Key Concepts & Formulae at a Glance
- A 2-D coordinate system uses two perpendicular lines — the horizontal x-axis and the vertical y-axis — to locate any point in a plane, the Cartesian plane.
- The point where the axes meet is the origin O, with coordinates (0, 0).
- A point's coordinates (x, y): x is its perpendicular distance from the y-axis (measured along the x-axis); y is its perpendicular distance from the x-axis (measured along the y-axis).
- The axes divide the plane into four quadrants: Quadrant I (+, +), Quadrant II (–, +), Quadrant III (–, –), Quadrant IV (+, –).
- Points on the x-axis have the form (x, 0); points on the y-axis have the form (0, y).
- If x = y, then (x, y) = (y, x). If x ≠ y, then (x, y) ≠ (y, x) — order matters.
- Distance between two points on a horizontal or vertical line is simply the absolute difference of their unequal coordinate.
This chapter opens with the history of coordinate systems — from the grid-planned streets of the Sindhu-Sarasvatī cities and Baudhāyana's East–West/North–South constructions, through Āryabhaṭa, Brahmagupta's formalisation of zero and negative numbers, Al-Bīrūnī, and finally Descartes' four-quadrant plane in 1637 CE.
The chapter then introduces Reiaan and Shalini, who use a pin-and-thread grid (Fig. 1.1) to map Reiaan's room so he can feel its layout — a physical, hands-on version of a coordinate system, using a scale of 1 cm : 1 foot.
Why can't the windows be marked on this map? Fig. 1.1 is a floor plan — a two-dimensional, top-down (bird's-eye) view of the room. It only records positions along the floor's length and breadth. A window is set into a wall at a certain height above the floor, so showing it needs a third piece of information (height) that a flat floor plan simply doesn't capture. To mark the windows, you would need a side view (elevation) of the walls, or a full 3-D coordinate system with a third axis for height.
The horizontal line is the x-axis and the vertical line is the y-axis. Their intersection, the origin O, has coordinates (0, 0). Distances to the right of O or upward from O are positive; distances to the left of O or downward from O are negative.
Points on the axes: B(4.5,0) on the x-axis, H(0,4) and G(0,-4.5) on the y-axis, and E(-2.9,0) on the x-axis.
A point P = (x, 0) lies on the x-axis (right of O if x > 0, left of O if x < 0). A point P = (0, y) lies on the y-axis (above O if y > 0, below O if y < 0).
Exercise Set 1.1
Based on Fig. 1.3 — Reiaan's room, with corners O(0,0), A(12,0), B(12,10), C(0,10); wardrobe corners W1(3,1), W2(7,1), W3(7,2), W4(3,2); bed corners S1(1,5), S2(7,5), S3(7,8), S4(1,8); room door D1R1; bathroom door B1B2.
Fig. 1.3: Reiaan's room — bed (blue), wardrobe (orange), room door D1R1 and bathroom door B1B2 (both in red).
1If D1R1 represents the door to Reiaan's room, how far is the door from the left wall (the y-axis)? How far is the door from the x-axis? What are the coordinates of D1? If R1 is the point (11.5, 0), how wide is the door? Is this a comfortable width? If B1 (0, 1.5) and B2 (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?
(i) The door D1R1 lies along the bottom wall of the room, which is the x-axis itself (y = 0), so its distance from the x-axis is 0. Its near end, D1, is 8 ft from the left wall (the y-axis), since D1 = (8, 0).
(ii) Coordinates of D1 = (8, 0).
(iii) Width of the door = distance D1R1 = x-coordinate of R1 − x-coordinate of D1 = \(11.5 - 8 = 3.5\) ft. A standard interior door is around 2.5–3 ft wide, and accessible doorways need at least about 2.7 ft (32 inches) of clear width, so a 3.5 ft door is generous — comfortable to walk through, and wide enough for a person using a wheelchair to enter easily.
(iv) Width of the bathroom door = distance B1B2 = \(4 - 1.5 = 2.5\) ft. Since \(2.5 \text{ ft} < 3.5 \text{ ft}\), the bathroom door is narrower than the room door.
Think and Reflect
TR1. What are the standard widths for a room door? Look around your home and in school. 2. Are the doors in your school suitable for people in wheelchairs?
1. Standard interior room doors are usually about 2.5 ft to 3 ft (30–36 inches) wide; main entrance doors are often wider, around 3.5 ft (42 inches). Measuring the doors around your own home and school is the best way to check these figures against real life.
