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

To find the reflected point A of (6, 7) in the line ( y = 2x ), we first determine the slope of the line, which is 2. The line's equation can be rewritten in the standard form as ( ax + by + c = 0 ), where ( a = 2, b = -1, c = 0 ).

Next, we need to find the foot of the perpendicular from the point (6, 7) to the line. The slope of the perpendicular line will be ( -\frac{1}{2} ). Thus, the equation of the line through the point ( (6, 7) ) is given by: [ y - 7 = -\frac{1}{2}(x - 6) ]

Solving for x and y, we substitute into the line's equation: [ y = -\frac{1}{2}x + 10 ]

Now, we find the intersection of ( y = -\frac{1}{2}x + 10 ) and ( y = 2x ): [ -\frac{1}{2}x + 10 = 2x ] [ 10 = \frac{5}{2}x ] [ x = 4 ]

Substituting back to find y: [ y = 2(4) = 8 ]

Thus, the foot of the perpendicular is at (4, 8). Now, the coordinates of point A can be found using the symmetry: [ A_x = 6 + (6 - 4) = 8 ] [ A_y = 7 + (8 - 7) = 8 ]

So, A is at (8, 6). The position vector of point A is therefore ( \begin{pmatrix} 8 \ 6 \end{pmatrix} ).