NCERT Solutions · Class 9 Mathematics Orienting Yourself: The Use of Coordinates

Three corners of a rectangle fix the fourth.Call the three given feet \(\displaystyle \mathrm{T}_1(8,9)\), \(\displaystyle \mathrm{T}_2(11,9)\) and \(\displaystyle \mathrm{T}_3(11,7)\).(i) The fourth foot.\(\displaystyle \mathrm{T}_1\) and \(\displaystyle \mathrm{T}_2\) have the same y-coordinate, 9. Points with the same y-coordinate lie on a line parallel to the x-axis, so \(\displaystyle \mathrm{T}_1\mathrm{T}_2\) is horizontal and\[\mathrm{T}_1\mathrm{T}_2 = 11 - 8 = 3\ \text{ft}. \]\(\displaystyle \mathrm{T}_2\) and \(\displaystyle \mathrm{T}_3\) have the same x-coordinate, $\displaystyle 11$, so \(\displaystyle \mathrm{T}_2\mathrm{T}_3\) is vertical and\[\mathrm{T}_2\mathrm{T}_3 = 9 - 7 = 2\ \text{ft}. \]These two sides meet at \(\displaystyle \mathrm{T}_2\) at a right angle (a horizontal line and a vertical line are perpendicular). The fourth foot is the corner diagonally opposite \(\displaystyle \mathrm{T}_2\), so it must take its x-coordinate from \(\displaystyle \mathrm{T}_1\) and its y-coordinate from \(\displaystyle \mathrm{T}_3\):\[\text{fourth foot} = (8,\ 7). \]Check it two ways. From \(\displaystyle \mathrm{T}_1(8,9)\) go $\displaystyle 2$ ft down (the same drop as \(\displaystyle \mathrm{T}_2 \to \mathrm{T}_3\)): you land on \(\displaystyle (8,7)\). From \(\displaystyle \mathrm{T}_3(11,7)\) go $\displaystyle 3$ ft left (the same shift as \(\displaystyle \mathrm{T}_2 \to \mathrm{T}_1\)): again \(\displaystyle (8,7)\). Both routes agree, and the four points now form a \(\displaystyle 3 \times 2\) rectangle with every side parallel to an axis.(ii) Is this a good spot?The table covers the floor patch \(\displaystyle 8 \le x \le 11\), \(\displaystyle 7 \le y \le 9\). From Fig. $\displaystyle 1.5$ the other things on the floor are: the bed, \(\displaystyle 0.5 \le x \le 6.5\), \(\displaystyle 5 \le y \le 8\); the wardrobe, \(\displaystyle 3 \le x \le 7\), \(\displaystyle 0 \le y \le 2\); the room door in the bottom wall from \(\displaystyle x=8\) to \(\displaystyle x=11.5\); and the bathroom door in the left wall from \(\displaystyle y=1.5\) to \(\displaystyle y=4\).So the table sits in the one large empty piece of floor. It is $\displaystyle 1.5$ ft clear of the bed, $\displaystyle 1$ ft from the right wall (\(\displaystyle x=12\)), $\displaystyle 1$ ft from the top wall (\(\displaystyle y=10\)), and $\displaystyle 7$ ft in front of the doorway, so nobody walking in collides with it. On those grounds, yes - it is a good spot.One honest reservation, though. The table is free-standing, floating $\displaystyle 1$ ft away from two walls. Reiaan finds his way by touch, and furniture in the middle of the floor is exactly what is hard to find and easy to walk into. Pushing the table right into the corner - feet at \(\displaystyle (9,8)\), \(\displaystyle (12,8)\), \(\displaystyle (12,10)\), \(\displaystyle (9,10)\), still \(\displaystyle 3\ \text{ft} \times 2\ \text{ft}\) - would make it easy to locate along a wall and would open up the walking space. This part of the question is a judgement, so a well-argued "yes, but push it into the corner" is as good an answer as a plain "yes".(iii) Width, length, height.The two side lengths were found above: $\displaystyle 3$ ft and $\displaystyle 2$ ft. The longer one is the length:\[\text{length} = 3\ \text{ft}, \qquad \text{width} = 2\ \text{ft}. \]The height cannot be found. Fig. $\displaystyle 1.5$ is a floor map: it records, for each point, only how far right and how far forward it is. Height is a third, independent direction, and a two-coordinate system simply carries no information about it. (This is the same reason the text gives for why windows cannot be marked on the map of Fig. 1.1.) To record height you would need a third coordinate, i.e. a $\displaystyle 3$-D coordinate system.Answer. (i) The fourth foot is at \(\displaystyle (8, 7)\). (ii) Yes - it is the only large clear patch of floor, well away from the bed, the wardrobe and both doorways; but for Reiaan a table pushed flush into the corner at \(\displaystyle (9,8), (12,8), (12,10), (9,10)\) would be safer and easier to find. (iii) Length $\displaystyle 3$ ft, width $\displaystyle 2$ ft; the height cannot be worked out, because a $\displaystyle 2$-D floor plan carries no information about the third dimension.