2. A doorway needs a clear width of roughly 2.7 ft (32 inches) or more, ideally 3 ft (36 inches), for a wheelchair to pass through comfortably. Whether your school's doors meet this depends on the actual doors — this is an open, observation-based question for you to check and discuss in class.
Fig. 1.4: Q(-5,3) lies in Quadrant II; S(3,-5) lies in Quadrant IV.
Think and Reflect
TR1. What is the x-coordinate of a point on the y-axis? 2. Is there a similar generalisation for a point on the x-axis? 3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer. 4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?
1. Every point on the y-axis has x-coordinate 0 — it has zero perpendicular distance from the y-axis, since it lies on it.
2. Yes — by the same reasoning, every point on the x-axis has y-coordinate 0.
3. Yes, but only when x = y. Points P(x, y) and Q(y, x) coincide exactly when their corresponding coordinates match: the x-coordinate of P (which is x) must equal the x-coordinate of Q (which is y), i.e. \(x = y\). When this holds, both points reduce to the same pair, e.g. P(3,3) = Q(3,3).
4. This is true. Two ordered pairs (x, y) and (y, x) are equal only when corresponding entries match: first entries equal (\(x=y\)) and second entries equal (\(y=x\)) — both conditions are really the same single condition, \(x=y\). So if \(x \ne y\), the pairs (x,y) and (y,x) genuinely represent two different points (e.g. (3,5) and (5,3) are different locations), and they coincide only in the special case x = y.
Exercise Set 1.2
On a graph sheet, mark the x-axis from (–7, 0) to (13, 0) and the y-axis from (0, –15) to (0, 12), using the scale 1 cm = 1 unit. Fig. 1.5 adds a bathroom (6 ft × 9 ft) to the left of Reiaan's bedroom from Fig. 1.3, with corners O(0,0), F(0,9), R(–6,9), P(–6,0), and a showering area SHWR.
Fig. 1.5: bedroom plus bathroom (light blue), with the showering area, washbasin and toilet space marked inside it.
1Place Reiaan's rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7). (i) Where will the fourth foot of the table be? (ii) Is this a good spot for the table? (iii) What is the width of the table? The length? Can you make out the height of the table?
Table ABCD placed clear of the bed, both inside the room boundary
Let the three given feet be A(8,9), B(11,9), C(11,7). AB is horizontal (both have y = 9), and BC is vertical (both have x = 11) — so ABC is already a right angle at B, and the table is a rectangle.
(i) The fourth foot D must share its x-coordinate with A (8) and its y-coordinate with C (7), completing the rectangle: \(D = (8, 7)\).
(ii) The table occupies x from 8 to 11 and y from 7 to 9. The bed (Fig. 1.3) occupies x from 1 to 7 and y from 5 to 8 — since the table's smallest x-value (8) is greater than the bed's largest x-value (7), the table doesn't overlap the bed. The whole table also lies comfortably inside the room boundary (0 ≤ x ≤ 12, 0 ≤ y ≤ 10). So yes, this is a good, clear spot for the table.
(iii) Width = \(BC = |9-7| = 2\) ft. Length = \(AB = |11-8| = 3\) ft. The height of the table cannot be determined — a floor plan is a top-down 2-D view and carries no information about how tall an object is; you'd need a side view (elevation) for that.
2If the bathroom door has a hinge at B1 and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?
The bathroom door has hinge at \(B_1 = (0, 1.5)\), and its width is 2.5 ft (from Exercise Set 1.1). Swinging fully open into the bedroom, the door sweeps out a quarter-circle arc of radius 2.5 ft, centred at \(B_1\).
The wardrobe's nearest corner to the hinge is W1 = (3, 1). The distance from the hinge to this corner is:
\(B_1W_1 = \sqrt{(3-0)^2+(1-1.5)^2} = \sqrt{9+0.25} = \sqrt{9.25} \approx 3.04 \text{ ft}\)
Since the wardrobe's nearest corner (≈3.04 ft away) is farther than the door's swing radius (2.5 ft), the door's arc doesn't reach the wardrobe — it will not hit it, with about 0.54 ft of clearance to spare.
If the door were made wider, the clearance would shrink: at a width beyond about 3.04 ft, the swinging door would begin to strike the wardrobe's corner. So if the door is widened, the wardrobe would need to be moved slightly farther from the bathroom wall (or the door could be hinged to swing the other way) to keep it clear.
3(i) What are the coordinates of the four corners O, F, R, and P of the bathroom? (ii) What is the shape of the showering area SHWR? Write the coordinates of its four corners. (iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.
