Textbook Exercises & Solutions

Given:

The opening runs along y = 0 from x = 8 to x = 11.5.

To Find:

Its separation from the walls represented by the axes.

Property:

Distance to y-axis is |x|; distance to x-axis is |y|.

Solution:

Every point of the opening lies on the x-axis, so its distance from that axis is 0.

Its nearest endpoint to the y-axis is $D_1$ at x = 8.

Results:

The opening starts 8 ft from the left wall and lies on the x-axis. Distances of individual points from the y-axis range from 8 to 11.5 ft.

Observation:

The horizontal value is 8; the vertical value is 0.

Results:

$D_1$$\left(8,0\right)$.

Given:

Both endpoints share y = 0.

To Find:

The width.

Formula:

Horizontal length is |$x_2$ − $x_1$|.

Solution:

Width = |11.5 − 8| = 3.5 ft.

This is the geometric opening. Actual ease of access also depends on clear space, door swing and the user’s needs; the sketch alone cannot guarantee it.

Results:

The model gives a 3.5 ft opening.

Given:

Bathroom endpoints have the same x-coordinate.

To Find:

Which opening is narrower.

Formula:

Vertical length is |$y_2$ − $y_1$|.

Solution:

Bathroom width = |4 − 1.5| = 2.5 ft.

3.5 − 2.5 = 1 ft.

Results:

The bathroom opening is narrower by 1 ft.

Given:

An axis-aligned rectangular footprint.

To Find:

Fourth foot, dimensions and what the plan tells us.

Property:

Opposite sides of a rectangle share coordinate levels.

Solution:

The missing corner combines x = 8 with y = 7: $\left(8,7\right)$.

Horizontal length = 11 − 8 = 3 ft; vertical width = 9 − 7 = 2 ft.

The footprint fits in the shown room, to the right of the bed. Practical suitability still needs access and clearances to be checked.

A floor plan does not give the table’s height.

Results:

Fourth foot $\left(8,7\right)$; footprint 3 ft × 2 ft; height cannot be inferred.

Given:

An inward-opening door pivots about $\left(0,1.5\right)$.

To Find:

Whether its sweep reaches the wardrobe.

Property:

Every position of the outer end is 2.5 ft from the hinge.

Solution:

The nearest wardrobe point is $\left(3,1.5\right)$, 3 ft from the hinge.

The existing 2.5 ft leaf cannot reach it.

A leaf widened to 3 ft can touch it; a larger leaf can intersect it during the swing. Consider relocating the wardrobe or changing the door arrangement.

Results:

No collision in this model for the existing door. Widening requires a fresh clearance check.

Observation:

Right boundary x = 0; left boundary x = −6. Bottom y = 0; top y = 9.

Results:

O$\left(0,0\right)$, F$\left(0,9\right)$, R$\left(-6,9\right)$, P$\left(-6,0\right)$.

Given:

The four boundary points in the given order.

To Find:

The shape.

Property:

Equal y-values give horizontal segments.

Solution:

SH and WR are horizontal and parallel.

SR is vertical; HW is oblique, so this pair is not parallel.

Results:

SHWR is a trapezium with one pair of opposite sides parallel.

Given:

Bathroom: −6 ≤ x ≤ 0, 0 ≤ y ≤ 9; shower is above y = 6.

To Find:

One possible placement.

Property:

Coordinate differences determine rectangle dimensions.

Solution:

Washbasin: $\left(-5,0\right)$, $\left(-2,0\right)$, $\left(-2,2\right)$, $\left(-5,2\right)$. Width 3, height 2.

Toilet: $\left(-5,2\right)$, $\left(-3,2\right)$, $\left(-3,5\right)$, $\left(-5,5\right)$. Width 2, height 3.

Their interiors do not overlap and both avoid the shower. Actual fixtures need practical access clearances.

Results:

This is one mathematical placement, not a unique plan.

Given:

PA is the 18 ft top boundary of the adjoining room.

To Find:

The four corners.

Property:

A horizontal 18 ft span and a downward 15 ft span define the rectangle.

Solution:

Keep x-levels −6 and 12.

The other y-level is −15.

Results:

Corners: $\left(-6,0\right)$, $\left(12,0\right)$, $\left(12,-15\right)$, $\left(-6,-15\right)$.

Given:

The room and the table dimensions.

To Find:

One set of table-foot coordinates.

Property:

A centred axis-aligned rectangle shares its diagonal midpoint with the room.

Solution:

Room centre: ($\frac{-6+12}{2}$, $\frac{-15+0}{2}$) = $\left(3,-7.5\right)$.

Put the 5 ft dimension horizontally: x = 3 ± 2.5, giving 0.5 and 5.5.

Put the 3 ft dimension vertically: y = −7.5 ± 1.5, giving −9 and −6.

Opposite-foot averages return $\left(3,-7.5\right)$.