How far a hinged door can sweep.Step $\displaystyle 1$: how long is the door leaf? The bathroom door runs from \(\displaystyle \mathrm{B}_1(0,1.5)\) to \(\displaystyle \mathrm{B}_2(0,4)\), both on the y-axis, so\[\mathrm{B}_1\mathrm{B}_2 = 4 - 1.5 = 2.5\ \text{ft}. \]Step $\displaystyle 2$: what region does it sweep? The hinge is at \(\displaystyle \mathrm{B}_1\), and the door is rigid, so as it opens into the bedroom the free edge \(\displaystyle \mathrm{B}_2\) travels along a circle of centre \(\displaystyle \mathrm{B}_1\) and radius $\displaystyle 2.5$ ft. Every point the door leaf can ever occupy is therefore at most $\displaystyle 2.5$ ft from \(\displaystyle \mathrm{B}_1\). When fully open (at right angles to the wall) its edge is at \(\displaystyle (2.5,\ 1.5)\).Step $\displaystyle 3$: how close is the wardrobe? From Fig. $\displaystyle 1.5$ the wardrobe has corners \(\displaystyle \mathrm{W}_1(3,0)\), \(\displaystyle \mathrm{W}_2(7,0)\), \(\displaystyle \mathrm{W}_3(7,2)\), \(\displaystyle \mathrm{W}_4(3,2)\), so it occupies \(\displaystyle 3 \le x \le 7\), \(\displaystyle 0 \le y \le 2\). Which of its points is nearest to \(\displaystyle \mathrm{B}_1(0,1.5)\)? Its left face is the segment \(\displaystyle x = 3\), \(\displaystyle 0 \le y \le 2\); the hinge height \(\displaystyle y = 1.5\) lies inside that range, so the nearest point of the wardrobe is \(\displaystyle (3,\ 1.5)\), straight across from the hinge:\[\text{distance} = \sqrt{(3-0)^2 + (1.5-1.5)^2} = \sqrt{9} = 3\ \text{ft}. \]Step $\displaystyle 4$: compare. The door reaches out $\displaystyle 2.5$ ft; the wardrobe starts $\displaystyle 3$ ft away. Since \(\displaystyle 2.5 < 3\), the door does not hit the wardrobe - but only by \(\displaystyle 3 - 2.5 = 0.5\) ft, i.e. six inches.If the door is made wider. A leaf of length \(\displaystyle w\) hinged at \(\displaystyle \mathrm{B}_1\) sweeps out a quarter-disc of radius \(\displaystyle w\). It reaches the wardrobe as soon as \(\displaystyle w > 3\) ft. So a bathroom door widened to $\displaystyle 3.5$ ft - the width of the room door, and the sensible width for wheelchair access - would strike the wardrobe, overlapping it by $\displaystyle 0.5$ ft. (It would just miss the bed: the nearest corner of the bed, \(\displaystyle \mathrm{S}_1(0.5,5)\), is \(\displaystyle \sqrt{0.5^2+3.5^2}=\sqrt{12.5}\approx 3.54\) ft from \(\displaystyle \mathrm{B}_1\).)Changes worth suggesting: Answer. No - the $\displaystyle 2.5$ ft door hinged at \(\displaystyle \mathrm{B}_1\) sweeps only $\displaystyle 2.5$ ft into the room, and the nearest point of the wardrobe is $\displaystyle 3$ ft away, so it clears it by $\displaystyle 0.5$ ft. But any door wider than $\displaystyle 3$ ft would hit the wardrobe, so if the door is widened (say to $\displaystyle 3.5$ ft to match the room door) the wardrobe must be shifted $\displaystyle 0.5$ ft to $\displaystyle 1$ ft to the right, or the door must be made to open into the bathroom or to slide.

