13 The points $$ A(3,5), B(-5,9) $$ and $$ C(k, k+2) $$ are the vertices of a triangle $$ A B C $$. (i) Given that the area of the triangle $$ A B C $$ is 18 units $$ { }^{2} $$, find the possible values of $$ k $$. (ii) Hence find the height of triangle $$ A B C $$ for which the base is $$ A B $$.

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1. The area of triangle $$ ABC $$ with vertices $$ A(3,5) $$, $$ B(-5,9) $$, and $$ C(k, k+2) $$ is given by  
   $$
   \text{Area} = \frac{1}{2} \left| x_A (y_B-y_C) + x_B (y_C-y_A) + x_C (y_A-y_B) \right|
   $$

2. Substituting the coordinates:
   $$
   \begin{aligned}
   \text{Area} &= \frac{1}{2} \left| 3(9 - (k+2)) + (-5)((k+2)-5) + k(5-9) \right| \
   &= \frac{1}{2} \left| 3(7-k) -5(k-3) -4k \right|
   \end{aligned}
   $$

3. Simplify the expression:
   $$
   \begin{aligned}
   3(7-k) &= 21 - 3k, \
   -5(k-3) &= -5k + 15, \
   k(5-9) &= -4k.
   \end{aligned}
   $$
   Adding these yields:
   $$
   21 - 3k - 5k + 15 - 4k = 36 - 12k.
   $$
   Thus, the area becomes:
   $$
   \text{Area} = \frac{1}{2} \left| 36 - 12k \right|
   $$

4. We are given that the area is 18 square units:
   $$
   \frac{1}{2} \left| 36 - 12k \right| = 18.
   $$
   Multiply both sides by 2:
   $$
   \left| 36 - 12k \right| = 36.
   $$

5. Solve the absolute value equation:
   - First case:
     $$
     36 - 12k = 36 \quad \Rightarrow \quad -12k = 0 \quad \Rightarrow \quad k = 0.
     $$
   - Second case:
     $$
     36 - 12k = -36 \quad \Rightarrow \quad -12k = -72 \quad \Rightarrow \quad k = 6.
     $$

6. Thus, the possible values of $$ k $$ are $$ k = 0 $$ and $$ k = 6 $$.

7. The length of the base $$ AB $$ between $$ A(3,5) $$ and $$ B(-5,9) $$ is given by:
   $$
   AB = \sqrt{(-5-3)^2 + (9-5)^2} = \sqrt{(-8)^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}.
   $$

8. The area of a triangle can also be written as:
   $$
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.
   $$
   With $$ \text{Area} = 18 $$ and $$ \text{base} = 4\sqrt{5} $$:
   $$
   18 = \frac{1}{2} \times 4\sqrt{5} \times \text{height}.
   $$
   Solve for the height:
   $$
   18 = 2\sqrt{5} \times \text{height} \quad \Rightarrow \quad \text{height} = \frac{18}{2\sqrt{5}} = \frac{9}{\sqrt{5}}.
   $$

9. Rationalizing the denominator:
   $$
   \text{height} = \frac{9\sqrt{5}}{5}.
   $$

Thus, the answers are:
- (i) $$ k = 0 $$ or $$ k = 6 $$.
- (ii) The height of the triangle, with base $$ AB $$, is $$ \frac{9\sqrt{5}}{5} $$ units (or equivalently $$ \frac{9}{\sqrt{5}} $$ units).
The possible values of $$ k $$ are 0 and 6. The height of triangle $$ ABC $$ with base $$ AB $$ is $$ \frac{9\sqrt{5}}{5} $$ units.

13 The points \( A(3,5), B(-5,9) \) and \( C(k, k+2) \) are the vertices of a triangle \( A B C \). (i) Given that the area of the triangle \( A B C \) is 18 units \( { }^{2} \), find the possible values of \( k \). (ii) Hence find the height of triangle \( A B C \) for which the base is \( A B \).

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