Bathroom layout: shower (top-left), washbasin and toilet along the bottom, all measured from origin O
(i) The bathroom measures 6 ft × 9 ft (as given), sharing its bottom-right corner with the bedroom's origin. Its four corners are: \(O=(0,0)\), \(F=(0,9)\), \(R=(-6,9)\), \(P=(-6,0)\).
(ii) The showering area SHWR sits in the upper part of the bathroom, sharing corner R with the bathroom outline. It is a rectangle with corners \(S=(-6,5)\), \(H=(-2,5)\), \(W=(-2,9)\), \(R=(-6,9)\) — a 4 ft × 4 ft square region.
(iii) In the remaining lower portion of the bathroom (from y = 0 to y = 5), place the washbasin (3 ft × 2 ft) against the left wall and the toilet (2 ft × 3 ft) next to it:
Washbasin corners: \((-6,0),\ (-3,0),\ (-3,2),\ (-6,2)\)
Toilet corners: \((-3,0),\ (-1,0),\ (-1,3),\ (-3,3)\)
These two spaces together use 5 ft of the bathroom's 6 ft width and stay below y = 5, so they don't overlap the showering area or each other.
4(i) Reiaan's room door leads from the dining room (18 ft length, 15 ft width), whose length extends from point P to point A. Sketch the dining room and mark the coordinates of its corners. (ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.
Dining room from P to A along the x-axis, extending downward into negative y; dining table centred inside
(i) P = (–6, 0) and A = (12, 0) both lie on the x-axis, and the distance between them is \(12-(-6)=18\) ft — exactly the given length of the dining room, confirming it runs along the x-axis from P to A. Since the bedroom and bathroom occupy the region above the x-axis (y ≥ 0), the dining room (width 15 ft) must extend downward, into negative y:
Dining room corners: \(P(-6,0)\), \(A(12,0)\), \((12,-15)\), \((-6,-15)\).
(ii) The centre of this 18 ft × 15 ft rectangle is the average of opposite corners: centre \(=\left(\dfrac{-6+12}{2}, \dfrac{0+(-15)}{2}\right) = (3,-7.5)\).
Placing a 5 ft × 3 ft table centred here (2.5 ft either side along x, 1.5 ft either side along y):
Table corners: \((0.5,-9),\ (5.5,-9),\ (5.5,-6),\ (0.5,-6)\)
Distances along or parallel to the axes are easy to find directly. For a segment that isn't parallel to either axis, we build a right triangle using the horizontal and vertical shifts, then apply the Baudhāyana–Pythagoras Theorem.
Triangle ADM in Quadrant I: A(3,4), D(7,1), M(9,6), with the construction point C(3,1) used to find AD.
Think and Reflect
TR1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis? 2. Can these distances help you find the distance AD?
1. Along the x-axis, the shift is \(7-3=4\) units. Along the y-axis, the shift is \(4-1=3\) units (moving downward from y = 4 to y = 1).
2. Yes — these two shifts are the legs of a right triangle whose hypotenuse is exactly the segment AD (using the construction point C(3,1) as the right-angle vertex). By the Baudhāyana–Pythagoras Theorem: \(AD=\sqrt{4^2+3^2}=\sqrt{25}=5\) units.
Triangle AMD reflected in the y-axis: A(3,4)→A'(-3,4), D(7,1)→D'(-7,1), M(9,6)→M'(-9,6). Side lengths are unchanged by the reflection.
Think and Reflect
TR1. What has remained the same and what has changed with this reflection? 2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)?
1. Reflecting in the y-axis negates every x-coordinate while leaving y-coordinates unchanged: A(3,4)→A'(-3,4), D(7,1)→D'(-7,1), M(9,6)→M'(-9,6). Since the distance formula depends only on the squares of the coordinate differences, and squaring removes the effect of a sign flip, all three side lengths stay exactly the same (AD = A'D' = 5, DM = D'M' = √29, MA = M'A' = √40). What changes is the triangle's position and orientation — it now sits as a mirror image in Quadrant II instead of Quadrant I.
2. Yes, the same reasoning applies. Reflecting in the x-axis instead negates the y-coordinates and leaves x-coordinates unchanged. Since squaring again removes the sign change, all side lengths would still be preserved. Only the triangle's position changes — it would now appear as a mirror image below the x-axis (in Quadrant IV) rather than above it.
End-of-Chapter Exercises
1What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
The two axes intersect at the origin, which is defined to have coordinates (0, 0).
2Point W has x-coordinate equal to –5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?
