Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade 11
Question:
The medians $AD$ and $BE$ of a triangle $ABC$ with vertices $A(0, b)$, $B(0, 0)$ and $C(a, 0)$ are perpendicular to each other if:
$a = \sqrt{2}b$
$a = -\sqrt{2}b$
$b = \sqrt{3}a$
$b = -\sqrt{3}a$
Step-by-Step Solution
Key Concept: Perpendicularity of two lines is expressed as the product of their slopes equaling $-1$.
With $AD \perp BE$ where $A(0,b)$, $B(0,0)$, $C(a,0)$, the point $E(\frac{a}{2}, \frac{b}{2})$ is the midpoint of $AC$. Using the perpendicularity condition, the slopes satisfy $\frac{b}{a} \cdot \frac{-b}{a/2} = -1$, which simplifies to $2b^2 = a^2$. Therefore $a = \pm\sqrt{2}b$.
Correct Answer: 1,2