If the co-ordinates of two points A and B are (3,4) and (5,-2) respectively, find the co-ordinates of any point P if PA=PB and Area of triangle PAB =10
PA=PB
PA2 = PB2
(x−3)2+(y−4)2=(x−5)2+(y+2)2
x2+9−6x+y2+16−8y=x2+25−10x+y2+4+4y
x - 3y - 1 = 0---------------(1)
now,
areaof△PAB=12∣∣ ∣∣521471741∣∣ ∣∣ = 10
12|3(−2−y)+5(y−4)+x(4+2)|
12|−6−3y+5y−20+6x|=10
12|6x+2y−26|=±20
6x + 2y - 26 = 20 or 6x + 2y - 6 = 0
3x + y - 23 = 0 or 3x + y - 3 = 0
3x + y - 23 = 0 ------------------(2)
3x + y - 3 = 0 ------------------(2)
Solving equation 1 and 2
x - 3y - 1 = 0
3x + y - 23 = 0
Multiplying 3 in equation (2) and add it to equation (1)
10x = 70 →x=7
Substituting x = 7 in equation 2, we get, y=2
Solving equation 1 and 3
x - 3y - 1 = 0
3x + y - 3 = 0
Multiply 3 in equation 3 and add it to equation 2,
10x = 10 →x=1
Substituting x = 1, in equation 2
We get, y = 0
Thus, the co-ordinates of P is (7,2) and (1,0)
So, option C and D are correct.