A line parallel to the y-axis is a vertical line, on which every point shares the same x-coordinate. Since W has x-coordinate –5, every point on this line — including H — also has x-coordinate –5. So \(H=(-5,y)\) for some value of y.
If \(y>0\), H lies in Quadrant II (–, +). If \(y<0\), H lies in Quadrant III (–, –). If \(y=0\), H lies on the x-axis itself, not inside a quadrant.
3Consider the points R (3, 0), A (0, –2), M (–5, –2) and P (–5, 2). If they are joined in the same order, predict: (i) Two sides of RAMP that are perpendicular to each other. (ii) One side of RAMP that is parallel to one of the axes. (iii) Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.
Quadrilateral RAMP: R(3,0), A(0,-2), M(-5,-2), P(-5,2).
(i) Side AM runs from A(0,–2) to M(–5,–2): both points share y = –2, so AM is horizontal. Side MP runs from M(–5,–2) to P(–5,2): both points share x = –5, so MP is vertical. A horizontal side and a vertical side are always perpendicular, so AM ⊥ MP.
(ii) AM (y = –2 throughout) is parallel to the x-axis; MP (x = –5 throughout) is parallel to the y-axis. Either is a valid answer.
(iii) M(–5,–2) and P(–5,2) share the same x-coordinate (–5), while their y-coordinates are equal in magnitude but opposite in sign (–2 and 2). This is exactly the condition for two points to be reflections of each other in the x-axis.
4Plot point Z (5, –6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. (Comment: Answers may differ from person to person.)
One valid right triangle IZN with the right angle at Z
Since answers can vary, here is one valid construction: choose \(I=(5,0)\), directly above Z on the x-axis, and \(N=(0,-6)\), directly left of Z on the y-axis. Then IZ is vertical and ZN is horizontal, so the right angle sits at Z.
\(IZ = |0-(-6)| = 6\) units
\(ZN = |5-0| = 5\) units
\(IN = \sqrt{(5-0)^2+(0-(-6))^2} = \sqrt{25+36} = \sqrt{61}\) units
5What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?
Without negative numbers, both coordinates of every point would have to be zero or positive. This restricts every locatable point to Quadrant I (together with the non-negative parts of the two axes) — just one quarter of the full plane.
Points in Quadrant II, III, or IV (any point with at least one negative coordinate) could not be assigned coordinates at all. So no, such a system would not allow us to locate every point on the 2-D plane — it was precisely Brahmagupta's formalisation of negative numbers that made the full four-quadrant Cartesian plane possible.
6*Are the points M (–3, –4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.
M(-3,-4), A(0,0) and G(6,8) all lie on the same straight line through the origin.
Method: Three points are collinear if the "steepness" (ratio of vertical shift to horizontal shift) from one point to another is the same for every pair sharing a common point — i.e. if the slope from A to M equals the slope from A to G.
Slope of AM \(=\dfrac{-4-0}{-3-0}=\dfrac{-4}{-3}=\dfrac{4}{3}\)
Slope of AG \(=\dfrac{8-0}{6-0}=\dfrac{8}{6}=\dfrac{4}{3}\)
Since both slopes equal \(\dfrac{4}{3}\), and both lines pass through the common point A, all three points lie on the same straight line.
7*Use your method (from Problem 6) to check if the points R (–5, –1), B (–2, –5) and C (4, –12) are on the same straight line. Now plot both sets of points and check your answers.
R(-5,-1), B(-2,-5) and C(4,-12): segment RB and segment BC point in visibly different directions — not collinear.
Slope of RB \(=\dfrac{-5-(-1)}{-2-(-5)}=\dfrac{-4}{3}\)
Slope of BC \(=\dfrac{-12-(-5)}{4-(-2)}=\dfrac{-7}{6}\)
Since \(\dfrac{-4}{3} \ne \dfrac{-7}{6}\) (that is, \(-1.33\overline{3} \ne -1.1\overline{6}\)), the slopes differ, so R, B and C do not lie on the same straight line.
8*Using the origin as one vertex, plot the vertices of: (i) A right-angled isosceles triangle. (ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
Two valid answers — any triangle with the stated properties is acceptable
(i) Choose O(0,0), (4,0) and (0,4). OA and OB are perpendicular (one along each axis) and equal in length (both 4 units), giving a right angle at O with two equal legs — a right-angled isosceles triangle.
(ii) Choose O(0,0) as the third vertex, with one vertex \((-3,-4)\) in Quadrant III and another \((3,-4)\) in Quadrant IV. Both of these are the same distance from O: \(\sqrt{3^2+4^2}=\sqrt{9+16}=5\) each — so the triangle formed with O is isosceles (two sides of length 5).
