Grade 11Discuss with colleagues and IITians

A and B are two points (3,4) and (5,-2). Find a point P such that PA=PB and area of ∆ PAB is 10

Abhishek

8 Years agoGrade 11

1 Answer

Arun

8 Years ago

P must be on the perpendicular bisector of the segment AB.

The midpoint of segment AB is M = (1/2)(A + B) = (4, 1).

The length of segment AB = √40 = 2√10

Let h be the length of segment MP, which is the altitude of triangle PAB at base AB.

Then the area of triangle PAB = (1/2) (2√10) (h) = 10.
So h = √10 = the length of segment MP

So segment MP is perpendicular to segment MA
and the length of segment MP = the length of segment MA = √10

To get from point M(4,1) to point A(3,4), you go left 1 and then up 3.
So to travel the same distance from M but in a direction that is perpendicular to MA,
you must go either
right 3 and up 1 to P = (7, 2)
or left 3 and down 1 to P = (1, 0)

So P = (7, 2) or (1, 0)