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).

Use what each axis is a set of.The x-axis is the set of all points whose distance from the x-axis is zero, that is, all points of the form \(\displaystyle (x, 0)\); every point on it has y-coordinate 0.The y-axis is the set of all points of the form \(\displaystyle (0, y)\); every point on it has x-coordinate 0.The point where the two axes cross lies on both lines at once. So it must satisfy both conditions at the same time: its y-coordinate is $\displaystyle 0$ (because it is on the x-axis) and its x-coordinate is $\displaystyle 0$ (because it is on the y-axis). Only one point can do that.\[\text{point of intersection} = (0,\ 0). \]This point is given a special name: the origin, O. It is the point from which all distances along both axes are measured, which is why both of its coordinates have to be zero - it is zero units from each axis.Answer: the x-coordinate is $\displaystyle 0$ and the y-coordinate is $\displaystyle 0$; the point of intersection of the two axes is the origin O$\displaystyle (0, 0)$.

A line parallel to the y-axis is a line of constant x.W has x-coordinate \(\displaystyle -5\), so \(\displaystyle \mathrm{W} = (-5,\ w)\) for some number \(\displaystyle w\) that we are not told.The y-axis is vertical, so a line parallel to the y-axis is vertical too. Travelling along a vertical line changes only how high you are; it never changes how far left or right of the y-axis you are. Since the x-coordinate of a point is its distance from the y-axis (with a sign for the side), every point on that line has the same x-coordinate as W, namely \(\displaystyle -5\).\[\mathrm{H} = (-5,\ k) \quad \text{for some number } k. \]So we can predict one coordinate exactly and not the other. The x-coordinate must be \(\displaystyle -5\). The y-coordinate \(\displaystyle k\) cannot be predicted, because the question tells us which line H is on but not where along it H sits - H could be anywhere on that vertical line.Which quadrants can H lie in? Its x-coordinate is \(\displaystyle -5\), which is negative, so H is on the left of the y-axis. Of the four quadrants, only two lie on the left: H can never be in Quadrant I or Quadrant IV, since those need a positive x-coordinate.Answer: \(\displaystyle \mathrm{H} = (-5, k)\) - its x-coordinate must be \(\displaystyle -5\), but its y-coordinate cannot be predicted. H lies in Quadrant II if its y-coordinate is positive and in Quadrant III if it is negative; if the y-coordinate is $\displaystyle 0$ then H is the point \(\displaystyle (-5,0)\) on the x-axis, which is in no quadrant.

Predict from the coordinates first, then plot to confirm.The four points are \(\displaystyle \mathrm{R}(3,0)\), \(\displaystyle \mathrm{A}(0,-2)\), \(\displaystyle \mathrm{M}(-5,-2)\), \(\displaystyle \mathrm{P}(-5,2)\), joined in that order, so the four sides are RA, AM, MP and PR.(i) Two sides perpendicular to each other.Look for equal coordinates. A\(\displaystyle (0,-2)\) and M\(\displaystyle (-5,-2)\) share the same y-coordinate*, \(\displaystyle -2\). Points with the same y-coordinate lie on a line parallel to the x-axis, so AM is horizontal, with \(\displaystyle \mathrm{AM} = |0-(-5)| = 5\). M\(\displaystyle (-5,-2)\) and P\(\displaystyle (-5,2)\) share the same x-coordinate*, \(\displaystyle -5\), so MP is vertical, with \(\displaystyle \mathrm{MP} = 2-(-2) = 4\).A horizontal line and a vertical line always meet at a right angle, and AM and MP meet at M. So \(\displaystyle \mathrm{AM} \perp \mathrm{MP}\), and the right angle is at M.The other two sides are slanting, and they are not perpendicular to anything here. You can test any corner with the Baudhāyana-Pythagoras theorem, since a right angle at a corner would force (leg)\(\displaystyle ^2\) + (leg)\(\displaystyle ^2\) = (opposite side)\(\displaystyle ^2\). For instance at R, \(\displaystyle \mathrm{RA}^2 = 3^2+2^2 = 13\) and \(\displaystyle \mathrm{RP}^2 = 8^2+2^2 = 68\), while \(\displaystyle \mathrm{AP}^2 = 5^2+4^2 = 41\); and \(\displaystyle 13 + 68 = 81 \neq 41\), so the angle at R is not a right angle. So M is the only right angle.(ii) One side parallel to one of the axes.From the same observation: AM is parallel to the x-axis (both A and M have \(\displaystyle y=-2\), so the whole side lies on the line \(\displaystyle y=-2\)). Equally, MP is parallel to the y-axis (both have \(\displaystyle x=-5\)), so either side may be quoted.(iii) Two points that are mirror images of each other in one axis.Reflecting in the x-axis sends \(\displaystyle (x,y)\) to \(\displaystyle (x,-y)\): the point flips to the other side, keeping the same x. Reflecting in the y-axis sends \(\displaystyle (x,y)\) to \(\displaystyle (-x,y)\). So look for two of the four points that agree in one coordinate and are opposite in the other.\[\mathrm{M}(-5,\,-2) \quad \text{and} \quad \mathrm{P}(-5,\,2): \ \text{same } x = -5,\ \text{opposite } y. \]So M and P are mirror images, and the mirror is the x-axis. (Check: M is $\displaystyle 2$ units below the x-axis and P is $\displaystyle 2$ units above it, both on the same vertical line \(\displaystyle x=-5\).)Now plot and verify. Mark R\(\displaystyle (3,0)\) on the positive x-axis, A\(\displaystyle (0,-2)\) on the negative y-axis, M\(\displaystyle (-5,-2)\) in Quadrant III and P\(\displaystyle (-5,2)\) in Quadrant II. Joining R \(\displaystyle \to\) A \(\displaystyle \to\) M \(\displaystyle \to\) P \(\displaystyle \to\) R you can see all three predictions at once: AM runs flat along \(\displaystyle y=-2\), MP runs straight up the line \(\displaystyle x=-5\), they meet in a square corner at M, and M and P sit at equal distances below and above the x-axis, so folding the paper along the x-axis lands one exactly on the other.Answer: (i) AM \(\displaystyle \perp\) MP, the right angle being at M. (ii) AM is parallel to the x-axis (and MP is parallel to the y-axis). (iii) M\(\displaystyle (-5,-2)\) and P\(\displaystyle (-5,2)\) are mirror images of each other in the x-axis.

