Question

two points A (3, 4) and B (5,-2 ) are given find the point P such that P is equal to PB and area of triangle ABC is equal to 20
*note that triangle ABC is not a right angled triangle
plzzz help me to solve this problem

Answer 1

Answer:

( 10, 3) or ( -2, -1).

Step-by-step explanation:

Hi,

Let the coordinates of point P be (h, k)

Given that PA = PB,

so PA² = PB²

(h - 3)² + (k - 4)² = (h - 5)² + (k + 2)²

4h - 12k - 4 = 0

h - 3k = 1------(1)

Also, given that Area of triangle PAB = 20

Area = 1/2|| 6h + 2k - 26 || = 20

3h  + k - 13 = ±20

3h  + k - 13 = 20 or

3h  + k - 13 = -20

3h + k = 33 -------(2)or

3h + k  = -7--------(3)

Solving (1) and (2), we get h = 10 and k = 3

Solving (1) and (3), we get h = -2 and k = -1

Hence P could be either ( 10, 3) or ( -2, -1).

Hope, it helps !