} from the origin, are
A. \sqrt{3}x+y-\sqrt{3}=0,\sqrt{3}x-y-\sqrt{3}=0
B. \sqrt{3}x+y+\sqrt{3}=0,\sqrt{3}x-y+\sqrt{3}=0
C. x+\sqrt{3}y-\sqrt{3}=0,x-\sqrt{3}y-\sqrt{3}=0
D. None of these.

Answers (1)

Let the equation of any line passing through the point (1,0) is y-y1=m(x-x1)

 y-0=m(x-1) 

y=mx-m 

 mx-m-y=0….........(i)

   Distance from the origin of the line is \frac{\sqrt{3}}{2}      Distance of point (x1,y1) from the equation Ax+By+C=0

 d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}

\frac{\sqrt{3}}{2}=\frac{\left | m*0+\left ( -1 \right )*0+\left ( -m \right ) \right |}{\sqrt{m^{2}+\left ( -1 \right )^{2}}}=\left |- \frac{m}{\sqrt{m^{2}+1}} \right |   

 Squaring both the sides we get  \frac{3}{4}=\frac{m^{2}}{m^{2}+1}

   3(m2+1)=4m2 

 3m2+3=4m2 

  4m2-3m2=3 

  m2=3 

 m = \pm \sqrt{3} 

  Putting the value of m= \sqrt{3} in equation (i) we get  \sqrt{3}x-y-\sqrt{3}=0 

  Now, putting the value of m= -\sqrt{3} in the same equation we get-\sqrt{3}x-y+\sqrt{3}=0 

Hence, the correct option is (a)

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question