Question

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

Answer 1

The equations of the given lines are
$y = 3x + 1 … (1)$
$2y = x + 3 … (2)$
$y = mx + 4 … (3)$

Slope of line (1), $m_1 = 3$
Slope of line (2), $m_2=\frac{1}{2}$
Slope of line (3),$m_3 = m$

It is given that lines (1) and (2) are equally inclined to line (3).
This means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).

$∴ \left|\frac{m_1-m_3}{1+m_1m_3}\right|=\left|\frac{m_2-m_3}{1+m_2m_3}\right|$

$⇒\left |\frac{3-m}{1+3m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{1}{2}m}\right|$

$⇒ \left|\frac{3-m}{1+3m}\right|=\left|\frac{1-2m}{m+2}\right|$

$⇒\frac{ 3-m}{1+3m}=±\left(\frac{1-2m}{m+2}\right)$

$⇒\frac{ 3-m}{1+3m}=\left(\frac{1-2m}{m+2}\right) or \frac{3-m}{1+3m}=-\left(\frac{1-2m}{m+2}\right)$

If $\frac{3-m}{1+3m}=\left(\frac{1-2m}{m+2}\right)$, then

$(3-m)(m+2)=(1-2m)(1+3m)$

$⇒ -m^2+m+6=1+m-6m^2$
$⇒ 5m^2+5=0$
$⇒ (m^2+1)=0$
$⇒ m=\sqrt{-1}$, which is not real.

Hence, this case is not possible.

If $\frac{3-m}{1+3m}=-\left(\frac{1-2m}{m+2}\right)$,then

$(3-m)(m+2)=-(1-2m)(1+3m)$
$⇒ -m^2+m+6=-(1+m-6m^2)$

$⇒ 7m^2-7=0$

$⇒ m=\frac{2±\sqrt{4-4(7)(-7)}}{2(7)}$

$⇒ m=\frac{2±2\sqrt{1+49}}{14}$

$⇒ m=\frac{1±5\sqrt2}{7}$

Thus, the required value of m is $\frac{1±5\sqrt2}{7}.$