Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade None
Question:
In a $\triangle ABC$, $A = (\alpha, \beta)$, $B(1,2)$, $C(2,3)$ and point $A$ lies on the line $y = 2x + 3$, where $a, b \in I$. If the area of $\triangle ABC$ be such that area of triangle lies in interval $[2, 3)$, then the possible value of $a + b$ is/are:
Step-by-Step Solution
Key Concept: The area of a triangle with vertices is found using the determinant formula, then the constraint on area restricts the parameter values.
The area is computed as the modulus of $\frac{1}{2}\begin{vmatrix} a & 2a+3 & 1 \\ 2 & 2 & 1 \\ 2 & 3 & 1 \end{vmatrix} = \frac{1}{2}|a + 2a + 3 - 1| = \frac{1}{2}|3a + 2|$. From $\frac{|a+2|}{2} \in [2,3]$, we get $a \in (-8,-6] \cup [2,4)$, giving the set of points $(\alpha, \beta) = \{(-7,-11), (-6,-9), (2,7), (3,9)\}$.
Correct Answer: 1,2,3,4