In Fig. 1.3, $D_1$ lies on the x-axis at $x = 8$, so its coordinates are $(8, 0)$. The distance from the y-axis is the x-coordinate, $8$ units, and the distance from the x-axis is the y-coordinate, $0$ units. The room-door segment runs from $D_1(8,0)$ to $R_1(11.5,0)$, so its width is $11.5 - 8 = 3.5$ units. The bathroom-door segment runs from $B_1(0,1.5)$ to $B_2(0,4)$, so its width is $4 - 1.5 = 2.5$ units. Since $2.5 \lt 3.5$, the bathroom door is narrower.
(i) The door is $8$ units from the left wall (the y-axis) and $0$ units from the x-axis.
(ii) $D_1 = (8, 0)$.
(iii) The width is $11.5 - 8 = 3.5$ units. A width of $3.5$ ft is comfortable for a room door and should allow a wheelchair to enter easily.
(iv) The bathroom door width is $4 - 1.5 = 2.5$ units, so it is narrower than the room door.
For a rectangle whose sides are parallel to the axes, opposite vertices have the same x- and y-coordinates. The given points $(8,9)$ and $(11,9)$ share the top side, while $(11,9)$ and $(11,7)$ share the right side. Therefore the missing corner must combine $x=8$ with $y=7$, giving $(8,7)$. The horizontal side is $3$ ft and the vertical side is $2$ ft. A plan view gives floor dimensions only, so the table height cannot be determined.
(i) The fourth foot will be at $(8, 7)$.
(ii) Yes, it is a good spot because the table fits in the open space near the upper-right part of the room without crossing the bed, wardrobe or walls.
(iii) The sides are $11 - 8 = 3$ ft and $9 - 7 = 2$ ft. From the top view, we can find only the length and width, not the height.
From Fig. 1.5, $B_1=(0,1.5)$ and $B_2=(0,4)$, so the bathroom-door width is $4-1.5=2.5$ ft. The wardrobe occupies the rectangle from $x=3$ to $x=7$ and from $y=0$ to $y=2$. A $2.5$ ft door opening into the room from the y-axis does not reach $x=3$, so it avoids the wardrobe. If the door is made wider, the clearance decreases; beyond $3$ ft it would reach the wardrobe line.
No, it will not hit the wardrobe in the present plan. The door hinged at $B_1(0,1.5)$ has width $2.5$ ft, so when it opens into the bedroom it reaches only up to about $x=2.5$ along the floor direction. The wardrobe begins at $x=3$, so there is a small clearance. If the door is made wider than $3$ ft, it may hit the wardrobe; a sliding door or an outward-opening door would be better.
The bathroom is the blue rectangle to the left of the y-axis in Fig. 1.5. Reading the axes gives the four outside corners as $(0,0)$, $(0,9)$, $(-6,9)$ and $(-6,0)$. The showering area has two horizontal parallel sides $RW$ and $SH$, but its side $WH$ is slanting, so it is a trapezium. The washbasin and toilet placements are not unique; the listed rectangles have dimensions $3 \times 2$ and $2 \times 3$ respectively and fit inside the bathroom.
(i) The bathroom corners are $O(0,0)$, $F(0,9)$, $R(-6,9)$ and $P(-6,0)$.
(ii) $SHWR$ is a trapezium. Its corners are approximately $S(-6,6)$, $H(-3,6)$, $W(-2,9)$ and $R(-6,9)$.
(iii) One suitable choice is: washbasin rectangle with corners $(-6,0)$, $(-3,0)$, $(-3,2)$ and $(-6,2)$; toilet rectangle with corners $(-6,2)$, $(-4,2)$, $(-4,5)$ and $(-6,5)$.
The segment from $P(-6,0)$ to $A(12,0)$ has length $12-(-6)=18$ ft, matching the dining-room length. Taking the room below this segment gives width $15$ ft, so the other two corners are $(-6,-15)$ and $(12,-15)$. The midpoint of the rectangle is $\left(\dfrac{-6+12}{2},\dfrac{0+(-15)}{2}\right)=(3,-7.5)$. A centred $5 \times 3$ table extends $2.5$ ft horizontally and $1.5$ ft vertically from this centre, giving the four listed corner coordinates.
(i) Since $P(-6,0)$ and $A(12,0)$ are $18$ ft apart, one possible dining room is the rectangle with corners $P(-6,0)$, $A(12,0)$, $(12,-15)$ and $(-6,-15)$.
(ii) The centre of this dining room is $(3,-7.5)$. Placing the $5$ ft side horizontally and the $3$ ft side vertically, the table feet are $(0.5,-6)$, $(5.5,-6)$, $(5.5,-9)$ and $(0.5,-9)$.
The point of intersection of the two axes is the origin, so its coordinates are $(0,0)$. The x-coordinate is $0$ and the y-coordinate is $0$.
A line parallel to the y-axis has a constant x-coordinate. Since W has x-coordinate $-5$, every point on the vertical line through W has the form $(-5,y)$. Negative x with positive y gives Quadrant II, and negative x with negative y gives Quadrant III.
Point H must have x-coordinate $-5$, so its coordinates are $(-5,y)$ for some real number $y$. If $y \gt 0$, H lies in Quadrant II; if $y \lt 0$, H lies in Quadrant III; if $y=0$, H lies on the negative x-axis and not in any quadrant.