Reading a rectangle and a trapezium straight off the grid.(i) The corners of the bathroom.The bathroom lies to the left of the y-axis, so its points have negative x-coordinates (that is the Quadrant II side). Reading Fig. $\displaystyle 1.5$:\[\mathrm{O}(0,0), \qquad \mathrm{F}(0,9), \qquad \mathrm{R}(-6,9), \qquad \mathrm{P}(-6,0). \]Check that this really is a rectangle: OF lies on \(\displaystyle x=0\) and PR on \(\displaystyle x=-6\), so both are parallel to the y-axis; FR lies on \(\displaystyle y=9\) and OP on \(\displaystyle y=0\), so both are parallel to the x-axis. A horizontal line meets a vertical line at a right angle, so all four angles are right angles. Its sides are\[\mathrm{OF} = 9 - 0 = 9\ \text{ft}, \qquad \mathrm{OP} = |0 - (-6)| = 6\ \text{ft}, \]so the bathroom is \(\displaystyle 6\ \text{ft} \times 9\ \text{ft} = 54\) sq ft.(ii) The shape of the showering area SHWR.Reading the four marked corners from Fig. $\displaystyle 1.5$:\[\mathrm{S}(-6,6), \qquad \mathrm{H}(-3,6), \qquad \mathrm{W}(-2,9), \qquad \mathrm{R}(-6,9). \]Now look at each side using the coordinates alone. So SHWR has exactly one pair of parallel sides (SH and RW), and those two are of different lengths, $\displaystyle 3$ ft and $\displaystyle 4$ ft. A quadrilateral with just one pair of parallel sides is a trapezium; here one of the other sides, SR, is perpendicular to both parallel sides, so it is a right trapezium. Its area is\[\tfrac{1}{2}(\mathrm{SH} + \mathrm{RW}) \times \mathrm{SR} = \tfrac{1}{2}(3+4)\times 3 = 10.5\ \text{sq ft}. \](iii) Marking off the washbasin and the toilet.There is no single right answer here - any placement is correct provided it lies inside the rectangle OFRP, does not overlap the showering area SHWR, and does not block the doorway \(\displaystyle \mathrm{B}_1\mathrm{B}_2\) (on the y-axis between \(\displaystyle y=1.5\) and \(\displaystyle y=4\)). Here is one workable choice, keeping both fixtures against the walls as in the picture:

Fixture	Size	Coordinates of the four corners
Washbasin	$\displaystyle 3$ ft \(\displaystyle \times\) $\displaystyle 2$ ft	\(\displaystyle (-6,0),\ (-3,0),\ (-3,2),\ (-6,2)\)
Toilet	$\displaystyle 2$ ft \(\displaystyle \times\) $\displaystyle 3$ ft	\(\displaystyle (-6,2.5),\ (-4,2.5),\ (-4,5.5),\ (-6,5.5)\)

Why these work: Answer. (i) \(\displaystyle \mathrm{O}(0,0)\), \(\displaystyle \mathrm{F}(0,9)\), \(\displaystyle \mathrm{R}(-6,9)\), \(\displaystyle \mathrm{P}(-6,0)\) - a $\displaystyle 6$ ft \(\displaystyle \times\) $\displaystyle 9$ ft rectangle. (ii) SHWR is a right trapezium: SH ($\displaystyle 3$ ft) is parallel to RW ($\displaystyle 4$ ft), SR ($\displaystyle 3$ ft) is perpendicular to both, and HW \(\displaystyle =\sqrt{10}\approx 3.16\) ft; its area is $\displaystyle 10.5$ sq ft. (iii) For example, washbasin at \(\displaystyle (-6,0), (-3,0), (-3,2), (-6,2)\) and toilet at \(\displaystyle (-6,2.5), (-4,2.5), (-4,5.5), (-6,5.5)\) - other placements are equally correct as long as they stay inside the bathroom, off the showering area, and clear of the doorway.

