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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Two lines passing through the point (2,3) intersects each other at an angle of $60^{\circ}$. If slope of one line is 2, find equation of the other line.

$\begin{array}{1 1}(A)\;(\sqrt 3-2)x+(2\sqrt 3-1)y=1+8\sqrt 3 \: and \: (2-\sqrt 3)x+(2\sqrt 3-1)y=1+8\sqrt 3 \\(B)\; (\sqrt 3+2)x+(2\sqrt 3-1)y=1+8\sqrt 3 \: and \: (2+\sqrt 3)x+(2\sqrt 3+1)y=-1+8\sqrt 3 \\(C)\; (2-\sqrt 3)x+(2\sqrt 3-1)y=1-8\sqrt 3 \: and \: (2+\sqrt 3)x+(1-2\sqrt 3)y=-1-8\sqrt 3. \\(D)\; (\sqrt 3-2)x+(2\sqrt 3+1)y=-1+8\sqrt 3\: and \: (2+\sqrt 3)x+(2\sqrt 3-1)y=1+8\sqrt 3 \end{array} $

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  • Angle between two line is given by $ \theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$ where $m_1$ and $m_2$ are the slopes of the two lines.
  • Equation of a line with slope $m$ and passing thru' $(x_1,y_1)$ is $(y-y_1)=m(x-x_1)$
Given point of intersection is (2,3) and the angle between the line is $ 60^{\circ}$
Also $m_1=2$
$ \therefore \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Substituting the values we get,
$ \tan 60^{\circ} = \bigg| \large\frac{2-m_2}{1+2m_2} \bigg|$
But $ \tan 60^{\circ} = \sqrt 3$
$ \sqrt 3 = \pm \large\frac{(2-m_2)}{1+2m_2}$
On rearranging we get,
$ \sqrt 3 (1+2m_2)= \pm (2-m_2)$
Taking the positive value we get,
$ \sqrt 3(1+2m_2)=2-m_2$
$ \Rightarrow m_2(2\sqrt 3+1)=2-\sqrt 3$
$ \Rightarrow m_2=\large\frac{2-\sqrt 3}{2\sqrt 3+1}$
Taking the negative value
$ \sqrt 3(1+2m_2)=-(2-m_2)$
$m_2(2\sqrt 3-1)=-2-\sqrt 3$
$ \Rightarrow m_2=-\large\frac{(2+\sqrt 3)}{2\sqrt 3-1}$
Case (i)
Considering the case when $m_2=\large\frac{2-\sqrt 3}{2\sqrt 3+1}$
Since the line passes through (2,3)
Equation of the line is
$(y-3)=\bigg[ \large\frac{2-\sqrt 3}{2\sqrt 3+1} \bigg]$$ (x-2)$
$(2\sqrt 3+1)(y-3)=(2-\sqrt 3)(x-2)$
$ \Rightarrow (\sqrt 3-2)x+(2\sqrt 3+1)y=-4+2\sqrt 3+6\sqrt 3+3$
$ \Rightarrow (\sqrt 3-2)x+(2\sqrt 3+1)y=-1+8 \sqrt 3$
Hence the equation of one line is
$(\sqrt 3-2)x+(2\sqrt 3+1)y=-1+8\sqrt 3$
Case (ii)
Considering the slope $m_2 = -\large\frac{(2+\sqrt 3)}{2\sqrt 3-1)}$
$ (y-3)=-\large\frac{(2+\sqrt 3)}{2\sqrt 3-1}$$(x-2)$
(i.e.,) $(y-3)(2\sqrt 3-1)=-(2+\sqrt 3)(x-2)$
(i.e.,) $(2+\sqrt 3)x+(2\sqrt 3-1)y=1+8\sqrt 3$
Hence the equation of the lines are
$ (\sqrt 3-2)x+(2\sqrt 3+1)y=-1+8\sqrt 3$
$ (2+\sqrt 3)x+(2\sqrt 3-1)y=1+8\sqrt 3$
answered May 13, 2014 by thanvigandhi_1
 

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