Joining $R(3,0)\to A(0,-2)\to M(-5,-2)\to P(-5,2)$ gives the sides $RA$, $AM$, $MP$, $PR$.
(i) $A(0,-2)$ and $M(-5,-2)$ have the same y-coordinate, so $AM$ is horizontal; $M(-5,-2)$ and $P(-5,2)$ have the same x-coordinate, so $MP$ is vertical. A horizontal line meets a vertical line at a right angle, so $AM \perp MP$.
(ii) Being horizontal, $AM$ is parallel to the x-axis (equivalently $MP$, being vertical, is parallel to the y-axis).
(iii) Reflecting a point in the x-axis keeps its x-coordinate and negates its y-coordinate: $(x,y)\mapsto(x,-y)$. Since $M(-5,-2)\mapsto(-5,2)=P$, the points $M$ and $P$ are mirror images of each other in the x-axis.
(i) $AM$ and $MP$ are perpendicular to each other (a horizontal side meets a vertical side).
(ii) $AM$ is parallel to the x-axis (and $MP$ is parallel to the y-axis).
(iii) $M(-5,-2)$ and $P(-5,2)$ are mirror images of each other in the x-axis — they share the same x-coordinate $(-5)$ and have opposite y-coordinates.
Point $Z(5,-6)$ is 5 units to the right of the y-axis and 6 units below the x-axis. Choose $I(5,0)$ directly above Z on the x-axis and $N(0,-6)$ directly left of Z on the horizontal line through Z. Then $ZI$ is vertical with length $6$, $ZN$ is horizontal with length $5$, and $\angle IZN$ is a right angle. By the Baudhāyana-Pythagoras theorem, $IN=\sqrt{6^2+5^2}=\sqrt{61}$ units.
One possible triangle is obtained by taking $I(5,0)$ and $N(0,-6)$. Then $ZI=6$ units, $ZN=5$ units and $IN=\sqrt{5^2+6^2}=\sqrt{61}$ units.
In the Cartesian plane, positive x-values mark positions to the right of the y-axis and negative x-values mark positions to the left. Similarly, positive y-values mark positions above the x-axis and negative y-values mark positions below it. If negative numbers were not available, points in Quadrants II, III and IV could not all be represented. So such a coordinate system would be incomplete for the whole 2-D plane.
Without negative numbers, the coordinate system could locate only points whose coordinates are non-negative, like points in the first quadrant and on the positive axes. It would not allow us to locate all points on a 2-D plane, because points to the left of the y-axis or below the x-axis need negative coordinates.
For points to lie on the same straight line, the change in y per change in x must be the same between consecutive pairs. From $M(-3,-4)$ to $A(0,0)$, the change is $(3,4)$. From $A(0,0)$ to $G(6,8)$, the change is $(6,8)=2(3,4)$. Since the second displacement is a multiple of the first, M, A and G lie on the same straight line.
Yes, the three points are on the same straight line. One method is to compare the ratios of the changes in coordinates: from M to A, $\Delta x=3$ and $\Delta y=4$; from A to G, $\Delta x=6$ and $\Delta y=8$. Since $\dfrac{4}{3}=\dfrac{8}{6}$, the directions are the same, so the points are collinear.
Using the method from Problem 6, compare the coordinate changes. $R(-5,-1)$ to $B(-2,-5)$ gives $(\Delta x,\Delta y)=(3,-4)$. $B(-2,-5)$ to $C(4,-12)$ gives $(\Delta x,\Delta y)=(6,-7)$. If the points were collinear, these direction ratios would be equal, but $-4/3 \neq -7/6$. Hence the three points are not collinear.
No, the points $R$, $B$ and $C$ are not on the same straight line. From R to B, $\Delta x=3$ and $\Delta y=-4$, so the ratio is $-\dfrac{4}{3}$. From B to C, $\Delta x=6$ and $\Delta y=-7$, so the ratio is $-\dfrac{7}{6}$. These are not equal.
For (i), $OA=3$ and $OB=3$, and the axes are perpendicular, so $\triangle OAB$ is right-angled and isosceles. For (ii), $C(-3,-4)$ lies in Quadrant III and $D(3,-4)$ lies in Quadrant IV. Also $OC=\sqrt{(-3)^2+(-4)^2}=5$ and $OD=\sqrt{3^2+(-4)^2}=5$, so $\triangle OCD$ is isosceles.
(i) One example is $O(0,0)$, $A(3,0)$ and $B(0,3)$.
(ii) One example is $O(0,0)$, $C(-3,-4)$ and $D(3,-4)$.
Use the midpoint rule: if $S(x_1,y_1)$ and $T(x_2,y_2)$ are endpoints, then the midpoint is $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$. Applying this to the four rows gives $(0,0)$, $(3,4)$, $(0,-5)$ and $(-1,2)$ respectively. Therefore only the first two listed M points are midpoints.
Case 1: Yes, because $M=\left(\dfrac{-3+3}{2},\dfrac{0+0}{2}\right)=(0,0)$.
Case 2: Yes, because $M=\left(\dfrac{2+4}{2},\dfrac{3+5}{2}\right)=(3,4)$.
Case 3: No, because the midpoint of $(0,0)$ and $(0,-10)$ is $(0,-5)$, not $(0,5)$.
Case 4: No, because the midpoint of $(-8,7)$ and $(6,-3)$ is $(-1,2)$, not $(0,-2)$.
When M is the midpoint of ST, each coordinate of M is the average of the corresponding coordinates of S and T.