Question

The equation of a line $L_{1}$ passing through the point $(2, 4)$ and making an angle $\tan^{-1}(2)$ with another line $x+2y=4$ is $ax+by+c=0$. If this line $L_{1}$ is neither horizontal nor vertical, then $\frac{b+c}{a}=$

• A. 1
• B. $\frac{14}{3}$
• C. 2
• D. 0

Hint:

Always check both positive and negative openings of the absolute value brackets when using the angle-between-lines formula.

Answer 1

The Correct Option is C

Step 1: Recall the angle between two lines.
If two lines have slopes $m_1$ and $m_2$, the angle between them satisfies \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|. \]
Step 2: Read the given line slope.
The line $x+2y=4$ can be written $y=-\frac12x+2$, so its slope is $m_2=-\frac12$.
Step 3: Use the given angle.
We are told the angle is $\tan^{-1}(2)$, so $\tan\theta=2$. Put the numbers in: \[ 2=\left|\frac{m_1+\tfrac12}{1-\tfrac12 m_1}\right|. \]
Step 4: Solve for the slope $m_1$.
Taking the plus case: $2\left(1-\tfrac12 m_1\right)=m_1+\tfrac12$, which gives $2-m_1=m_1+\tfrac12$, so $m_1=\tfrac34$. The minus case gives no real line, and a vertical line is ruled out by the question.
Step 5: Write the line through $(2,4)$.
\[ y-4=\tfrac34(x-2)\ \Rightarrow\ 3x-4y+10=0. \] So $a=3,\ b=-4,\ c=10$.
Step 6: Find the asked ratio.
\[ \frac{b+c}{a}=\frac{-4+10}{3}=\frac{6}{3}=2. \] The value $2$ is the third option. \[ \boxed{2} \]