Standard +0.3 This is a straightforward coordinate geometry question requiring midpoint formula, perpendicular bisector properties, and basic rhombus properties. All steps are routine: find M using midpoint formula, use perpendicularity to find the line through M and A, then solve for A on x-axis. Finding C uses the property that diagonals bisect each other. Slightly easier than average due to standard techniques and clear structure.
5
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The diagram shows a rhombus \(A B C D\). The points \(B\) and \(D\) have coordinates \(( 2,10 )\) and \(( 6,2 )\) respectively, and \(A\) lies on the \(x\)-axis. The mid-point of \(B D\) is \(M\). Find, by calculation, the coordinates of each of \(M , A\) and \(C\).
Correct method leading to A - the equation may not be seen - \(y = 0\) may be used with gradient.
Eqn of \(AC\): \(y - 6 = \frac{1}{2}(x - 4)\)
M1
\(\Rightarrow x = -8\) when \(y = 0\), \(A(-8, 0)\)
A1
\(\Rightarrow C = (16, 12)\) by vector move etc.
M1 A1 [6]
Any valid method - vectors, midpoint backwards, or solution of 2 sim eqns.
$M(4, 6)$ | B1 | CAO
$m$ of $BD = -2$ | M1 | Use of $m_1 m_2 = -1$
$M$ of $AC = \frac{1}{2}$ | M1 | Correct method leading to A - the equation may not be seen - $y = 0$ may be used with gradient.
Eqn of $AC$: $y - 6 = \frac{1}{2}(x - 4)$ | M1 |
$\Rightarrow x = -8$ when $y = 0$, $A(-8, 0)$ | A1 |
$\Rightarrow C = (16, 12)$ by vector move etc. | M1 A1 [6] | Any valid method - vectors, midpoint backwards, or solution of 2 sim eqns.
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The diagram shows a rhombus $A B C D$. The points $B$ and $D$ have coordinates $( 2,10 )$ and $( 6,2 )$ respectively, and $A$ lies on the $x$-axis. The mid-point of $B D$ is $M$. Find, by calculation, the coordinates of each of $M , A$ and $C$.
\hfill \mbox{\textit{CAIE P1 2005 Q5 [6]}}