Build the right angle out of the directions of the axes.(The question says answers may differ from person to person - this is one correct choice; yours may be different and still be right.)There is a way to be certain of a right angle without measuring anything: make one side of the triangle parallel to the x-axis and the other side parallel to the y-axis. Those two directions are perpendicular, because the axes are drawn at right angles to each other. Two points have a horizontal join exactly when their y-coordinates are equal, and a vertical join exactly when their x-coordinates are equal.Plot \(\displaystyle \mathrm{Z}(5,-6)\). Its x-coordinate is positive and its y-coordinate is negative, so Z lies in Quadrant IV: $\displaystyle 5$ units right of the y-axis, $\displaystyle 6$ units below the x-axis.Now put the right angle at Z: Since ZI is vertical and ZN is horizontal, \(\displaystyle \angle \mathrm{IZN} = 90^\circ\), so IZN is right-angled at Z.The three side lengths.Both ZI and ZN join points that share a coordinate, so each is just a difference:\[\mathrm{ZI} = |-2 - (-6)| = 4\ \text{units}, \qquad \mathrm{ZN} = |8 - 5| = 3\ \text{units}. \]For the hypotenuse IN, use the distance formula:\[\mathrm{IN} = \sqrt{(8-5)^2 + \big(-6-(-2)\big)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5\ \text{units}. \]Check with the Baudhāyana-Pythagoras theorem: \(\displaystyle 3^2 + 4^2 = 9 + 16 = 25 = 5^2\). It fits, as it must.Other answers are equally correct. For example, taking \(\displaystyle \mathrm{I}(5,0)\) and \(\displaystyle \mathrm{N}(11,-6)\) gives \(\displaystyle \mathrm{ZI}=6\), \(\displaystyle \mathrm{ZN}=6\) and \(\displaystyle \mathrm{IN}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\); the triangle is still right-angled at Z. The lengths will differ from person to person - what must be true for everybody is that the squares of the two shorter sides add up to the square of the longest side.Answer: with \(\displaystyle \mathrm{Z}(5,-6)\), \(\displaystyle \mathrm{I}(5,-2)\) and \(\displaystyle \mathrm{N}(8,-6)\), the triangle is right-angled at Z with \(\displaystyle \mathrm{ZI}=4\), \(\displaystyle \mathrm{ZN}=3\) and \(\displaystyle \mathrm{IN}=5\) units (and \(\displaystyle 3^2+4^2=5^2\)). Any other choice of I and N giving a right angle - most easily by keeping one side vertical and the other horizontal - is equally acceptable.

What the minus sign is actually doing.In the system we have built, a coordinate carries two pieces of information at once: how far from an axis, and which side. The size of the number gives the distance; the sign gives the side - right or left of the y-axis for x, above or below the x-axis for y.Now suppose negative numbers did not exist. Then every coordinate would have to be $\displaystyle 0$ or positive, and a pair \(\displaystyle (x,y)\) with \(\displaystyle x \ge 0\) and \(\displaystyle y \ge 0\) names exactly those points that are on or to the right of the y-axis and on or above the x-axis. Those points make up\[\text{Quadrant I} \ +\ \text{the positive halves of the two axes} \ +\ \text{the origin}, \]which is only one quarter of the plane. It would be a single-quadrant system: a corner, not a whole plane.So points such as \(\displaystyle \mathrm{Q}(-5,3)\) in Quadrant II, or a point $\displaystyle 5$ units left and $\displaystyle 3$ units below O, would simply have no name. Two different points - one $\displaystyle 3$ units right of O and one $\displaystyle 3$ units left of O, both on the x-axis - would both have to be called "$\displaystyle 3$", and we could not tell them apart.No, such a system could not locate all the points of a $\displaystyle 2$-D plane.There are two ways round it, and they are worth noticing.1. Move the origin. Put O at one corner of the region you care about, so that everything you want to describe already lies up and to the right. This is exactly how a seating chart, a chess board, or the street grid of the Sindhu-Sarasvati cities works when measured from one corner. It works, but only for a bounded piece of the plane, and you must know in advance where the corner is. 2. Attach a direction word to every number, for example "$\displaystyle 5$ left, $\displaystyle 3$ down". This does work for the whole plane - but writing a direction beside each number is clumsy, and you cannot then calculate with the coordinates.A negative sign is just the second idea written compactly, in a form you can also do arithmetic with. That is why Brahmagupta's treatment of negative numbers as genuine algebraic quantities had to come before the four-quadrant plane could exist at all.Answer: without negative numbers we would have only the first quadrant - a quarter-plane of points on or above the x-axis and on or to the right of the y-axis. It would let us locate points in that corner only, so no, it would not allow us to locate all the points of a $\displaystyle 2$-D plane; the negative sign is what records which side of each axis a point lies on.

Collinearity by distances: the longest must equal the sum of the other two.The method (no plotting needed). Take any three points. If they are not on one line they form a triangle, and in a triangle every side is strictly shorter than the sum of the other two - a detour through a third corner is always longer than going straight. If they are on one line, one of them lies between the other two, and the journey from the first to the last passes through the middle one, so the longest distance is exactly the sum of the other two. So:\[\text{three points are collinear} \iff \text{(largest distance)} = \text{(sum of the other two distances)}. \]All three distances come from the Baudhāyana-Pythagoras distance formula, so this is pure arithmetic.Applying it to M\(\displaystyle (-3,-4)\), A\(\displaystyle (0,0)\), G\(\displaystyle (6,8)\).\[\mathrm{MA} = \sqrt{\big(0-(-3)\big)^2 + \big(0-(-4)\big)^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5. \]\[\mathrm{AG} = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36+64} = \sqrt{100} = 10. \]\[\mathrm{MG} = \sqrt{\big(6-(-3)\big)^2 + \big(8-(-4)\big)^2} = \sqrt{9^2 + 12^2} = \sqrt{81+144} = \sqrt{225} = 15. \]The largest is MG. And\[\mathrm{MA} + \mathrm{AG} = 5 + 10 = 15 = \mathrm{MG}. \]They add up exactly, so the three points are collinear, with A lying between M and G.A second method, often quicker: compare the steps. Instead of distances, look at how you travel from one point to the next. The second step is the first step doubled: same direction, just twice as far. Since you never turn at A, all three points lie on one straight line. (If the two steps had not been multiples of each other - say \(\displaystyle (3,4)\) then \(\displaystyle (6,7)\) - the direction would have changed at A and the points would not be collinear.) This method uses only whole-number arithmetic, with no square roots at all.Answer: yes, M, A and G lie on the same straight line, with A between M and G. Method: use the distance formula to get MA \(\displaystyle =5\), AG \(\displaystyle =10\), MG \(\displaystyle =15\) and check that the largest equals the sum of the other two, \(\displaystyle 5+10=15\); or, more simply, check that the step from M to A, \(\displaystyle (3,4)\), and the step from A to G, \(\displaystyle (6,8)\), are multiples of each other, so the direction of travel never changes.

