Cartesian plane — axes, quadrants, ordered pairs: Exercises

Before we start, three words we will lean on the whole page — each defined the moment it is used:

An ordered pair $(x, y)$ = an address: $x$ says how far right(+)/left(−), $y$ says how far up(+)/down(−). Order matters.
A quadrant = one of the four regions the two axes cut the plane into. A point is in a quadrant only if BOTH its numbers are non-zero.
The distance between two points = the straight-line length between them, which we will get from a right triangle (see below).

Figure — Cartesian plane — axes, quadrants, ordered pairs

The picture above is the map we return to all page long: the four quadrants, their sign patterns, and the origin $O (0, 0)$ in the middle. Notice the numbering goes counterclockwise (anti-clockwise) starting top-right — that is the one accent (red) arrow.

Level 1 — Recognition

Recall Solution

The rule: look at the sign of each number. Both non-zero → a quadrant. Any zero → an axis.

Point	$x$ -sign	$y$ -sign	Where
$A (4, 7)$	$+$	$+$	Quadrant I
$B (- 3, 2)$	$-$	$+$	Quadrant II
$C (- 5, - 1)$	$-$	$-$	Quadrant III
$D (6, - 8)$	$+$	$-$	Quadrant IV
$E (0, - 4)$	$0$	$-$	On the negative $y$ -axis (no quadrant)
$F (9, 0)$	$+$	$0$	On the positive $x$ -axis (no quadrant)

WHY $E$ and $F$ get no quadrant: a quadrant is a region, and $E, F$ sit exactly on a border (a coordinate is $0$ ).

Recall Solution

Left = negative $x$ → $x = - 7$ . Up = positive $y$ → $y = + 2$ . $P = (- 7, 2)$ WHAT IT LOOKS LIKE: from $O$ , slide left along the $x$ -axis to $- 7$ , then rise $2$ . That lands in Quadrant II.

Level 2 — Application

Recall Solution

WHY a right triangle? Two points don't come with a length attached — but if we drop a horizontal and a vertical leg between them, we get a right angle, and Pythagoras hands us the hypotenuse (the straight distance).

Horizontal leg $Δ x = 4 - 1 = 3$ . Vertical leg $Δ y = 6 - 2 = 4$ . $d = (Δ x)^{2} + (Δ y)^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 = 5.$ WHAT IT LOOKS LIKE: the classic $3$ – $4$ – $5$ right triangle (figure below). See Distance formula.

Recall Solution

$Δ x = 3 - (- 2) = 5$ . $Δ y = - 13 - (- 1) = - 12$ . WHY the squaring saves us: $Δ y$ came out negative, but $(- 12)^{2} = 144$ — squaring erases sign, so distance is always $\geq 0$ , as a length must be. $d = 5^{2} + (- 12)^{2} = 25 + 144 = 169 = 13.$

Level 3 — Analysis

Recall Solution

WHY set $y = 0$ ? Every point on the $x$ -axis has the form $(a, 0)$ — its $y$ -coordinate is exactly $0$ . That is the defining condition. $2 k - 8 = 0 \Rightarrow k = 4.$ Then $x = k + 3 = 7$ , so $R = (7, 0)$ . (It sits on the positive $x$ -axis.)

Recall Solution

Quadrant II means $a < 0$ and $b > 0$ .

$(- a, - b)$ : since $a < 0$ , $- a > 0$ ; since $b > 0$ , $- b < 0$ . So $(+, -)$ = Quadrant IV. (Negating both flips a point through the origin — QII $\to$ QIV.)
$(b, a)$ : here $b > 0$ and $a < 0$ , so $(+, -)$ = Quadrant IV as well. (Swapping coordinates reflects across the line $y = x$ .) WHAT IT LOOKS LIKE: $(- a, - b)$ is the point spun $18 0^{\circ}$ about $O$ ; $(b, a)$ is the mirror image across the diagonal.

