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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

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

Grade:11

1 Answers

Arun
25750 Points
6 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)

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