Results:

Feet: $\left(0.5,-6\right)$, $\left(5.5,-6\right)$, $\left(5.5,-9\right)$, $\left(0.5,-9\right)$. A rotated orientation is also possible.

Reasoning:

The two zero marks coincide at the origin.

Results:

x = 0, y = 0; the point is O$\left(0,0\right)$.

Given:

A vertical line through W.

To Find:

The possible coordinates and quadrants of H.

Property:

Vertical movement keeps x unchanged.

Solution:

H has coordinates $\left(-5,y\right)$; y is not fixed by the information.

If y > 0, H is in II. If y < 0, H is in III.

If y = 0, H is on the negative x-axis, not in a quadrant.

Results:

H$\left(-5,y\right)$, in II or III when off the axis.

Given:

Vertices joined in the given order.

To Find:

Perpendicular sides, an axis-parallel side and a reflection pair.

Property:

Horizontal and vertical lines are perpendicular; x-axis reflection changes y to −y.

Solution:

AM is horizontal because y = −2 at both ends. MP is vertical because x = −5 at both ends.

Thus AM ⟂ MP. AM is parallel to x; MP is parallel to y.

M and P have equal x and opposite y, so they reflect in the x-axis.

Results:

AM and MP are perpendicular. M and P are mirror images across x.

Given:

Z$\left(5,-6\right)$.

To Find:

One right triangle IZN and its lengths.

Formula:

Baudhāyana–Pythagoras: hypotenuse squared equals the sum of leg squares.

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Solution:

Choose I$\left(0,0\right)$, N$\left(5,0\right)$. IN is horizontal and NZ vertical, so the right angle is at N.

IN = 5; NZ = 6.

${\mathrm{IZ}}^2$ = $5^2$ + $6^2$ = 61.

Results:

IN = 5, NZ = 6, IZ = $\sqrt{61}$ units. Other valid constructions can give different lengths.

Given:

A single fixed Cartesian frame.

To Find:

Whether non-negative coordinate values suffice for the whole plane.

Property:

Signs distinguish opposite directions from the reference point.

Solution:

With x ≥ 0 and y ≥ 0, only the upper-right region and its bounding rays can be described.

Moving the origin can help with a bounded set of points, but no fixed origin makes every point of the entire plane non-negative.

A different notation could encode directions separately, but that restores the information supplied by signs.

Results:

Negative coordinates are needed to cover the whole plane with the usual fixed Cartesian system.

Given:

Two perpendicular main roads and parallel streets at uniform spacing.

To Find:

A model and the number of crossings named $\left(4,3\right)$ and $\left(3,4\right)$.

Property:

Two nonparallel straight lines have one intersection.

Solution:

Draw the main roads at right angles through the model centre. Draw the stated streets parallel to them, with adjacent streets 1 cm apart.

Label the streets consistently with the textbook convention; the first entry selects a north–south street and the second an east–west street.

$\left(4,3\right)$ selects one pair of streets, hence one crossing. $\left(3,4\right)$ selects another pair, also one crossing.

Results:

One intersection for each label; the two locations differ.

Given:

Origin at the screen’s bottom-left corner.

To Find:

Containment and intersection.

Property:

A circle’s bounds are centre ± radius. Two circles cross twice if |$r_B$ − $r_A$| < AB < $r_A$ + $r_B$.

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Solution:

Circle A: x from 20 to 180; y from 70 to 230. Circle B: x from 150 to 350; y from 130 to 330. All bounds fit the screen.

AB = $\sqrt{{\left(250-100\right)}^2+{\left(230-150\right)}^2}$ = $\sqrt{28900}$ = 170 pixels.

The radius difference is 20 and the sum is 180. Since 20 < 170 < 180, the circles intersect twice.

Results:

Both circles lie inside the screen and intersect in two points.

Given:

The four vertices in order.

To Prove:

ABCD is a square, then determine its area.

Property / Theorem / Formula:

Equal sides plus a right angle establish a square; use the converse of Baudhāyana–Pythagoras.

Proof:

${\mathrm{AB}}^2$ = (−3)${}^2$ + $1^2$ = 10. ${\mathrm{BC}}^2$ = (−1)${}^2$ + (−3)${}^2$ = 10.

${\mathrm{CD}}^2$ = $3^2$ + (−1)${}^2$ = 10. ${\mathrm{DA}}^2$ = $1^2$ + $3^2$ = 10.

${\mathrm{AC}}^2$ = (−4)${}^2$ + (−2)${}^2$ = 20 = ${\mathrm{AB}}^2$ + ${\mathrm{BC}}^2$, so ∠ABC is a right angle.

All four sides are $\sqrt{10}$ and the angle is right. Area = ($\sqrt{10}$)${}^2$ = 10.

Results:

ABCD is a square of area 10 square units.