Step-by-Step Solution
Key Concept: Case study on coordinate geometry.
(a) Let the fountain be at point $(x,0)$. Write the equation obtained by setting its distance from G equal to its distance from C. [1 Mark]
$(x-7)^2+36=(x+3)^2+16$. [1.0 Mark]
(b) Solve this equation to find the $x$-coordinate of the fountain. [1 Mark]
$x^2-14x+49+36=x^2+6x+9+16\Rightarrow-14x+85=6x+25\Rightarrow-20x=-60\Rightarrow x=3$. [1.0 Mark]
(c) Find the distance from the fountain to the gate G. [1 Mark]
Fountain is at $(3,0)$. Distance to $G(7,6)$: $\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\approx7.21$ m. [1.0 Mark]
(d) Verify that this distance equals the distance from the fountain to the bench C. [1 Mark]
Distance from $(3,0)$ to $C(-3,4)$: $\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$, which matches the distance to G found in part (c). Verified. [1.0 Mark]
Correct Answer: $(x-7)^2+36=(x+3)^2+16$. [1.0 Mark] | $x^2-14x+49+36=x^2+6x+9+16\Rightarrow-14x+85=6x+25\Rightarrow-20x=-60\Rightarrow x=3$. [1.0 Mark] | Fountain is at $(3,0)$. Distance to $G(7,6)$: $\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\approx7.21$ m. [1.0 Mark] | Distance from $(3,0)$ to $C(-3,4)$: $\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$, which matches the distance to G found in part (c). Verified. [1.0 Mark]