Coordinates & the Cartesian plane

Fluency

Quadrant 1
Quadrant 2
Quadrant 3
Quadrant 4
On the $y$ -axis (not in a quadrant)
On the $x$ -axis
$A$ in Q1; $B$ in Q2; $C$ in Q3; $D$ in Q4. (Check positions on a plotted grid.)
$(5, -2)$
$(-3, 0)$
$(0, 0)$
$(4, -4)$
Quadrant 2 (the point becomes $(-2, 3)$ )
Quadrant 3 (the point becomes $(-4, -5)$ )
Any point of the form $(0, n)$ with $n < 0$ , e.g. $(0, -3)$
Any point with both coordinates negative, e.g. $(-1, -1)$

Reasoning

$y = 2x$ . Each $y$ -value is double its $x$ -value; the points lie on the line $y = 2x$ .
They all lie on the $x$ -axis (every $y$ -coordinate is $0$ ).
$6$ units $^2$ . Method: right-angled triangle with legs $4$ and $3$ ; area $= \dfrac{1}{2}(4)(3)$ .
Perimeter $16$ units; area $15$ units $^2$ . Method: length $= 6 - 1 = 5$ ; height $= 4 - 1 = 3$ ; $P = 2(5 + 3) = 16$ ; $A = 5 \times 3 = 15$ .
$y$ -values: $-5,\ -3,\ -1,\ 1,\ 3$ .
Yes. When $x = 3$ , $y = 2(3) - 1 = 5$ , matching.
The $y$ -coordinate changes sign.
$(-2, 1)$ .

Reasoning

Ravi treated the $x$ -coordinate as positive. For $(-2, 3)$ you move $2$ units left (because $x$ is negative), then $3$ up. The point belongs in quadrant 2, not quadrant 1.
The four quadrants are the open regions between the axes - they exclude the axes themselves. Since $(0, 5)$ has $x = 0$ , it lies on the $y$ -axis, not inside any quadrant.
Almost. If both coordinates are strictly positive, the point is in quadrant 1. But if one of them is $0$ (e.g. $(3, 0)$ or $(0, 3)$ ), the point sits on an axis, not in the quadrant. So the correct statement is “every point with strictly positive coordinates is in quadrant 1”.
Any points where the two coordinates are equal, e.g. $(0, 0)$ , $(1, 1)$ , $(-2, -2)$ .

Problem solving

$300$ m east, $200$ m north; straight-line distance $\approx 360.6$ m. Method: Pythagoras $\sqrt{300^2 + 200^2}$ .
$(2, 3)$ ; distance $\sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.6$ units. Method: $4 - 2 = 2$ east; $3$ north.
Fourth vertex $(1, 5)$ ; perimeter $20$ units. Method: width $6$ (from $1$ to $7$ ), height $4$ (from $1$ to $5$ ); $P = 2(6 + 4)$ .
Centre at $(4, 2.5)$ . Method: average opposite corners, e.g. $(0+8)/2, (0+5)/2$ .
Midpoint $(4, 5)$ .

Tier 1: basic skills