Show that $R$ is an equivalence relation.

Determine the equivalence class containing $({x_1},{y_1})$ and interpret it geometrically.

$\frac{{{x_1}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_1}}} \Rightarrow ({x_1},{y_1})R({x_1},{y_1})$ so $R$ is reflexive     R1

$({x_1},{y_1})R({x_2},{y_2}) \Rightarrow \frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}} \Rightarrow \frac{{{x_2}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_2}}} \Rightarrow ({x_2},{y_2})R({x_1},{y_1})$     M1A1

so $R$ is symmetric

$({x_1},{y_1})R({x_2},{y_2})$ and  $({x_2},{y_2})R({x_3},{y_3}) \Rightarrow \frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}}$ and $\frac{{{x_2}}}{{{x_3}}} = \frac{{{y_3}}}{{{y_2}}}$     M1

multiplying the two equations,     M1

$\Rightarrow \frac{{{x_1}}}{{{x_3}}} = \frac{{{y_3}}}{{{y_1}}} \Rightarrow ({x_1},{y_1})R({x_3},{y_3})$ so $R$ is transitive     A1

thus $R$ is an equivalence relation     AG

[6 marks]

$(x,y)R({x_1},{y_1}) \Rightarrow \frac{{{x_{}}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_{}}}} \Rightarrow xy = {x_1}{y_1}$     (M1)

the equivalence class is therefore $\left\{ {(x,y)|xy = {x_1}{y_1}} \right\}$     A1

geometrically, the equivalence class is (one branch of) a (rectangular) hyperbola     A1

[3 marks]