The same two tests - and this time the plot cannot decide it.The points are \(\displaystyle \mathrm{R}(-5,-1)\), \(\displaystyle \mathrm{B}(-2,-5)\), \(\displaystyle \mathrm{C}(4,-12)\).Distance test.\[\mathrm{RB} = \sqrt{\big(-2-(-5)\big)^2 + \big(-5-(-1)\big)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = \sqrt{25} = 5. \]\[\mathrm{BC} = \sqrt{\big(4-(-2)\big)^2 + \big(-12-(-5)\big)^2} = \sqrt{6^2+(-7)^2} = \sqrt{36+49} = \sqrt{85}. \]\[\mathrm{RC} = \sqrt{\big(4-(-5)\big)^2 + \big(-12-(-1)\big)^2} = \sqrt{9^2+(-11)^2} = \sqrt{81+121} = \sqrt{202}. \]Numerically \(\displaystyle \sqrt{85} \approx 9.2195\) and \(\displaystyle \sqrt{202} \approx 14.2127\), so\[\mathrm{RB} + \mathrm{BC} \approx 5 + 9.2195 = 14.2195, \qquad \mathrm{RC} \approx 14.2127. \]These are not equal, so the points are not collinear - but they differ by only about \(\displaystyle 0.007\) of a unit, far less than the thickness of a pencil line.We can settle it exactly, without decimals. Suppose \(\displaystyle 5 + \sqrt{85} = \sqrt{202}\) were true. Squaring both sides, \[25 + 10\sqrt{85} + 85 = 202 \ \Longrightarrow\ 10\sqrt{85} = 92 \ \Longrightarrow\ \sqrt{85} = 9.2 \ \Longrightarrow\ 85 = 84.64, \] which is false. So \(\displaystyle \mathrm{RB} + \mathrm{BC} \neq \mathrm{RC}\), for certain.Step test (cleaner, and it needs no square roots at all). If the direction had not changed at B, then a step twice as long sideways (\(\displaystyle 6 = 2\times 3\)) would have to be twice as long vertically too, that is \(\displaystyle (6,-8)\). It is \(\displaystyle (6,-7)\), one unit short. So the direction does change at B, and R, B, C are not collinear.Now plot both sets and check. R\(\displaystyle (-5,-1)\) and B\(\displaystyle (-2,-5)\) are in Quadrant III and C\(\displaystyle (4,-12)\) is in Quadrant IV. On the graph the three look as good as straight - the bend at B is a single unit spread over a journey of more than $\displaystyle 14$ units, so the eye cannot see it. Plotting M\(\displaystyle (-3,-4)\), A\(\displaystyle (0,0)\), G\(\displaystyle (6,8)\) from Problem $\displaystyle 6$, by contrast, gives a line that is genuinely straight, matching \(\displaystyle 5+10=15\) exactly.That contrast is the real lesson of this pair of questions: a picture can suggest collinearity, but only the calculation can prove it or disprove it.Answer: no - R, B and C are not on the same straight line. RB \(\displaystyle =5\), BC \(\displaystyle =\sqrt{85}\approx 9.22\), RC \(\displaystyle =\sqrt{202}\approx 14.21\), and \(\displaystyle \mathrm{RB}+\mathrm{BC}\approx 14.22 \neq \mathrm{RC}\); equivalently, the step \(\displaystyle (3,-4)\) from R to B would have to be followed by \(\displaystyle (6,-8)\), but the actual step from B to C is \(\displaystyle (6,-7)\). They form a very flat triangle, which is exactly why plotting alone cannot settle it.

Let the axes do the work.(Both parts have many correct answers; these are worked examples, and a different choice that meets the conditions is equally right.)(i) A right-angled isosceles triangle with a vertex at the origin.Such a triangle needs a right angle and two equal sides. The easiest way to guarantee the right angle is to lay the two equal sides along the axes, since the axes are perpendicular to each other by construction. Take\[\mathrm{O}(0,0), \qquad \mathrm{A}(4,0), \qquad \mathrm{B}(0,4). \] Check: \(\displaystyle 4^2 + 4^2 = 32 = (4\sqrt2)^2\). So OAB is right-angled at O and isosceles.Another valid answer, with the right angle not at the origin: \(\displaystyle \mathrm{O}(0,0)\), \(\displaystyle (3,3)\), \(\displaystyle (6,0)\). Here the two sides from \(\displaystyle (3,3)\) both have length \(\displaystyle \sqrt{3^2+3^2}=\sqrt{18}\), the third side is $\displaystyle 6$, and \(\displaystyle 18+18=36=6^2\), so the right angle is at \(\displaystyle (3,3)\).(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.The origin is the third vertex. Quadrant III needs both coordinates negative; Quadrant IV needs x positive and y negative. A neat way to force the two sides from O to be equal is to choose the two vertices as mirror images in the y-axis: reflecting in the y-axis does not change a point's distance from O, so the two distances come out equal automatically. Take\[\mathrm{O}(0,0), \qquad \mathrm{P}(-3,-4)\ \text{in Quadrant III}, \qquad \mathrm{Q}(3,-4)\ \text{in Quadrant IV}. \]\[\mathrm{OP} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9+16} = 5, \qquad \mathrm{OQ} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5. \]\[\mathrm{PQ} = \sqrt{\big(3-(-3)\big)^2 + \big(-4-(-4)\big)^2} = \sqrt{36+0} = 6. \]Since \(\displaystyle \mathrm{OP} = \mathrm{OQ} = 5\), the triangle is isosceles. It is a genuine triangle (not three points in a line), because P and Q lie on the line \(\displaystyle y = -4\) while O is $\displaystyle 4$ units above it.Answer: (i) for example \(\displaystyle \mathrm{O}(0,0)\), \(\displaystyle (4,0)\) and \(\displaystyle (0,4)\) - legs of $\displaystyle 4$ units along the two axes, right angle at O, hypotenuse \(\displaystyle 4\sqrt2\). (ii) for example \(\displaystyle \mathrm{O}(0,0)\), \(\displaystyle (-3,-4)\) in Quadrant III and \(\displaystyle (3,-4)\) in Quadrant IV, giving \(\displaystyle \mathrm{OP}=\mathrm{OQ}=5\) and \(\displaystyle \mathrm{PQ}=6\). Many other answers are correct; in (ii) any two points that are mirror images in the y-axis, one below-left and one below-right, will do.

