Find the gradient of $L_1$.

Find the coordinates of point $\text{C}$.

Find the equation of $L_2$. Give your answer in the form $ax+by+d=0$, where $a, b, d\in \mathrm{\mathbb{Z}}$.

Find the value of $k$.

Find the distance between points $\text{M}$ and $\text{N}$.

Given that the length of $\text{AM}$ is $\sqrt{45}$, find the area of triangle $\text{ANC}$.

$\frac{2-\left(-1\right)}{-3-\left(-9\right)}$       (M1)

Note: Award (M1) for correct substitution into the gradient formula.

$=\frac{1}{2} \left(\frac{3}{6}, 0.5\right)$       (A1)(G2)

[2 marks]

$-3=\frac{-9+x}{2} \left(-6+9=x\right)$  and  $2=\frac{-1+y}{2} \left(4+1=y\right)$       (M1)

Note: Award (M1) for correct substitution into the midpoint formula for both coordinates.

OR

       (M1)

Note: Award (M1) for a sketch showing the horizontal displacement from $\text{M}$ to $\text{C}$ is $6$ and the vertical displacement is $3$ and the coordinates at $\text{M}$.

OR

$-3+6=3$  and  $2+3=5$       (M1)

Note: Award (M1) for correct equations seen.

$\left(3, 5\right)$      (A1)(G1)(G1)

Note: Accept $x=3, y=5$. Award at most (M1)(A0) or (G1)(G0) if parentheses are missing.

[2 marks]

gradient of the normal $=-2$      (A1)(ft)

Note: Follow through from their gradient from part (a).

$y-2=-2\left(x+3\right)$  OR  $2=-2\left(-3\right)+c$        (M1)

Note: Award (M1) for correct substitution of $\text{M}$ and their gradient of normal into straight line formula.

$2x+y+4=0$ (accept integer multiples)        (A1)(ft)(G3)

[3 marks]

$2\left(k\right)+4+4=0$      (M1)

Note: Award (M1) for substitution of $y=4$ into their equation of normal line or substitution of $\text{M}$ and $\left(k, 4\right)$ into equation of gradient of normal.

$k=-4$        (A1)(ft)(G2)

Note: Follow through from part (c).

[2 marks]

$\sqrt{{\left(-4+3\right)}^2+{\left(4-2\right)}^2}$      (M1)

Note: Award (M1) for correctly substituting point $\text{M}$ and their $\text{N}$ into distance formula.

$\sqrt{5} \left(2.24, 2.23606\dots \right)$        (A1)(ft)

Note: Follow through from part (d).

[2 marks]

$\frac{1}{2}\times \left(2\times \sqrt{45}\right)\times \sqrt{5}$      (M1)

Note: Award (M1) for their correct substitution into area of a triangle formula. Award (M0) for their $\frac{1}{2}\times \left(\sqrt{45}\right)\times \sqrt{5}$ without any evidence of multiplication by $2$ to find length $\text{AC}$. Accept any other correct method to find the area.

$15$        (A1)(ft)(G2)

Note: Accept $15.02637\dots$ from use of a $3$ sf value for $\sqrt{5}$. Follow through from part (e).

[2 marks]