Coordinates and geometric problems on axes

(a) $-6$

1. Coordinates are written in the order (x, y).
2. The first coordinate is $-6.$

• B1 Correct answer: $-6$

(a) $(7,-9)$

1. $$2\times 3-(-1)$$
2. $$2\times (-2)-5$$
3. The diagonals of a rectangle bisect each other, so M is the midpoint of AC.
4. C = (2 $\times$ 3 $-$ ($-1$), 2 $\times$ ($-2$) $-$ 5) = (7, $-9$).

• P1 Establishing $2\times 3-(-1)$ or an equivalent valid method.
• P1 Establishing $2\times (-2)-5$ or an equivalent valid method.
• A1 Correct answer: $(7,-9)$

(a) $(1, -1)$

1. $D$ is below $A$ (same $x$) and level with $C$ (same $y$): $(1, -1)$.

• B1 The correct answer, $(1, -1)$.

(b) $(4, 1)$

1. Average the $x$-coordinates and the $y$-coordinates.
   $$\left(\frac{1 + 7}{2}, \frac{3 + (-1)}{2}\right) = (4, 1)$$

• M1 Averaging either pair of coordinates correctly.
• A1 The correct answer, $(4, 1)$.

(c) $24$ square units

1. Width: $7 - 1 = 6$. Height: $3 - (-1) = 4$. Area: $6 \times 4 = 24$.

• M1 Finding both side lengths, 6 and 4.
• A1 The correct answer, $24$.

(a) $(2,-1)$

1. Average the x coordinates and the y coordinates separately.
   $$\frac{-2+6}{2}$$
2. Therefore $(2,-1)$.

• M1 Average the x coordinates and the y coordinates separately.
• A1 Correct answer: $(2,-1)$

(b) $(4,-4)$

1. Double the midpoint coordinates and subtract C.
   $$2\times (-1)-2$$
2. Therefore $(4,-4)$.

• P1 Double the midpoint coordinates and subtract C.
• A1 Correct answer: $(4,-4)$

(a) $(-3,-6)$

1. Use A’s x-coordinate and C’s y-coordinate.
   $$x=-3$$
2. Therefore $(-3,-6)$.

• M1 Use A’s x-coordinate and C’s y-coordinate.
• A1 Correct answer: $(-3,-6)$

(b) $100$

1. Multiply the positive horizontal and vertical differences.
   $$(7-(-3))(4-(-6))$$
2. Therefore $100$.

• M1 Multiply the positive horizontal and vertical differences.
• A1 Correct answer: $100$

(a) $(5.5,5)$

1. Average the x-coordinates and the y-coordinates.
   $$\frac{7+4}{2}$$
2. Therefore $(5.5,5)$.

• M1 Average the x-coordinates and the y-coordinates.
• A1 Correct answer: $(5.5,5)$

(b) $18$ square units

1. Use base AB = 6 and the perpendicular height 6.
   $$6\times 6/2$$
2. Therefore $18$ square units.

• M1 Use base AB = 6 and the perpendicular height 6.
• A1 Correct answer: $18$ square units

(a) $(7, -6)$

1. From $P$ to $M$: $x$ goes up by 5 and $y$ goes down by 5.
2. From $M$ to $Q$ is the same step again: $Q = (2 + 5, -1 - 5) = (7, -6)$.

• M1 Finding the step from $P$ to $M$, $(+5, -5)$, or using $\frac{-3 + x}{2} = 2$.
• A1 The correct answer, $(7, -6)$.

(b) $(-1, 2)$

1. From $P$ to $Q$ the change is $(7 - (-3), -6 - 4) = (10, -10)$.
2. $R$ is $\frac{1}{5}$ of the way along, because the ratio has $1 + 4 = 5$ parts.
3. $$R = (-3 + 2,\ 4 - 2) = (-1, 2)$$

• P1 Finding the change from $P$ to $Q$, $(10, -10)$.
• P1 Taking $\frac{1}{5}$ of the change: $(2, -2)$.
• A1 The correct answer, $(-1, 2)$.

(a) $36$ square units

1. $$7-(-2)$$
2. $$9-1$$
3. $$9\times 8/2$$
4. AB is horizontal and has length 7 $-$ ($-2$) = 9 units.
5. The perpendicular height from C to the line AB is 9 $-$ 1 = 8 units.
6. Area = $\frac{1}{2}$ $\times$ 9 $\times$ 8 = 36 square units.

• P1 Establishing $7-(-2)$ or an equivalent valid method.
• P1 Establishing $9-1$ or an equivalent valid method.
• P1 Establishing $9\times 8/2$ or an equivalent valid method.
• A1 Correct answer: $36$ square units