The midpoint is the average of the two endpoints.The test to use. M is the midpoint of ST only if two things hold: M lies on the segment ST, and \(\displaystyle \mathrm{SM} = \mathrm{MT}\). Checking \(\displaystyle \mathrm{SM}=\mathrm{MT}\) alone is not enough - a point can be equally far from S and from T while sitting well off the line.The safest single check is arithmetic: work out the average of the endpoints, \[\left(\frac{x_\mathrm{S}+x_\mathrm{T}}{2},\ \frac{y_\mathrm{S}+y_\mathrm{T}}{2}\right), \] and see whether it is the given M. If it is, M is the midpoint; if it is not, M is not.Row by row.Row $\displaystyle 1$: S\(\displaystyle (-3,0)\), T\(\displaystyle (3,0)\). Average \(\displaystyle =\left(\frac{-3+3}{2}, \frac{0+0}{2}\right) = (0,0)\), which is M. All three points are on the x-axis, and \(\displaystyle \mathrm{SM}=3=\mathrm{MT}\). Yes.Row $\displaystyle 2$: S\(\displaystyle (2,3)\), T\(\displaystyle (4,5)\). Average \(\displaystyle =\left(\frac{2+4}{2}, \frac{3+5}{2}\right) = (3,4)\), which is M. Notice also that the step S \(\displaystyle \to\) M is \(\displaystyle (1,1)\) and the step M \(\displaystyle \to\) T is \(\displaystyle (1,1)\) - identical, so M is on ST and exactly halfway. Yes.Row $\displaystyle 3$: S\(\displaystyle (0,0)\), T\(\displaystyle (0,-10)\). Average \(\displaystyle =\left(0, \frac{0+(-10)}{2}\right) = (0,-5)\), but M is \(\displaystyle (0,5)\). No. All three are on the y-axis, but M is $\displaystyle 5$ units above S while T is $\displaystyle 10$ units below it, so M is not even between S and T; here \(\displaystyle \mathrm{SM}=5\) and \(\displaystyle \mathrm{MT}=15\).Row $\displaystyle 4$: S\(\displaystyle (-8,7)\), T\(\displaystyle (6,-3)\). Average \(\displaystyle =\left(\frac{-8+6}{2}, \frac{7+(-3)}{2}\right) = (-1,2)\), but M is \(\displaystyle (0,-2)\). No. (The distances are nowhere near equal either: \(\displaystyle \mathrm{SM}=\sqrt{8^2+9^2}=\sqrt{145}\approx 12.04\) while \(\displaystyle \mathrm{MT}=\sqrt{6^2+1^2}=\sqrt{37}\approx 6.08\).)

S	M	T	Is M the midpoint of ST?	Reason
\(\displaystyle (-3,0)\)	\(\displaystyle (0,0)\)	\(\displaystyle (3,0)\)	Yes	\(\displaystyle \left(\frac{-3+3}{2},\frac{0+0}{2}\right)=(0,0)=\mathrm{M}\); \(\displaystyle \mathrm{SM}=\mathrm{MT}=3\) on the x-axis
\(\displaystyle (2,3)\)	\(\displaystyle (3,4)\)	\(\displaystyle (4,5)\)	Yes	\(\displaystyle \left(\frac{2+4}{2},\frac{3+5}{2}\right)=(3,4)=\mathrm{M}\); equal steps \(\displaystyle (1,1)\) then \(\displaystyle (1,1)\)
\(\displaystyle (0,0)\)	\(\displaystyle (0,5)\)	\(\displaystyle (0,-10)\)	No	the midpoint is \(\displaystyle (0,-5)\), not \(\displaystyle (0,5)\); M is not even between S and T (\(\displaystyle \mathrm{SM}=5\), \(\displaystyle \mathrm{MT}=15\))
\(\displaystyle (-8,7)\)	\(\displaystyle (0,-2)\)	\(\displaystyle (6,-3)\)	No	the midpoint is \(\displaystyle (-1,2)\), not \(\displaystyle (0,-2)\); also \(\displaystyle \mathrm{SM}=\sqrt{145}\), \(\displaystyle \mathrm{MT}=\sqrt{37}\)

The connection between the coordinates. In the two "Yes" rows, each coordinate of M is the average of the corresponding coordinates of S and T:\[x_\mathrm{M} = \frac{x_\mathrm{S}+x_\mathrm{T}}{2}, \qquad y_\mathrm{M} = \frac{y_\mathrm{S}+y_\mathrm{T}}{2}. \]Why this must be so: travelling from S to T, the total sideways change is \(\displaystyle x_\mathrm{T}-x_\mathrm{S}\). Halfway along, you have made half of it, so \[x_\mathrm{M} = x_\mathrm{S} + \tfrac{1}{2}\left(x_\mathrm{T}-x_\mathrm{S}\right) = \tfrac{1}{2}\left(x_\mathrm{S}+x_\mathrm{T}\right), \] and the same argument on the vertical change gives the formula for \(\displaystyle y_\mathrm{M}\).Answer: Row $\displaystyle 1$ - Yes. Row $\displaystyle 2$ - Yes. Row $\displaystyle 3$ - No (the midpoint is \(\displaystyle (0,-5)\)). Row $\displaystyle 4$ - No (the midpoint is \(\displaystyle (-1,2)\)). The connection is that the midpoint's coordinates are the averages of the endpoints' coordinates: \(\displaystyle x_\mathrm{M}=\frac{x_\mathrm{S}+x_\mathrm{T}}{2}\) and \(\displaystyle y_\mathrm{M}=\frac{y_\mathrm{S}+y_\mathrm{T}}{2}\).

