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 \(\frac{1±5\sqrt2}{7}.\)