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Step 1: Use the Euler line relationship to find the coordinates of A(a, b)
In any triangle, the centroid (G), circumcentre (O), and orthocentre (H) lie on a common line called the Euler line. They satisfy the relation
$G = \dfrac{H + 2\,O}{3}.$
Here,
• $A(a, b)$ is the centroid (G),
• $B(3, 4)$ is the circumcentre (O),
• $C(-6, -8)$ is the orthocentre (H).
Substituting into $A = \dfrac{C + 2B}{3}$ gives:
$
A(a, b)
= \dfrac{(-6, -8) + 2(3, 4)}{3}
= \dfrac{(-6, -8) + (6, 8)}{3}
= \dfrac{(0, 0)}{3}
= (0, 0).
$
So $a = 0$ and $b = 0.$
Step 2: Determine the coordinates of P
The point $P$ is given by $P(2a + 3,\, 7b + 5)$. Substituting $a = 0$ and $b = 0$:
$
P(2(0) + 3,\, 7(0) + 5) = (3,\, 5).
$
Step 3: Identify the direction parallel to the line $x - 2y - 1 = 0$
The line $x - 2y - 1 = 0$ can be written as
$
x = 2y + 1.
$
A suitable direction vector parallel to this line is
$
\vec{v} = (2,\, 1).
$
Step 4: Find the intersection of the line through P parallel to $\vec{v}$ with $2x + 3y - 4 = 0$
Let $Q$ be the point on $2x + 3y - 4 = 0$ that lies on a line starting at $P(3, 5)$ and moving in the direction $\vec{v} = (2,\, 1)$. We can write
$
Q = P + \lambda\,\vec{v} = (3 + 2\lambda,\, 5 + \lambda).
$
Since $Q$ lies on $2x + 3y - 4 = 0$, we substitute these coordinates into that line equation:
\[
2(3 + 2\lambda) + 3(5 + \lambda) - 4 = 0.
\]
Simplify:
\[
6 + 4\lambda + 15 + 3\lambda - 4 = 0
\quad\Longrightarrow\quad
17 + 7\lambda = 0
\quad\Longrightarrow\quad
\lambda = -\frac{17}{7}.
\]
Therefore:
\[
Q = \Bigl(3 + 2\bigl(-\frac{17}{7}\bigr),\, 5 + \bigl(-\frac{17}{7}\bigr)\Bigr)
= \Bigl(3 - \frac{34}{7},\, 5 - \frac{17}{7}\Bigr)
= \Bigl(\frac{21 - 34}{7},\, \frac{35 - 17}{7}\Bigr)
= \Bigl(-\frac{13}{7},\, \frac{18}{7}\Bigr).
\]
Step 5: Calculate the distance from P to Q along $\vec{v}$
The vector from $P$ to $Q$ is
\[
\overrightarrow{PQ}
= Q - P
= \Bigl(-\frac{13}{7} - 3,\; \frac{18}{7} - 5\Bigr)
= \Bigl(-\frac{13}{7} - \frac{21}{7},\; \frac{18}{7} - \frac{35}{7}\Bigr)
= \Bigl(-\frac{34}{7},\, -\frac{17}{7}\Bigr)
= -\frac{17}{7}\,(2,\, 1).
\]
The magnitude of this vector is
\[
\|\overrightarrow{PQ}\|
= \left|\, -\frac{17}{7}\right|\sqrt{2^2 + 1^2}
= \frac{17}{7}\sqrt{5}.
\]
Step 6: State the final answer
Therefore, the distance of the point $P(3, 5)$ from the line $2x + 3y - 4 = 0$, measured parallel to the line $x - 2y - 1 = 0$, is
$
\frac{17 \sqrt{5}}{7}.
$
