Coordinate Geometry
Maximum of |PA-PB| on a line
MJMT_Full_Test_11
Grade 12
Question:
The coordinate of point $P$ on the line $3x+2y+10=0$ such that $|PA-PB|$ is maximum, where $A=(4,2)$ and $B=(2,4)$, is
$(-22, 28)$
$(22, -28)$
$(22, 28)$
$(-22, -28)$
Step-by-Step Solution
Key Concept: $|PA-PB|$ is maximum when $P$, $A$, $B$ are collinear (i.e., $P$ lies on line $AB$ extended) AND $P$ is also on the given line.
Line through $A,B$: $x+y=6$. Intersect with $3x+2y+10=0$: $P=(-22,28)$.
Correct Answer: 1