Orienting Yourself: The Use of Coordinates

The entrance starts at D₁, eight grid units to the right of the vertical axis.

Its entire segment sits on the horizontal axis; there is no vertical separation.

Read horizontal position first, then vertical position.

D₁ is at horizontal position 8 and vertical position 0.

The endpoints have the same y-coordinate, so subtract their x-coordinates.

11.5 − 8 = 3.5. The floor-plan scale uses feet.

This width gives generous space in the sketch. Actual wheelchair access also depends on clear opening, turning space, thresholds and the chair; measure these before deciding.

Bathroom opening: 4 − 1.5 = 2.5 ft.

Room opening: 3.5 ft. The difference is 3.5 − 2.5 = 1 ft.

The two vertical edges must use x = 8 and x = 11.

The two horizontal edges must use y = 7 and y = 9. Combine the unused pair.

In the given plan, the table occupies x from 8 to 11 and y from 7 to 9.

This keeps it within the bedroom and away from the door segments. Consider the walking route and usable space too.

Horizontal span: 11 − 8 = 3 ft.

Vertical span: 9 − 7 = 2 ft.

A floor plan gives two dimensions and contains no vertical-height measurement.

The hinge is (0, 1.5); the door length is 4 − 1.5 = 2.5 ft.

The nearest wardrobe edge is at x = 3. Its closest point to the hinge is (3, 1.5), 3 ft away.

The door sweeps a radius of 2.5 ft, leaving 0.5 ft before that edge. A width of 3 ft would reach it.

For a wider door, move the wardrobe or change the opening direction, checking the space on the other side.

The bathroom spans x = −6 to 0 and y = 0 to 9.

Combine these boundary values, using the labels from the textbook.

Read S(−6, 6), H(−3, 6), W(−2, 9) and R(−6, 9).

SH and WR are horizontal, hence parallel. SR is vertical while HW is slanted, so the other pair is not parallel.

One possible basin rectangle uses x = −6 to −3 and y = 0 to 2: dimensions 3 ft by 2 ft.

One possible toilet rectangle uses x = −6 to −4 and y = 2 to 5: dimensions 2 ft by 3 ft.

Their interiors do not overlap and both lie below the shower. These are example allocations, not the only answer.

The top edge runs from P(−6, 0) to A(12, 0), giving 18 ft.

Place the 15 ft depth below this edge, at y = −15.

Room centre: ((−6 + 12)/2, (0 − 15)/2) = (3, −7.5).

With the 5 ft side horizontal, half-length is 2.5 and half-width is 1.5.

Use x = 3 ± 2.5 and y = −7.5 ± 1.5. A rotated placement is another valid choice.

The horizontal axis has y = 0 and the vertical axis has x = 0.

At their shared point both conditions hold.

Moving parallel to the y-axis changes only y, leaving x fixed at −5.

Positive y places the point in Quadrant II; negative y in Quadrant III. At y = 0 it lies on an axis.

A and M share y = −2, so AM is horizontal.

M and P share x = −5, so MP is vertical: AM and MP meet at a right angle.

M and P have equal x-values and opposite y-values; reflection across the x-axis exchanges them.

Choose I(5, 0) and N(0, −6), with Z(5, −6).

IZ is vertical with length 6, and ZN is horizontal with length 5.

IN² = 6² + 5² = 61. The right angle is at Z.

With a fixed origin and only nonnegative coordinates, x cannot describe locations to the left and y cannot describe locations below.

Only the first quadrant, the nonnegative parts of the axes and the origin remain available.

The vector from A to M is (−3, −4), while the vector from A to G is (6, 8).

(6, 8) = −2(−3, −4). The vectors point along the same line, in opposite directions.

Distance check: MA = 5, AG = 10 and MG = 15; MA + AG = MG.

Use the official textbook value C(4, −12).

RB has vector (3, −4); BC has vector (6, −7). Twice (3, −4) is (6, −8), not (6, −7).

Equivalently, the determinant is 3 × (−7) − (−4) × 6 = 3, which is nonzero.

The distance method also gives RB = 5, BC = √85 and RC = √202; no distance is the sum of the other two.

For (i), choose O(0, 0), A(4, 0), B(0, 4). OA and OB are perpendicular and both have length 4.

