The line $3x - 6y + 2 = 0$ contains the point $\left( k, \frac{2k+2}{3} \right)$ where $k \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2021

For the point $P(s, t)$ to be on the line $x - 2y - 8 = 0$, we substitute:

$s - 2t - 8 = 0 \Rightarrow s = 2t + 8.$

Next, we find the distance from point $P(s, t)$ to the line $4x + 3y + 6 = 0$:

The distance $D$ from point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by:

$D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.$

Here, $A = 4$, $B = 3$, $C = 6$, and $P(s, t) = (s, t)$:

$D = \frac{|4s + 3t + 6|}{\sqrt{4^2 + 3^2}} = \frac{|4s + 3t + 6|}{5}.$

Setting this equal to $1$:

$\frac{|4s + 3t + 6|}{5} = 1 \Rightarrow |4s + 3t + 6| = 5.$

Therefore, we have two cases:

1. $4s + 3t + 6 = 5$
2. $4s + 3t + 6 = -5$

Solving these equations will yield values for $s$ and $t$.

To find $|AD|$, we first determine the coordinates of point $D$. Let the coordinates of $D$ be $(x_D, y_D)$. Given the ratio $|AD| : |DC| = 2:1$, we can express this in terms of the coordinates of points $A$ and $C$:

$D = \left(\frac{1}{3}(2 * 16 + 1 * 4), \frac{1}{3}(2 * 11 + 1 * 2)\right) = \left(\frac{32 + 4}{3}, \frac{22 + 2}{3}\right) = \left(\frac{36}{3}, \frac{24}{3}\right) = (12, 8).$

Now finding the distance $|AD|$:

$|AD| = \sqrt{(12 - 4)^2 + (8 - 2)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10.$

Since $|AB| = 33$ and both segments are horizontal, the $y$-coordinate of both points will be constant and equal to that of point $A$: $y_A = 2$. Thus if we let the coordinates of $B$ be $(x_B, 2)$, we know:

$|AB| = |x_B - 4| = 33 \Rightarrow x_B = 4 + 33 = 37 \text{ or } x_B = 4 - 33 = -29.$

So $B$ can be either $(37, 2)$ or $(-29, 2)$.

To find $E$, we note that $D$ lies on $[AC]$, which leads us to $E$ lying on $[CB]$. Since $D$ is at $(12, 8)$ end it needs to have the same y-coordinate as $C$:

$E = (x_E, 11).$

Now we can use the same approach as for $D$ and find the relationship on segment $CB$ to achieve the value needed.