The given equation of lines are 3x – y + 1 = 0 ... (i) x – 2y + 3 = 0 ... (ii) Clearly slope of (i) is 3 and slope of (ii) is $\frac{1}{2}$. Since, we have given that both these lines inclined to the same line y = mx + 4, whose slope is m, then we must have $\left|\frac{m-3}{1+3m}\right|$ = $\left|\frac{m-\frac{1}{2}}{1+m\frac{1}{2}}\right|$ ⇒ $\left|\frac{m-3}{1+3m}\right|$ = $\left|\frac{2m-1}{2+m}\right|$ ⇒ $\frac{m-3}{1+3m}$ = ± $\left(\frac{2m-1}{2+m}\right)$ Now we have two cases Case (I) : $\frac{m-3}{1+3m}$ = $\frac{2m-1}{2+m}$ ⇒ (m – 3) (2 + m) = (1 + 3m) (2m – 1) ⇒ 2m + $m^2$ – 6 – 3m = 2m – 1 + $6m^2$ – 3m ⇒ – $5m^2$ – 5 = 0 ⇒ $5m^2$ + 5 = 0 ⇒ m2 = – 1, which is not possible. Case (ii) : $\frac{m-3}{1+3m}$ = - $\left(\frac{2m-1}{2+m}\right)$ ⇒ (m – 3) (2 + m) = – (2m – 1) (1 + 3m) ⇒ 2m + $m^2$ – 6 – 3m = – (2m + $6m^2$ – 1 – 3m) ⇒ $6m^2$ – m – 1 + $m^2$ – 6 – m = 0 ⇒ $7m^2$ – 2m – 7 = 0 ⇒ m = $\frac{2\pm\left((-2)^2-4\times7\times(-7)\right)}{2\times7}$ = $\frac{2\pm\sqrt{4+196}}{14}$ = $\frac{2\pm\sqrt{200}}{14}$ = $\frac{2\pm10\sqrt{2}}{14}$ Hence, m = $\frac{1\pm5\sqrt{2}}{7}$