The coordinates of 2 given points are 3,4 and 5,-2. Find the coordinates of point P such pa`square` = pb`square` and area of triangle PAB = 10`` sq. units.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the determinant formula that helps you to evaluate the area of triangle such that:

`A = (1/2)|[(3,4,1),(5,-2,1),(x_P,y_P,1)]|`

Since the problem provides the value of the area yields:

`10 = (1/2)*|(-6 + 5y_P + 4x_P + 2x_P - 3y_P - 20)|`

`20 = 6x_P + 2y_P - 26 => 10 = 3x_P + y_P - 13 => 3x_P + y_P = 23`

You need to use the information provided by the problem that `PA = PB` , such that:

`{(PA = sqrt((x_A - x_P)^2 + (y_A - y_P)^2)),(PB = sqrt((x_B - x_P)^2 + (y_B - y_P)^2)):}`

`(3 - x_P)^2 + (4 - y_P)^2 = (5 - x_P)^2 + (-2 - y_P)^2`

`9 - 6x_P + (x_P)^2 + 16 - 8y_P + (y_P)^2 = 25 - 10x_P + (x_P)^2 + 4 + 4y_P + (y_P)^2`

Reducing like terms yields:

`25 - 6x_P - 8y_P = 29 - 10x_P + 4y_P`

`4x_P - 12y_P = 4 => x_P - 3y_P = 1`

`x_P = 3y_P + 1`

You need to substitute` 3y_P + 1` for `x_P` in `3x_P + y_P = 23` such that:

`3(3y_P + 1) + y_P = 23 => 9y_P + 3 + y_P = 23`

`10y_P = 20 => y_P = 2`

`x_P = 7`

Hence, evaluating the coordinates of the point P, under the given conditions, yields `x_P = 7 ;y_P = 2, P(7,2).`