Question

The coordinates of the point ‘P’ on the line 2x + 3y + 1 = 0, such that |PA – PB| is maximum, where A is (2, 0) and B is (0, 2), is

• A. (5, –3)
• B. (7, –5)
• C. (9, –7)
• D. (11, –9)

Answer 1

The correct option is B (7, –5)

$|PA–PB|\le |AB|$
Maximum value of |PA – PB| is |AB|, which is possible only when P, A and B are collinear.
Let P be denoted by (x, y) then equation of AB is x + y = 2. ….(i)
Now solving (i) and 2x + 3y + 1 = 0
We get x = 7, y = –5
$P^{\circ }(7,-5)$