Read the midpoint rule backwards.From Problem $\displaystyle 9$, if M is the midpoint of AB then each coordinate of M is the average of the corresponding coordinates of A and B:\[x_\mathrm{M} = \frac{x_\mathrm{A}+x_\mathrm{B}}{2}, \qquad y_\mathrm{M} = \frac{y_\mathrm{A}+y_\mathrm{B}}{2}. \]Here we are told the answer of each average and one of the two numbers being averaged, so we can solve for the missing one.With \(\displaystyle \mathrm{M}(-7,1)\), \(\displaystyle \mathrm{A}(3,-4)\) and \(\displaystyle \mathrm{B}(x,y)\):x-coordinate. \[\frac{3+x}{2} = -7 \ \Longrightarrow\ 3+x = -14 \ \Longrightarrow\ x = -17. \]y-coordinate. \[\frac{-4+y}{2} = 1 \ \Longrightarrow\ -4+y = 2 \ \Longrightarrow\ y = 6. \]So \(\displaystyle \mathrm{B}(-17,\ 6)\).Check by going forwards again. The midpoint of A\(\displaystyle (3,-4)\) and B\(\displaystyle (-17,6)\) is \[\left(\frac{3+(-17)}{2},\ \frac{-4+6}{2}\right) = \left(\frac{-14}{2},\ \frac{2}{2}\right) = (-7,\ 1) = \mathrm{M}. \ \checkmark \]A second check, using distances: \(\displaystyle \mathrm{AM} = \sqrt{(-7-3)^2+(1-(-4))^2} = \sqrt{100+25} = \sqrt{125}\), and \(\displaystyle \mathrm{MB} = \sqrt{(-17-(-7))^2+(6-1)^2} = \sqrt{100+25} = \sqrt{125}\). Equal, as they must be.There is also a quick way to see the answer without algebra: to get from A to M you move \(\displaystyle -7-3 = -10\) in x and \(\displaystyle 1-(-4) = +5\) in y. Since M is halfway, you must repeat exactly the same step to reach B: \(\displaystyle -7-10 = -17\) and \(\displaystyle 1+5 = 6\).Answer: \(\displaystyle \mathrm{B} = (-17,\ 6)\).

Split the journey into three equal steps instead of two.The idea. The midpoint formula works because going from A to B is one journey: the total change is \(\displaystyle \big(x_\mathrm{B}-x_\mathrm{A},\ y_\mathrm{B}-y_\mathrm{A}\big)\), and the midpoint is where you have completed half of that change in x and half of it in y. Trisection points are the same idea with thirds. P, the point closer to A, is reached after one third of the journey; Q is reached after two thirds. So:\[\text{compute the total step } (\Delta x,\ \Delta y),\ \text{take one third of it, and add that step on twice.} \]Doing it for A\(\displaystyle (4,7)\) and B\(\displaystyle (16,-2)\).Total step from A to B: \[\Delta x = 16 - 4 = 12, \qquad \Delta y = -2 - 7 = -9. \]One third of the step: \[\left(\frac{12}{3},\ \frac{-9}{3}\right) = (4,\ -3). \]Add it once to reach P, and once more to reach Q: \[\mathrm{P} = (4+4,\ 7-3) = (8,\ 4), \qquad \mathrm{Q} = (8+4,\ 4-3) = (12,\ 1). \]And adding it a third time gives \(\displaystyle (16, -2) = \mathrm{B}\), which confirms the arithmetic.Written as a formula, this says \[\mathrm{P} = \left(\frac{2x_\mathrm{A}+x_\mathrm{B}}{3},\ \frac{2y_\mathrm{A}+y_\mathrm{B}}{3}\right), \qquad \mathrm{Q} = \left(\frac{x_\mathrm{A}+2x_\mathrm{B}}{3},\ \frac{y_\mathrm{A}+2y_\mathrm{B}}{3}\right), \] which for our points gives \(\displaystyle \mathrm{P}=\left(\frac{8+16}{3},\frac{14-2}{3}\right)=(8,4)\) and \(\displaystyle \mathrm{Q}=\left(\frac{4+32}{3},\frac{7-4}{3}\right)=(12,1)\), as before.Checking with midpoints - the link the question asks for. If P and Q really cut AB into three equal parts, then P must be the midpoint of A and Q, and Q must be the midpoint of P and B. Test both with the averaging rule:\[\text{midpoint of A}(4,7)\text{ and Q}(12,1) = \left(\frac{4+12}{2},\ \frac{7+1}{2}\right) = (8,\ 4) = \mathrm{P}\ \checkmark \]\[\text{midpoint of P}(8,4)\text{ and B}(16,-2) = \left(\frac{8+16}{2},\ \frac{4+(-2)}{2}\right) = (12,\ 1) = \mathrm{Q}\ \checkmark \]Checking with distances. Each of the three pieces should be the same length: \[\mathrm{AP} = \sqrt{4^2+(-3)^2} = 5, \quad \mathrm{PQ} = \sqrt{4^2+(-3)^2} = 5, \quad \mathrm{QB} = \sqrt{4^2+(-3)^2} = 5, \] and the whole segment is \(\displaystyle \mathrm{AB} = \sqrt{12^2+(-9)^2} = \sqrt{144+81} = \sqrt{225} = 15 = 3\times 5\). The three parts are equal and they add up to AB, so P and Q really are the points of trisection, with P nearer A.Answer: take one third of the total step from A to B and add it on twice. For A\(\displaystyle (4,7)\) and B\(\displaystyle (16,-2)\) the total step is \(\displaystyle (12,-9)\), one third of it is \(\displaystyle (4,-3)\), so \(\displaystyle \mathrm{P} = (8,\ 4)\) and \(\displaystyle \mathrm{Q} = (12,\ 1)\), with \(\displaystyle \mathrm{AP}=\mathrm{PQ}=\mathrm{QB}=5\) and \(\displaystyle \mathrm{AB}=15\).