Level 4 — Synthesis

Recall Solution

Step 1 — side lengths. $A B = (3 - (- 2))^{2} + (1 - 1)^{2} = 25 + 0 = 5 (horizontal, same y) .$ $B C = (3 - 3)^{2} + (5 - 1)^{2} = 0 + 16 = 4 (vertical, same x) .$ $A C = (3 - (- 2))^{2} + (5 - 1)^{2} = 25 + 16 = 41 .$ Step 2 — Pythagoras check. $A B^{2} + B C^{2} = 25 + 16 = 41 = A C^{2} .$ ✓ Right angle at $B$ . WHY at $B$ : $A B$ is horizontal and $B C$ is vertical, and they share vertex $B$ — a horizontal and a vertical leg meet at $9 0^{\circ}$ . Step 3 — area. With the two legs as base and height: $Area = \frac{1}{2} \times A B \times B C = \frac{1}{2} \times 5 \times 4 = 10 sq units .$ See Area of triangle.

Recall Solution

WHY compute all three sides? "Isosceles" means at least two sides equal — so we measure all three and compare. $P Q = 4^{2} + 3^{2} = 25 = 5.$ $P R = 3^{2} + (- 4)^{2} = 25 = 5.$ $QR = (3 - 4)^{2} + (- 4 - 3)^{2} = 1 + 49 = 50 .$ $P Q = P R = 5$ while $QR = 50 \approx 7.07$ . Two sides equal → isosceles. ✓

Level 5 — Mastery

Recall Solution

Step 1 — name the unknown point. Any point on the $y$ -axis has $x = 0$ , so call it $M (0, y)$ . That single unknown $y$ is what we solve for. Step 2 — set the two distances equal. "Equidistant" = same distance to both. To avoid square roots, set the squared distances equal (equal distances $\Leftrightarrow$ equal squared distances, since both are $\geq 0$ ): $M A^{2} = M B^{2}$ $(0 - 6)^{2} + (y - 5)^{2} = (0 - (- 4))^{2} + (y - 3)^{2}$ $36 + y^{2} - 10 y + 25 = 16 + y^{2} - 6 y + 9.$ Step 3 — solve. The $y^{2}$ cancels: $61 - 10 y = 25 - 6 y \Rightarrow 36 = 4 y \Rightarrow y = 9.$ Answer: $M = (0, 9)$ . Check: $M A^{2} = 36 + (9 - 5)^{2} = 36 + 16 = 52$ ; $M B^{2} = 16 + (9 - 3)^{2} = 16 + 36 = 52$ . ✓ Equal. Related idea: Distance formula.

Recall Solution

WHY distances? Three points are collinear exactly when the two shorter gaps add up to the longest gap — i.e. the "detour" through the middle point is no detour at all. $A B = (3 - 1)^{2} + (8 - 4)^{2} = 4 + 16 = 20 = 25 .$ $B C = (5 - 3)^{2} + (12 - 8)^{2} = 4 + 16 = 20 = 25 .$ $A C = (5 - 1)^{2} + (12 - 4)^{2} = 16 + 64 = 80 = 45 .$ Since $A B + B C = 25 + 25 = 45 = A C$ , the three points are collinear, with $B$ between $A$ and $C$ . ✓ See Colinearity.

Recall Solution

Step 1 — see the shape. $P Q$ is horizontal (same $y = 1$ ), length $∣5 - 1∣ = 4$ . $QR$ is vertical (same $x = 5$ ), length $∣4 - 1∣ = 3$ . So $P Q ⊥ QR$ — a proper corner at $Q$ . Step 2 — find $S$ . In a rectangle, $S$ sits opposite $Q$ . Going from $R$ we must move left by $4$ (mirror of $P \to Q$ ) keeping height, landing at $S = (5 - 4, 4) = (1, 4)$ . Check: $P S$ is vertical ( $x = 1$ ), length $3$ ✓, and $S R$ is horizontal ( $y = 4$ ), length $4$ ✓. $S = (1, 4) .$ Step 3 — area. $Area = 4 \times 3 = 12 sq units .$

Recall Quick self-test (cloze)

A point with $x = 0$ lies on the == $y$ -axis==. The distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is ::: $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ Three points are collinear when the two shorter distances sum to ::: the longest distance. Quadrant IV has sign pattern ::: $(+, -)$

Where this leads: Distance formula, Midpoint formula, Section formula, Slope of a line, Area of triangle, Colinearity, and eventually Polar coordinates, the Complex plane and Vectors.