1. Problem 12(a)i: Write down the gradient of the line $y=5x+2$. The gradient (slope) is the coefficient of $x$, which is $5$. 2. Problem 12(a)ii: Find $y$ when $x=7$. Substitute $7$ into the equation: $$y=5(7)+2=35+2=37.$$ So, $y=37$. 3. Problem 12(b): Make $x$ the subject of $y=5x+2$. Start with: $$y=5x+2$$ Subtract $2$ from both sides: $$y-2=5x$$ Divide both sides by $5$: $$x=\frac{y-2}{5}$$ 4. Problem 13: Rectangle A has vertices at $(2,1)$ and $(5,9)$. We rotate rectangle A 90° anticlockwise about $(5,9)$ to get rectangle B. The vertex $P$ is the image of vertex $(2,1)$ after the rotation. Step 1: Translate point $(2,1)$ relative to center $(5,9)$: $$ (2-5, 1-9) = (-3, -8) $$ Step 2: Apply 90° anticlockwise rotation formula: $$ (x', y') = (-y, x) $$ So, $$ (-y, x) = (-(-8), -3) = (8, -3) $$ Step 3: Translate back to original coordinate system: $$ (8+5, -3+9) = (13, 6) $$ Therefore, the coordinates of vertex $P$ are $$\boxed{(13, 6)}.$$