The projection of the vector \(egin{pmatrix} 6 \\ 7 \end{pmatrix}\) onto the line \(y = 2x\) is \(egin{pmatrix} 4 \\ 8 \end{pmatrix}\) - HSC - SSCE Mathematics Extension 1 - Question 9 - 2020 - Paper 1

To find the reflection of the point (6, 7) across the line (y = 2x), we first need the intersection of the line and the perpendicular from the point to the line. The slope of the line is 2, so the perpendicular slope will be (-\frac{1}{2}).

The equation of the line through (6, 7) with slope (-\frac{1}{2}) is:

$y - 7 = -\frac{1}{2}(x - 6)$

The line equation simplifies to:

$y = -\frac{1}{2}x + 10$.

To find the intersection, we set the equations equal to each other:

$2x = -\frac{1}{2}x + 10 \Rightarrow \frac{5}{2}x = 10 \Rightarrow x = 4\,\, y = 2(4) = 8$.

Now we have the intersection point (4, 8). The reflection point A can be found by using:

$A = 2 \cdot \text{Intersection Point} - (6, 7)$

Calculating:

$A = 2 \begin{pmatrix} 4 \\ 8 \end{pmatrix} - \begin{pmatrix} 6 \\ 7 \end{pmatrix} = \begin{pmatrix} 8 \\ 16 \end{pmatrix} - \begin{pmatrix} 6 \\ 7 \end{pmatrix} = \begin{pmatrix} 2 \\ 9 \end{pmatrix}$.

Thus, the position vector of point A is (\begin{pmatrix} 2 \ 9 \end{pmatrix}).