For (ii), choose O(0, 0), C(−4, −3), D(4, −3). C lies in Quadrant III and D in Quadrant IV.

OC = OD = √(16 + 9) = 5. Other choices with equal side lengths are also valid.

A midpoint averages the x-coordinates and independently averages the y-coordinates.

Row 1: ((−3 + 3)/2, 0) = (0, 0), matching M: yes.

Row 2: ((2 + 4)/2, (3 + 5)/2) = (3, 4), matching M: yes.

Row 3: (0, (0 − 10)/2) = (0, −5), not (0, 5): no.

Row 4: ((−8 + 6)/2, (7 − 3)/2) = (−1, 2), not (0, −2): no.

Since M is the midpoint, B = 2M − A, coordinate by coordinate.

x = 2(−7) − 3 = −17; y = 2(1) − (−4) = 6.

Check: ((3 − 17)/2, (−4 + 6)/2) = (−7, 1).

For general endpoints A and B, P = A + (B − A)/3 and Q = A + 2(B − A)/3.

Here B − A = (12, −9), so each third-step is (4, −3).

Starting at (4, 7), one step gives (8, 4); two steps give (12, 1).

Midpoint connection: P is the midpoint of AQ and Q is the midpoint of PB. These equations give the same one-third and two-thirds formulas.

Squared distances from O: A gives 1² + (−8)² = 65; B gives (−4)² + 7² = 65; C gives (−7)² + (−4)² = 65.

All three therefore lie on the circle of radius √65.

D gives (−5)² + 6² = 61 < 65: inside. E gives 0² + 9² = 81 > 65: outside.

Choose the convention D is the midpoint of BC, E of CA, and F of AB. The question does not assign the sides, so another convention may permute the vertex labels.

A = E + F − D = (6 + 0 − 5, 5 + 3 − 1) = (1, 7).

B = F + D − E = (0 + 5 − 6, 3 + 1 − 5) = (−1, −1).

C = D + E − F = (5 + 6 − 0, 1 + 5 − 3) = (11, 3).

Checking the three averages reproduces D(5, 1), E(6, 5) and F(0, 3).

Draw the two main roads perpendicular, then parallel streets at 1 cm intervals to represent 200 m.

Number the north–south streets and east–west streets consistently with the question; the ordered pair identifies one of each.

One vertical line and one horizontal line cross exactly once. Reversing the pair selects a different crossing.

Circle A: x ranges from 100 − 80 = 20 to 180; y from 150 − 80 = 70 to 230.

Circle B: x ranges from 250 − 100 = 150 to 350; y from 230 − 100 = 130 to 330. Both fit within 0 ≤ x ≤ 800 and 0 ≤ y ≤ 600.

Centre distance = √((250 − 100)² + (230 − 150)²) = √28900 = 170.

The radii differ by 20 and sum to 180. Since 20 < 170 < 180, the circumferences cross at two points.

Successive side vectors are AB(−3, 1), BC(−1, −3), CD(3, −1), DA(1, 3). Each squared length is 10.

Adjacent vectors AB and BC have dot product (−3)(−1) + (1)(−3) = 0, so the angle is 90°.

Equal sides and a right angle establish a square. Area is side² = 10.

Measure several actual clear openings at home and school, and record the units. Door widths vary; the task asks for observations.

For wheelchair access, consider the chair’s width, clear opening with the door open, approach and turning space, and any threshold. Ask the wheelchair user what works for them.

A point on the y-axis needs no horizontal movement from it.

A point on the x-axis needs no vertical movement from it.

Ordered pairs match only when their first entries match and their second entries match.

For (x, y) and (y, x), both requirements reduce to x = y.

For example, (2, 2) remains unchanged, but (2, 5) and (5, 2) locate different points.

From (3, 4) to (7, 1), horizontal displacement is +4 and vertical displacement is −3.

Their magnitudes are 4 and 3. A right triangle with these legs has hypotenuse √(4² + 3²) = 5.

Across the y-axis, (x, y) becomes (−x, y); lengths and angles stay the same.

Across the x-axis, (x, y) becomes (x, −y); lengths and angles again stay the same.

For A(3, 4), M(9, 6), D(7, 1), the y-axis images are (−3, 4), (−9, 6), (−7, 1); the x-axis images are (3, −4), (9, −6), (7, −1).