A circle is the set of points at one fixed distance from its centre - so compare distances from O.(i) Showing A, B, C lie on one circle centred at O.If three points are all the same distance from O, then a circle drawn with centre O and that distance as radius passes through all three. So compute OA, OB and OC with the distance formula. A useful trick: compare the squares of the distances. That avoids square roots altogether and avoids any rounding, and since distances are never negative, equal squares mean equal distances.\[\mathrm{OA}^2 = (1-0)^2 + (-8-0)^2 = 1 + 64 = 65, \] \[\mathrm{OB}^2 = (-4-0)^2 + (7-0)^2 = 16 + 49 = 65, \] \[\mathrm{OC}^2 = (-7-0)^2 + (-4-0)^2 = 49 + 16 = 65. \]All three squares are $\displaystyle 65$, so \[\mathrm{OA} = \mathrm{OB} = \mathrm{OC} = \sqrt{65} \approx 8.06. \]Since A, B and C are all exactly \(\displaystyle \sqrt{65}\) units from O, they all lie on the circle K with centre O\(\displaystyle (0,0)\) and radius \(\displaystyle \sqrt{65}\).\[\text{radius of K} = \sqrt{65} \approx 8.06\ \text{units}. \](Notice the three points are in three different quadrants - A in IV, B in II, C in III - which is a nice reminder that the distance formula does not care about signs; only the squares matter.)(ii) Where D and E lie.The rule is simply: a point is inside K if its distance from O is less than the radius, on K if it equals the radius, and outside if it is greater. Again compare squares against \(\displaystyle 65\).For D\(\displaystyle (-5,6)\): \[\mathrm{OD}^2 = (-5)^2 + 6^2 = 25 + 36 = 61. \] \(\displaystyle 61 < 65\), so \(\displaystyle \mathrm{OD} = \sqrt{61} \approx 7.81 < \sqrt{65} \approx 8.06\). D lies inside the circle (though only just - about a quarter of a unit in from the edge).For E\(\displaystyle (0,9)\): \[\mathrm{OE}^2 = 0^2 + 9^2 = 81. \] \(\displaystyle 81 > 65\), so \(\displaystyle \mathrm{OE} = 9 > \sqrt{65} \approx 8.06\). E lies outside the circle. (E is on the y-axis, $\displaystyle 9$ units above O, while the circle crosses the y-axis at \(\displaystyle (0,\sqrt{65})\), about $\displaystyle 8.06$ units up - so E is roughly $\displaystyle 0.94$ units beyond it.)Answer: (i) \(\displaystyle \mathrm{OA}^2=\mathrm{OB}^2=\mathrm{OC}^2=65\), so A, B and C are all \(\displaystyle \sqrt{65}\) units from the origin and therefore lie on the circle K with centre O and radius \(\displaystyle \sqrt{65}\approx 8.06\) units. (ii) \(\displaystyle \mathrm{OD}^2=61<65\), so D\(\displaystyle (-5,6)\) lies inside K; \(\displaystyle \mathrm{OE}^2=81>65\), so E\(\displaystyle (0,9)\) lies outside K.

Turn each midpoint into an equation, then add the equations up.Let the vertices be \(\displaystyle \mathrm{A}(x_1,y_1)\), \(\displaystyle \mathrm{B}(x_2,y_2)\), \(\displaystyle \mathrm{C}(x_3,y_3)\), and take the usual labelling: D is the midpoint of BC, E of CA and F of AB.From the midpoint rule \(\displaystyle x_\mathrm{M} = \frac{x_\mathrm{S}+x_\mathrm{T}}{2}\), it is neater to clear the fraction at once and write "sum of the two endpoints \(\displaystyle =\) twice the midpoint".The x-coordinates. \[x_2 + x_3 = 2 \times 5 = 10 \qquad (\text{from D}(5,1)) \] \[x_3 + x_1 = 2 \times 6 = 12 \qquad (\text{from E}(6,5)) \] \[x_1 + x_2 = 2 \times 0 = 0 \qquad (\text{from F}(0,3)) \]Add all three. Each of \(\displaystyle x_1, x_2, x_3\) appears exactly twice on the left: \[2(x_1+x_2+x_3) = 10+12+0 = 22 \ \Longrightarrow\ x_1+x_2+x_3 = 11. \]Now subtract each original equation from this total: \[x_1 = 11 - (x_2+x_3) = 11 - 10 = 1, \quad x_2 = 11 - (x_3+x_1) = 11 - 12 = -1, \quad x_3 = 11 - (x_1+x_2) = 11 - 0 = 11. \]The y-coordinates, exactly the same way. \[y_2+y_3 = 2, \qquad y_3+y_1 = 10, \qquad y_1+y_2 = 6. \] \[2(y_1+y_2+y_3) = 2+10+6 = 18 \ \Longrightarrow\ y_1+y_2+y_3 = 9, \] \[y_1 = 9-2 = 7, \qquad y_2 = 9-10 = -1, \qquad y_3 = 9-6 = 3. \]So \[\mathrm{A}(1,\,7), \qquad \mathrm{B}(-1,\,-1), \qquad \mathrm{C}(11,\,3). \]Check all three midpoints. \[\text{midpoint of BC} = \left(\frac{-1+11}{2},\ \frac{-1+3}{2}\right) = (5,1) = \mathrm{D}\ \checkmark \] \[\text{midpoint of CA} = \left(\frac{11+1}{2},\ \frac{3+7}{2}\right) = (6,5) = \mathrm{E}\ \checkmark \] \[\text{midpoint of AB} = \left(\frac{1+(-1)}{2},\ \frac{7+(-1)}{2}\right) = (0,3) = \mathrm{F}\ \checkmark \]A remark on the labelling. The question does not say which midpoint belongs to which side, and it does not need to. Working through the algebra above shows that each vertex comes out as \[(\text{sum of the other two midpoints}) - (\text{the opposite midpoint}), \] for instance \(\displaystyle \mathrm{A} = \mathrm{E}+\mathrm{F}-\mathrm{D} = (6+0-5,\ 5+3-1) = (1,7)\). Pairing the midpoints with the sides differently only swaps the names A, B, C around; the three vertices themselves always come out as the same three points.Answer: \(\displaystyle \mathrm{A}(1,\,7)\), \(\displaystyle \mathrm{B}(-1,\,-1)\) and \(\displaystyle \mathrm{C}(11,\,3)\).

