Question

Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1+\theta_2)$ is

• A. $\dfrac{\pi}{6}$
• B. $\dfrac{\pi}{2}$
• C. $\dfrac{\pi}{12}$
• D. $\dfrac{\pi}{3}$

Hint:

When the product of slopes equals 1, the sum of the angles made with the $x$-axis is $\dfrac{\pi}{2}$.

Answer 1

The Correct Option is B

Step 1: Write equation of a variable line through $P(2,3)$. 
Let the slope be $m=\tan\theta$. Equation of the line is \[ y-3=m(x-2) \] Step 2: Find distance of intersection point from $P$. 
The given line is $x+y-6=0$. Distance from point of intersection to $P$ is given as \[ \sqrt{\frac{2}{3}} \] Using distance between two intersecting lines formula, \[ \frac{|2m-3+6|}{\sqrt{m^2+1}\sqrt{2}}=\sqrt{\frac{2}{3}} \] Step 3: Simplify the equation. 
\[ \frac{|2m+3|}{\sqrt{2(m^2+1)}}=\sqrt{\frac{2}{3}} \] Squaring both sides, \[ \frac{(2m+3)^2}{2(m^2+1)}=\frac{2}{3} \] \[ 3(2m+3)^2=4(m^2+1) \] Step 4: Solve for $m$. 
\[ 12m^2+36m+27=4m^2+4 \] \[ 8m^2+36m+23=0 \] This gives two slopes corresponding to $\theta_1$ and $\theta_2$. 
Step 5: Use angle sum property. 
For slopes $m_1,m_2$, \[ \tan(\theta_1+\theta_2)=\frac{m_1+m_2}{1-m_1m_2} \] Here, denominator becomes zero, hence \[ \theta_1+\theta_2=\frac{\pi}{2} \]