Browse Questions

# If the lines $y = 3x +1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$, find the value of $m$.

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

Toolbox:
• 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 $\large\frac{1 \pm 5\sqrt 2}{7}$