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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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If the lines $ y = 3x +1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$, find the value of $m$.

$\begin {array} {1 1} (A)\;\large\frac{-1 \pm 5 \sqrt 2}{7} & \quad (B)\;\large\frac{1 \pm 5 \sqrt 2}{7} \\ (C)\;\large\frac{-1 \pm 2\sqrt 5}{7} & \quad (D)\;\large\frac{1 \pm 2\sqrt 5}{7} \end {array}$

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  • Angle between two lines whose slopes are $m_1$ and $m_2$ are $ \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
The equation of the given lines are
$ \quad y=3x+1$-------(1)
$ \quad 2y=x+3$--------(2)
$ \quad y=mx+4$--------(3)
Let the slope of line (1) be $m_1=3$
Let the slope of line (2) be $m_2= \large\frac{1}{2}$
Let the slope of line (3) be $m_3=m$
Since it is given that the line (1) and (2) are equally inclined to line (3),
$ \bigg| \large\frac{m_1-m_3}{1+m_1m_3} \bigg|$$ = \bigg| \large\frac{m_2-m_3}{1+m_2m_3} \bigg|$
Substituting the values we get,
$\bigg| \large\frac{ 3-m}{1+3m} \bigg|$$ = \bigg| \large\frac{\Large\frac{1}{2}-m}{1+ \Large\frac{1}{2}m} \bigg|$
$ \Rightarrow \bigg| \large\frac{3-m}{1+3m} \bigg|$$ = \bigg| \large\frac{1-2m}{2+m} \bigg|$
i.e., $ \large\frac{3-m}{1+3m} = \pm \bigg( \large\frac{1-2m}{2+m} \bigg)$
Consider the case when
$ \large\frac{3-m}{1+3m} = \bigg( \large\frac{1-2m}{2+m} \bigg)$
On simplifying we get,
$ (3-m)(2+m)= (1-2m)(1+3m)$
$ \Rightarrow -m^2+m+6 = 1 + m -6m^2$
(i.e., ) $5m^2+5=0$
$ \Rightarrow m = \sqrt{-1}$
which is not real.
Hence it is not admissable.
Let us consider the case when
$ \large\frac{3-m}{1+3m}$$ = \large\frac{-(1-2m)}{m+2}$
$ \Rightarrow (3-m)(m+2) = -(1-2m)(1+3m)$
$ \Rightarrow -m^2+m+6 = -(1+m-6m^2)$
i.e., $7m^2-2m-7=0$
Let us apply the quadratic formula to solve the above equation.
$ \large\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
Substituting the values for $a=7, b=-2$ and $c=-7$ we get,
$ m = \large\frac{-(-2) \pm \sqrt{4-4(-7)(7)}}{2 \times 7}$
$ = \large\frac{2 \pm \sqrt{4+196}}{14}$
$ = \large\frac{2 \pm \sqrt{200}}{14}$
$ = \large\frac{2 \pm 10\sqrt 2}{14}$
$ = \large\frac{1 \pm 5\sqrt 2}{7}$
Hence the value of $m$ is $ \large\frac{1 \pm 5\sqrt 2}{7}$
answered May 22, 2014 by thanvigandhi_1
 

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