Straight LineHardBloom L3
Question
Let $A(1,1)$ and $B(3,3)$ be two fixed points and $P$ be a variable point such that area of $\bigtriangleup PAB$ remains constant equal to 1 for all position of $P$, then locus of $P$ is given by:
Options
A.$2y = 2x + 1$
B.$2y = 2x - 1$
C.$y = x + 1$
D.$y = x - 1$
Solution
Sol. [Key Concept: Area $\bigtriangleup \ PAB = 1$ with $A(1,1),B(3,3)$ ⇒ $P$ on a line parallel to $AB$ at fixed distance.]
$P$ will be on line parallel to $y = x$ at a perpendicular distance of $\frac{1}{\sqrt{2}}$
locus of $P$ will be $y - x = 1$ or $y - x = - 1$
✘ (a) $2y = 2x + 1$ — wrong perpendicular distance.
✘ (b) $2y = 2x - 1$ — wrong perpendicular distance.
✔ (c) $y = x + 1$ is parallel to $AB$ at the right distance.
✔ (d) $y = x - 1$ is the parallel on the other side.
Create a free account to view solution
View Solution FreeMore Straight Line Questions
A line is perpendicular to 3x + y = 3 and passes through a point (2, 2) Its y intercept is -...The equations of the lines on which the perpendiculars from the origin make 30o angle with x-axis and which form a trian...If pair of lines x2 − 2mxy − y2 = 0 and x2 − 2nxy − y2 = 0 are such that each bisects the angles...Let the line $L_{1}:x + 3 = 0$ intersect the lines $L_{2}:x - y = 0$ and $L_{3}:3x + y = 0$ at the points A and B , resp...Let $ABC$ be a triangle. Let $A$ be the point $(1,2),y = x$ is the perpendicular bisector of $AB$ and $x - 2y + 1 = 0$ i...