(iii) Co-ordinates of P’ = (3,5)
Co-ordinates of P’’ = (-3,-5)
Here x1 = 3, y1 = 5 , x2 = -3, y2 = -5
Let Q(x,y) be the midpoint of P’P’’
By midpoint formula,
x = (x1+x2)/2
y = (y1+y2)/2
x = (3+-3)/2 = 0/2 = 0
y = (5+-5)/2 = 0/2 = 0
Hence the co-ordinate of midpoint of P’P’’ is (0,0) .
(iv) Co-ordinates of P = (3,-5)
Co-ordinates of P’’ = (-3,-5)
Here x1 = 3, y1 = -5 , x2 = -3, y2 = -5
Let R(x,y) be the midpoint of PP’’
By midpoint formula,
x = (x1+x2)/2
y = (y1+y2)/2
x = (3+-3)/2 = 0/2 = 0
y = (-5+-5)/2 = -10/2 = -5
So the co-ordinate of midpoint of PP’’ is (0,-5) .
Here x co-ordinate is zero.
Hence the point lies on Y-axis.