Question

If the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$, then $m =$

• A. $\frac{1+3\sqrt{2}}{7}$
• B. $\frac{1-3\sqrt{2}}{7}$
• C. $\frac{1\pm3\sqrt{2}}{7}$
• D. $\frac{1\pm5\sqrt{2}}{7}$

Answer 1

Answer: (D)

$\frac{1\pm5\sqrt{2}}{7}$

Answer 2

Let the slopes of the given lines $y = 3x + 1$, $2y = x + 3$, and $y = mx + 4$ be $m_1$, $m_2$, and $m_3$ respectively.

From the equations, we have: $m_1 = 3$

$2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2}$, so $m_2 = \frac{1}{2}$

$m_3 = m$

The condition that the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$ means that the angle between the first line and the third line is equal to the angle between the second line and the third line.

Let $\theta_1$ be the angle between the line with slope $m_1$ and the line with slope $m_3$. Let $\theta_2$ be the angle between the line with slope $m_2$ and the line with slope $m_3$.

The angle $\theta$ between two lines with slopes $m_a$ and $m_b$ is given by $\tan \theta = \left| \frac{m_a - m_b}{1 + m_a m_b} \right|$.

According to the problem, $\theta_1 = \theta_2$, which implies $\tan \theta_1 = \tan \theta_2$.

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

Substitute the values of $m_1$ and $m_2$:

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

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

This equation gives two possibilities:

Case 1: $\frac{3 - m}{1 + 3m} = \frac{1 - 2m}{2 + m}$

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

$6 + 3m - 2m - m^2 = 1 + 3m - 2m - 6m^2$

$6 + m - m^2 = 1 + m - 6m^2$

$6 - m^2 = 1 - 6m^2$

$5m^2 = -5$

$m^2 = -1$

This equation has no real solution for $m$.

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

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

$6 + 3m - 2m - m^2 = 2m + 6m^2 - 1 - 3m$

$6 + m - m^2 = -m + 6m^2 - 1$

Rearrange the terms to form a quadratic equation:

$6m^2 + m^2 - m - m - 1 - 6 = 0$

$7m^2 - 2m - 7 = 0$

We solve this quadratic equation for $m$ using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = 7$, $b = -2$, $c = -7$.

$m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)}$

$m = \frac{2 \pm \sqrt{4 + 196}}{14}$

$m = \frac{2 \pm \sqrt{200}}{14}$

$m = \frac{2 \pm \sqrt{100 \times 2}}{14}$

$m = \frac{2 \pm 10\sqrt{2}}{14}$

Factor out 2 from the numerator:

$m = \frac{2(1 \pm 5\sqrt{2})}{14}$

$m = \frac{1 \pm 5\sqrt{2}}{7}$

The possible values of $m$ are $\frac{1 + 5\sqrt{2}}{7}$ and $\frac{1 - 5\sqrt{2}}{7}$.