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Equations of line L1 is \sqrt{3}y-x+1=0 and Equation of line L2 is y-\sqrt{3}x+1=0 then find the equation of Line which is equally inclinde with both line and passing through intersection point .

Option: 1

y=-x+1+\sqrt{3}


Option: 2

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


Option: 3

y=x+1-\sqrt{3}


Option: 4

Option (i) and option (iii) both


Answers (1)

best_answer

 

 

Line Equally Inclined with two lines -

Line Equally Inclined with two lines

If the two lines with slope m1 and m2 are equally inclined to a line having slope m, then 

Two lines with slope m1 and m2 intersect at point A

 

As from the fig

Hence,

-

\sqrt{3}y-x+1=0 \Rightarrow m_1=\frac{1}{\sqrt{3}}\\ y-\sqrt{3}x+1=0 \Rightarrow m_2=\sqrt{3}\\ y-\sqrt{3}x+1=0

\begin{array}{l}{\text { If the two lines with slope } m_{1} \text { and } m_{2} \text { are equally inclined to a line having slope } m, \text { then }} \\ {\left(\frac{m_{1}-m}{1+m_{1} m}\right)=-\left(\frac{m_{2}-m}{1+m_{2} m}\right)}\end{array}\\

\left(\frac{\mathbf{\frac{1}{\sqrt{3}}}-\mathbf{m}}{1+\mathbf{\frac{1}{\sqrt{3}}} \mathbf{m}}\right)=-\left(\frac{\mathbf{\sqrt{3}}-\mathbf{m}}{1+\mathbf{\sqrt{3}} \mathbf{m}}\right)\Rightarrow m^2=1\\ m=\pm 1\\

\text{For Intersection point }\\ \sqrt{3}y-x+1=0\\ y-\sqrt{3}x+1=0 \\ x=\frac{\sqrt{3}-1}{2}\\ y=\frac{3}{2}(1-\sqrt{3})\\ \text{Equation of lines}\\ y-\frac{3}{2}(1-\sqrt{3})=x-\frac{\sqrt{3}-1}{2}\\ L_1: \ y=x+2-2\sqrt{3}\\ y-\frac{3}{2}(1-\sqrt{3})=-1(x-\frac{\sqrt{3}-1}{2})\\ L_2: y=-x+1-\sqrt{3}

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