Question

origin
PA=PB and the coordinates of point A
B are (2,0) and (-2,4) respectively
and if P lies on the y axis then calculate
e coordinates of point P.​

Answer 1

$\color{red}\huge{\underline{\underline{GIVEN\dag}}}$

• $A(2,0)$
• $B(-2,4)$
• $PA=PB$

$\color{red}\huge{\underline{\underline{TO\:FIND\dag}}}$

• $P(x,y)$
• Since 'P' lies on y axis,x-coordinate of 'P' is 0
• $P(0,y)$

$\color{red}\huge{\underline{\underline{SOLUTION\dag}}}$

$PA=PB$

${(0-2)}^{2}+{(y-0)}^{2}={(0+2)}^{2}+{(y-4)}^{2}$

${(-2)}^{2}+{(y)}^{2}={(2)}^{2}+{(y-4)}^{2}$

$4 + {y}^{2} = 4 + {y}^{2} - 8y + 16$

$8y = 16$

$\therefore y = 2$

Therefore,the co-ordinates of 'P' is,

$\orange{ \boxed{P(0,2)}}$

HOPE THAT HELPS!!

Answer 2

Step-by-step explanation:

$GIVEN$

• A ( 2, 0).
• B ( - 2,4)
• PA = PB

$TO \: \: FIND$

• P (x, y).
• Since ' P' line on Y axis x - coordinate of ' P' is 0
• P ( 0, y )

$SOLUTION$

$PA = PB$

$⟾( {0 - 2)}^{2} + {(y - 0)}^{2} = ( {0 + 2)}^{2} + (y - {4)}^{2}$

$⟾ {( - 2)}^{2} + ( {y)}^{2} = {(2)}^{2} + {(y - 4)}^{2}$

$⟾4 + {y}^{2} = 4 + {y}^{2} - 8y + 16$

$⟾8y = 16$

$⟾y = \frac{16}{8}$

$∴ \: \: \: y = 2$

Therefore, the co- ordinates of ' p' is P ( 0, 2)