Read the city plan as a coordinate grid.The two main roads cross at the centre of the city, one running North–South, the other East–West. Every other street runs parallel to one of these two, so the whole city is a rectangular grid: a family of N–S lines and a family of E–W lines, with neighbouring parallel lines \(\displaystyle 200\) m apart. Read "there are $\displaystyle 10$ streets in each direction" as ten lines running N–S and ten running E–W, the two main roads being among them.(i) Drawing the model.The scale is \(\displaystyle 1\) cm \(\displaystyle =200\) m. So a real gap of \(\displaystyle 200\) m must be drawn as \[\frac{200\ \text{m}}{200\ \text{m per cm}} = 1\ \text{cm}. \] Every pair of neighbouring streets is therefore \(\displaystyle 1\) cm apart on paper. Then: Ten equally spaced lines leave \(\displaystyle 10-1=9\) gaps between them, so the grid measures \(\displaystyle 9\) cm by \(\displaystyle 9\) cm on paper. Since \(\displaystyle 9\) cm stands for \(\displaystyle 9\times 200 = 1800\) m, the model represents a square of city \(\displaystyle 1800\) m \(\displaystyle \times\) \(\displaystyle 1800\) m — a useful check that your drawing fits on a notebook page.(ii) Naming the street intersections.Number the ten N–S streets \(\displaystyle 1, 2, 3, \dots, 10\) in order from west to east, and number the ten E–W streets \(\displaystyle 1, 2, 3, \dots, 10\) in order from south to north. Then the label \(\displaystyle (m, n)\) means:\[(m,\,n) \;=\; \text{the crossing of the } m\text{-th N–S street with the } n\text{-th E–W street}. \]For example, \(\displaystyle (2, 5)\) is the single point where the 2nd street in the N–S direction meets the 5th street in the E–W direction. This is exactly how coordinates work: the first entry tells you how far east you have gone, the second how far north.Two facts follow straight from this, and they are what the counting parts of the question turn on: CANNOT SOLVE: needs the rest of the stem — the sentence "Using this convention, find:" is where the printed sub-parts begin, and those sub-parts are missing from the text supplied, so there is nothing specific left to count.Answer. (i) A \(\displaystyle 9\) cm \(\displaystyle \times\) \(\displaystyle 9\) cm grid of \(\displaystyle 10\) vertical and \(\displaystyle 10\) horizontal lines drawn \(\displaystyle 1\) cm apart, with the two main roads drawn thicker. (ii) With the streets numbered \(\displaystyle 1\) to \(\displaystyle 10\) in each direction, \(\displaystyle (m,n)\) names the crossing of the \(\displaystyle m\)-th N–S street with the \(\displaystyle n\)-th E–W street; there are \(\displaystyle 100\) such crossings, and \(\displaystyle (m,n)\) and \(\displaystyle (n,m)\) are different crossings whenever \(\displaystyle m\neq n\). The specific items to be found are missing from the question text.

Compare distances with radii.Because the origin sits at the bottom-left corner, the screen is the rectangle \[0 \le x \le 800, \qquad 0 \le y \le 600, \] and its four edges are the lines \(\displaystyle x=0\) (left), \(\displaystyle x=800\) (right), \(\displaystyle y=0\) (bottom) and \(\displaystyle y=600\) (top).(i) Does either circle spill off the screen?A circle with centre \(\displaystyle (h,k)\) and radius \(\displaystyle r\) reaches exactly \(\displaystyle r\) pixels away from its centre in every direction. So it stays completely inside an edge precisely when the centre is more than \(\displaystyle r\) away from that edge. From the equations of the edges,\[\text{distance to left edge}=h,\quad \text{to right edge}=800-h,\quad \text{to bottom}=k,\quad \text{to top}=600-k. \]Circle A, centre \(\displaystyle A(100,150)\), radius \(\displaystyle 80\): \[h = 100,\quad 800-h = 700,\quad k = 150,\quad 600-k = 450. \] The smallest of these is \(\displaystyle 100\), and \(\displaystyle 100 > 80\). So circle A clears every edge.The same thing said with extreme points: the circle runs from \(\displaystyle x=100-80=20\) to \(\displaystyle x=100+80=180\), and from \(\displaystyle y=150-80=70\) to \(\displaystyle y=150+80=230\). Both ranges sit inside \(\displaystyle [0,800]\) and \(\displaystyle [0,600]\).Circle B, centre \(\displaystyle B(250,230)\), radius \(\displaystyle 100\): \[h = 250,\quad 800-h = 550,\quad k = 230,\quad 600-k = 370. \] The smallest is \(\displaystyle 230\), and \(\displaystyle 230 > 100\). So circle B clears every edge too. Its extreme points are \(\displaystyle x\) from \(\displaystyle 150\) to \(\displaystyle 350\) and \(\displaystyle y\) from \(\displaystyle 130\) to \(\displaystyle 330\) — again inside the screen.So no part of either circle lies outside the screen.(ii) Do the two circles meet?Whether two circles meet depends only on how far apart their centres are compared with their radii, so first find \(\displaystyle AB\).Go from \(\displaystyle A(100,150)\) horizontally across to the point \(\displaystyle (250,150)\), then vertically up to \(\displaystyle B(250,230)\). The first leg is along a horizontal line and the second along a vertical line, so they meet at a right angle, and the two legs are \[250-100 = 150 \quad\text{and}\quad 230-150 = 80 . \] By the Pythagoras theorem in that right triangle, \[AB^{2} = 150^{2} + 80^{2} = 22500 + 6400 = 28900, \] \[AB = \sqrt{28900} = 170 \text{ pixels.} \]Now compare with the radii \(\displaystyle r_A = 80\) and \(\displaystyle r_B = 100\): \[r_B - r_A = 20, \qquad r_A + r_B = 180, \qquad 20 < 170 < 180 . \] Between those two limits the circles must cross, and they cross at two points.A check you can do without that rule. Walk along the straight line from \(\displaystyle A\) towards \(\displaystyle B\). Circle B's near point on this line is at distance \(\displaystyle 170-100=70\) from \(\displaystyle A\), and \(\displaystyle 70 < 80\), so that point is inside circle A. Circle B's far point on this line is at distance \(\displaystyle 170+100 = 270\) from \(\displaystyle A\), and \(\displaystyle 270 > 80\), so that point is outside circle A. A circle with some points inside A and some outside A has to cross A's boundary — twice, once on each side. This agrees with the comparison above.Answer. (i) No — both icons are drawn completely on the screen; circle A stays at least \(\displaystyle 100-80=20\) pixels clear of the nearest edge and circle B at least \(\displaystyle 230-100=130\) pixels clear. (ii) Yes — the distance between the centres is \(\displaystyle AB = 170\) pixels, and since \(\displaystyle 20 < 170 < 180\), the two circular icons overlap, meeting at two points.

