Answer :
Answer: No
Step-by-step explanation:
H=9 ; 8)
B=(5 ;4) (ball)
R=(7 ; 0) (hit point)
B' = symetric of B axis perpendicular of x in R
B' = (7 + (7 - 5); 4) = (9 ; 4)
Equation BR: y - 4 = (0 - 4)/(7 - 5)(x - 5)==> y = -2x + 14
Equation RB': y - 4 = (4 - 0) (9 - 7)(x - 9) ==> y = 2 x -14
Is H a point of RB'? y = 2x- 14 : 8= ? 2 * 9 - 14==> 8= ? 4
No!
Hope it helps :)