Which side of the wall must the dining room be on?(i) Sketching the dining room.First decide where it goes. Reiaan's room door \(\displaystyle \mathrm{D}_1\mathrm{R}_1\) lies in the wall along the x-axis, and Reiaan's room itself fills \(\displaystyle 0 \le y \le 10\), i.e. everything above that wall. A room you enter through that door must therefore lie on the other side of the wall - below the x-axis, where y-coordinates are negative. (That is why the exercise asked you to mark the y-axis all the way down to \(\displaystyle (0,-15)\): the numbers you need are waiting there.)Now use the measurements. The length runs from P to A, and both are already on the x-axis: \(\displaystyle \mathrm{P}(-6,0)\) and \(\displaystyle \mathrm{A}(12,0)\). Since they share the y-coordinate $\displaystyle 0$,\[\mathrm{PA} = 12 - (-6) = 18\ \text{ft}, \]which is exactly the stated length of $\displaystyle 18$ ft - a good confirmation that PA is indeed the length side. The width, $\displaystyle 15$ ft, is measured at right angles to PA, i.e. straight down from the x-axis, from \(\displaystyle y = 0\) to \(\displaystyle y = -15\).So the dining room is the rectangle whose corners are\[\mathrm{P}(-6,\,0), \qquad \mathrm{A}(12,\,0), \qquad (12,\,-15), \qquad (-6,\,-15). \]Sketch it as a rectangle sitting under the x-axis, $\displaystyle 18$ units wide (from \(\displaystyle x=-6\) to \(\displaystyle x=12\)) and $\displaystyle 15$ units deep (from \(\displaystyle y=0\) down to \(\displaystyle y=-15\)), sharing the whole edge PA with the bathroom-and-bedroom wall above.(ii) The dining table at the centre.The centre of a rectangle is the midpoint of either diagonal. Using the diagonal from \(\displaystyle \mathrm{P}(-6,0)\) to \(\displaystyle (12,-15)\):\[\text{centre} = \left(\frac{-6+12}{2},\ \frac{0+(-15)}{2}\right) = (3,\ -7.5). \]The table is $\displaystyle 5$ ft by $\displaystyle 3$ ft. Put its $\displaystyle 5$ ft side parallel to the x-axis (parallel to the length of the room). Then from the centre you must go half of $\displaystyle 5$, i.e. $\displaystyle 2.5$ ft, each way in the x-direction, and half of $\displaystyle 3$, i.e. $\displaystyle 1.5$ ft, each way in the y-direction:\[x = 3 \pm 2.5 \ \Rightarrow\ x = 0.5 \text{ and } 5.5, \qquad y = -7.5 \pm 1.5 \ \Rightarrow\ y = -6 \text{ and } -9. \]\[\text{Feet of the table: } (0.5,\,-6),\ (5.5,\,-6),\ (5.5,\,-9),\ (0.5,\,-9). \]Check: the midpoint of the diagonal from \(\displaystyle (0.5,-6)\) to \(\displaystyle (5.5,-9)\) is \(\displaystyle \left(\frac{0.5+5.5}{2}, \frac{-6-9}{2}\right) = (3,-7.5)\), the centre of the room. And the sides are \(\displaystyle 5.5-0.5 = 5\) ft and \(\displaystyle -6-(-9) = 3\) ft, as required.If instead you turn the table so that its $\displaystyle 5$ ft side is parallel to the y-axis, the feet come out as \(\displaystyle (1.5,-5)\), \(\displaystyle (4.5,-5)\), \(\displaystyle (4.5,-10)\), \(\displaystyle (1.5,-10)\) - also exactly central, so that answer is equally correct.Answer. (i) The dining room lies below the x-axis, with corners \(\displaystyle \mathrm{P}(-6,0)\), \(\displaystyle \mathrm{A}(12,0)\), \(\displaystyle (12,-15)\) and \(\displaystyle (-6,-15)\) - $\displaystyle 18$ ft long along PA and $\displaystyle 15$ ft wide. (ii) The centre is \(\displaystyle (3,-7.5)\), so the table's feet are at \(\displaystyle (0.5,-6)\), \(\displaystyle (5.5,-6)\), \(\displaystyle (5.5,-9)\), \(\displaystyle (0.5,-9)\) (or \(\displaystyle (1.5,-5)\), \(\displaystyle (4.5,-5)\), \(\displaystyle (4.5,-10)\), \(\displaystyle (1.5,-10)\) if the table is turned the other way).