Straight line – Position of Point – JEE Main 24 January 2025 Shift 2

Solution:

We identify the vertices of the triangle formed by the intersection of the given lines:
$L_1: x+y=11$
$L_2: x+2y=16$
$L_3: 2x+3y=29$
Solving $L_1$ and $L_2$: $y=5, x=6 \Rightarrow P(6,5)$
Solving $L_1$ and $L_3$: $y=7, x=4 \Rightarrow Q(4,7)$
Solving $L_2$ and $L_3$: $y=3, x=10 \Rightarrow R(10,3)$


The triangle vertices are $Q(4,7)$, $P(6,5)$, and $R(10,3)$.
We need to find the range of $a$ for the point $(\frac{11}{2}, a)$ lying inside the triangle.
The x-coordinate is $x = 5.5$.
Since $x=5.5$ lies between $x=4$ (Vertex Q) and $x=6$ (Vertex P), the vertical line $x=5.5$ intersects the sides $QP$ and $QR$.

Equation of side $QP$ (which is $L_1$):
$x + y = 11$
At $x = 5.5$: $5.5 + y = 11 \Rightarrow y = 5.5$
So, the lower bound (or upper bound depending on geometry) is $5.5$.

Equation of side $QR$ (which is $L_3$):
$2x + 3y = 29$
At $x = 5.5$: $2(5.5) + 3y = 29 \Rightarrow 11 + 3y = 29 \Rightarrow 3y = 18 \Rightarrow y = 6$
So, the other bound is $6$.

Thus, for the point to be inside the triangle, $a \in [5.5, 6]$.
The smallest value $a_{min} = 5.5 = \frac{11}{2}$.
The largest value $a_{max} = 6$.

Product $= a_{min} \times a_{max} = \frac{11}{2} \times 6 = 33$.

Ans. (3)