Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade 11
Question:
Given $A \equiv (1,1)$ and $AB$ is any line through it cutting the $x$-axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y$-axis in $C$, then the equation of locus of mid-point $P$ of $BC$ is:
x + y = 1
x + y = 2
x + y = 2xy
2x + 2y = 1
Step-by-Step Solution
Key Concept: The locus of the foot of perpendicular from a fixed point to a variable line through another fixed point satisfies a linear equation.
The line $y - 1 = m(x - 1)$ has perpendicular line $y - 1 = -\frac{1}{m}(x - 1)$. Finding their intersection at $(1, 1)$ and using the constraint that both lines pass through specific intercepts, the locus of the foot of the perpendicular is derived. The foot traces the path where $x + y = 1$, which is the locus equation.
Correct Answer: 1