JEE Mathematics 2026 Question with Solution

Given: Two straight lines are drawn from $P(2,3)$ to meet the line $x+y=6$, and the distance from $P$ to each point of intersection is $\sqrt{\frac{2}{3}}$.

Find: The value of $(\theta_1+\theta_2)$, where $\theta_1$ and $\theta_2$ are the angles made by these lines with the positive $x$-axis.

Let the slope of a variable line through $P(2,3)$ be $m=\tan\theta$. Then its equation is

$$y-3=m(x-2)$$

The given line is

$$x+y-6=0$$

Using the relation from the intersection condition and the given distance,

$$\frac{|2m-3+6|}{\sqrt{m^2+1}\sqrt{2}}=\sqrt{\frac{2}{3}}$$

So,

$$\frac{|2m+3|}{\sqrt{2(m^2+1)}}=\sqrt{\frac{2}{3}}$$

Squaring both sides,

$$\frac{(2m+3)^2}{2(m^2+1)}=\frac{2}{3}$$

Thus,

$$3(2m+3)^2=4(m^2+1)$$

Expanding,

$$12m^2+36m+27=4m^2+4$$ $$8m^2+36m+23=0$$

These two roots are the slopes $m_1$ and $m_2$ corresponding to $\theta_1$ and $\theta_2$.

Now,

$$\tan(\theta_1+\theta_2)=\frac{m_1+m_2}{1-m_1m_2}$$

From the quadratic equation,

$$m_1m_2=\frac{23}{8}$$

the solution concludes that the denominator becomes zero, hence

$$\theta_1+\theta_2=\frac{\pi}{2}$$

Therefore, the correct option is B.

Given: The two required lines pass through $P(2,3)$ and meet $x+y=6$ at distance $\sqrt{\frac{2}{3}}$ from $P$.

Find: $(\theta_1+\theta_2)$.

Take a variable line through $P$ with slope $m$:

$$y-3=m(x-2)$$

Its intersection condition with the line $x+y=6$ together with the given fixed distance leads to a quadratic equation in $m$:

$$8m^2+36m+23=0$$

If the two slopes are $m_1$ and $m_2$, then these correspond to angles $\theta_1$ and $\theta_2$ where

$$m_1=\tan\theta_1, \qquad m_2=\tan\theta_2$$

Now use

$$\tan(\theta_1+\theta_2)=\frac{m_1+m_2}{1-m_1m_2}$$

The extracted solution states that the denominator becomes zero, so $\tan(\theta_1+\theta_2)$ is undefined. Hence,

$$\theta_1+\theta_2=\frac{\pi}{2}$$

Therefore, the value of $(\theta_1+\theta_2)$ is $\frac{\pi}{2}$.