8 \includegraphics[max width=\textwidth, alt={}, center]{3b527397-7781-41e9-8218-57277cc977bf-3_599_716_1071_717} The diagram shows points \(A , B\) and \(C\) lying on the line \(2 y = x + 4\). The point \(A\) lies on the \(y\)-axis and \(A B = B C\). The line from \(D ( 10 , - 3 )\) to \(B\) is perpendicular to \(A C\). Calculate the coordinates of \(B\) and \(C\).

$m$ of $AC = \frac{1}{2}$ | M1 | Use of $m_1m_2 = -1$
Perpendicular gradient $= -2$ | M1 | Correct method for eqn of line
Eqn $BD$: $y + 3 = -2(x - 10)$ | A1 | In any form
(or $y + 2x = 17$) | | 
| | 
Sim. eqns $BD$ with given eqn | M1 | Correct method of solution
$\rightarrow B(6,5)$ | A1 | co
| | 
Vector move (step) $\rightarrow C(12, 8)$ | M1 A1√ [7] | Any valid method. √ for his $B$

8\\
\includegraphics[max width=\textwidth, alt={}, center]{3b527397-7781-41e9-8218-57277cc977bf-3_599_716_1071_717}

The diagram shows points $A , B$ and $C$ lying on the line $2 y = x + 4$. The point $A$ lies on the $y$-axis and $A B = B C$. The line from $D ( 10 , - 3 )$ to $B$ is perpendicular to $A C$. Calculate the coordinates of $B$ and $C$.

\hfill \mbox{\textit{CAIE P1 2009 Q8 [7]}}