All four sides equal, then one right angle by the converse of the Pythagoras theorem.Plotting. \(\displaystyle A(2,1)\) has both coordinates positive, so it is in the first quadrant. \(\displaystyle B(-1,2)\) has \(\displaystyle x\) negative and \(\displaystyle y\) positive — second quadrant. \(\displaystyle C(-2,-1)\) has both negative — third quadrant. \(\displaystyle D(1,-2)\) has \(\displaystyle x\) positive and \(\displaystyle y\) negative — fourth quadrant. Joining \(\displaystyle A \to B \to C \to D \to A\) takes you once round the origin in the anticlockwise sense, so the four segments do not cross one another and \(\displaystyle ABCD\) really is a quadrilateral.Lengths from coordinates. For any two points, go across and then up (or down): the horizontal step and the vertical step form the legs of a right triangle whose hypotenuse is the segment itself. So \[(\text{length})^{2} = (\text{horizontal step})^{2} + (\text{vertical step})^{2}. \]

Side	horizontal step	vertical step	(length)\(\displaystyle ^2\)
\(\displaystyle AB\): \(\displaystyle (2,1)\to(-1,2)\)	\(\displaystyle |-1-2| = 3\)	\(\displaystyle |2-1| = 1\)	\(\displaystyle 9+1 = 10\)
\(\displaystyle BC\): \(\displaystyle (-1,2)\to(-2,-1)\)	\(\displaystyle |-2+1| = 1\)	\(\displaystyle |-1-2| = 3\)	\(\displaystyle 1+9 = 10\)
\(\displaystyle CD\): \(\displaystyle (-2,-1)\to(1,-2)\)	\(\displaystyle |1+2| = 3\)	\(\displaystyle |-2+1| = 1\)	\(\displaystyle 9+1 = 10\)
\(\displaystyle DA\): \(\displaystyle (1,-2)\to(2,1)\)	\(\displaystyle |2-1| = 1\)	\(\displaystyle |1+2| = 3\)	\(\displaystyle 1+9 = 10\)

So \[AB = BC = CD = DA = \sqrt{10}. \] All four sides are equal, so \(\displaystyle ABCD\) is at least a rhombus. That is not yet enough — a rhombus can be squashed — so we need an angle.Finding a right angle. The diagonal \(\displaystyle AC\) runs from \(\displaystyle (2,1)\) to \(\displaystyle (-2,-1)\): horizontal step \(\displaystyle 4\), vertical step \(\displaystyle 2\), so \[AC^{2} = 4^{2} + 2^{2} = 16 + 4 = 20 . \] Now look at triangle \(\displaystyle ABC\): \[AB^{2} + BC^{2} = 10 + 10 = 20 = AC^{2}. \] By the converse of the Pythagoras theorem, a triangle whose two shorter sides satisfy this relation is right-angled at the vertex between them. So \(\displaystyle \angle ABC = 90^{\circ}\).A rhombus with one right angle has all four angles right: opposite angles of a rhombus are equal and adjacent angles add to \(\displaystyle 180^{\circ}\), so \(\displaystyle \angle B = 90^{\circ}\) forces \(\displaystyle \angle A = \angle C = \angle D = 90^{\circ}\).Four equal sides and four right angles: \(\displaystyle ABCD\) is a square.Area. \[\text{Area} = (\text{side})^{2} = \left(\sqrt{10}\right)^{2} = 10 \text{ square units.} \]A second, independent check — through the diagonals.Midpoint of \(\displaystyle AC\): \[\left(\frac{2+(-2)}{2},\ \frac{1+(-1)}{2}\right) = (0,0). \] Midpoint of \(\displaystyle BD\): \[\left(\frac{-1+1}{2},\ \frac{2+(-2)}{2}\right) = (0,0). \] The two diagonals have the same midpoint, the origin, so they bisect each other — that alone makes \(\displaystyle ABCD\) a parallelogram. Their lengths are \[AC^{2} = 4^{2}+2^{2} = 20, \qquad BD^{2} = 2^{2}+4^{2} = 20, \] so \(\displaystyle AC = BD = \sqrt{20} = 2\sqrt5\): equal diagonals, which upgrades the parallelogram to a rectangle. A rectangle with equal sides is a square — the same conclusion as before.For a square, area \(\displaystyle =\tfrac12 \times d_1 \times d_2\) where \(\displaystyle d_1,d_2\) are the diagonals, so \[\text{Area} = \tfrac12 \times \sqrt{20} \times \sqrt{20} = \tfrac12 \times 20 = 10, \] matching the \(\displaystyle 10\) found from the side. Two different routes give the same number, which is a good sign the answer is right.A remark worth noticing. Each vertex is the same distance from the origin: \[2^2+1^2 = 5,\quad (-1)^2+2^2 = 5,\quad (-2)^2+(-1)^2 = 5,\quad 1^2+(-2)^2 = 5, \] so all four points lie on the circle of radius \(\displaystyle \sqrt5\) about \(\displaystyle O\), and \(\displaystyle A, C\) are opposite ends of one diameter while \(\displaystyle B, D\) are opposite ends of another. That is exactly the picture of a square inscribed in a circle, with the centre of the circle at the centre of the square.Answer. Yes, \(\displaystyle ABCD\) is a square — all four sides measure \(\displaystyle \sqrt{10}\) units and \(\displaystyle AB^2+BC^2=AC^2\) gives a right angle at \(\displaystyle B\) (equivalently, the diagonals \(\displaystyle AC\) and \(\displaystyle BD\) are equal, each \(\displaystyle 2\sqrt5\), and bisect each other at the origin). Its area is \(\displaystyle 10\) square units.