(ii) O(0,0), (–3,–4) in Quadrant III, (3,–4) in Quadrant IV — isosceles, with both slanted sides equal to 5 units.
9*The table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.
M is the midpoint of ST exactly when its coordinates are the average of the corresponding coordinates of S and T: \(M=\left(\dfrac{x_S+x_T}{2},\dfrac{y_S+y_T}{2}\right)\).
| S | M | T | Midpoint of ST? | Reason |
|---|---|---|---|---|
| (–3, 0) | (0, 0) | (3, 0) | Yes | \(\left(\frac{-3+3}{2},\frac{0+0}{2}\right)=(0,0)=M\) |
| (2, 3) | (3, 4) | (4, 5) | Yes | \(\left(\frac{2+4}{2},\frac{3+5}{2}\right)=(3,4)=M\) |
| (0, 0) | (0, 5) | (0, –10) | No | \(\left(\frac{0+0}{2},\frac{0-10}{2}\right)=(0,-5)\ne(0,5)\) |
| (–8, 7) | (0, –2) | (6, –3) | No | \(\left(\frac{-8+6}{2},\frac{7-3}{2}\right)=(-1,2)\ne(0,-2)\) |
10*Use the connection you found to find the coordinates of B given that M (–7, 1) is the midpoint of A (3, –4) and B (x, y).
Using the midpoint connection from Problem 9: \(-7=\dfrac{3+x}{2}\) and \(1=\dfrac{-4+y}{2}\).
\(3+x=-14 \Rightarrow x=-17\)
\(-4+y=2 \Rightarrow y=6\)
11*Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, –2).
P is one-third of the way from A to B, and Q is two-thirds of the way from A to B. So \(P=A+\dfrac{1}{3}(B-A)\) and \(Q=A+\dfrac{2}{3}(B-A)\).
\(B-A = (16-4,\ -2-7) = (12,-9)\)
\(P = (4,7)+\dfrac{1}{3}(12,-9) = (4+4,\ 7-3) = (8,4)\)
\(Q = (4,7)+\dfrac{2}{3}(12,-9) = (4+8,\ 7-6) = (12,1)\)
12*(i) Given the points A (1, –8), B (–4, 7) and C (–7, –4), show that they lie on a circle K whose centre is the origin O (0, 0). What is the radius of circle K? (ii) Given the points D (–5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.
Circle K, radius √65, centred at O, passing through A, B and C. D lies inside it; E lies outside it.
(i) A point lies on a circle centred at O exactly when its distance from O equals the circle's radius. Computing OA, OB, OC:
\(OA=\sqrt{1^2+(-8)^2}=\sqrt{1+64}=\sqrt{65}\)
\(OB=\sqrt{(-4)^2+7^2}=\sqrt{16+49}=\sqrt{65}\)
\(OC=\sqrt{(-7)^2+(-4)^2}=\sqrt{49+16}=\sqrt{65}\)
All three distances equal \(\sqrt{65}\), so A, B and C all lie on a circle of radius \(\sqrt{65}\) centred at O.
(ii) \(OD=\sqrt{(-5)^2+6^2}=\sqrt{25+36}=\sqrt{61}\). Since \(\sqrt{61} < \sqrt{65}\), D lies inside the circle.
\(OE=\sqrt{0^2+9^2}=\sqrt{81}=9\). Since \(9 = \sqrt{81} > \sqrt{65} \approx 8.06\), E lies outside the circle.
13*The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C.
△ABC (solid) with its midpoint triangle △DEF (dashed)
Take D, E, F as the midpoints of BC, CA and AB respectively. Each vertex can be recovered from the other two midpoints minus the opposite one — for instance, since \(E=\frac{A+C}{2}\) and \(F=\frac{A+B}{2}\), adding them gives \(E+F=A+\frac{B+C}{2}=A+D\), so \(A=E+F-D\). The same pattern gives B and C.
\(A = E+F-D = (6,5)+(0,3)-(5,1) = (6+0-5,\ 5+3-1) = (1,7)\)
\(B = D+F-E = (5,1)+(0,3)-(6,5) = (5+0-6,\ 1+3-5) = (-1,-1)\)
\(C = D+E-F = (5,1)+(6,5)-(0,3) = (5+6-0,\ 1+5-3) = (11,3)\)
Check: midpoint of BC \(=\left(\frac{-1+11}{2},\frac{-1+3}{2}\right)=(5,1)=D\) ✓; midpoint of CA \(=\left(\frac{11+1}{2},\frac{3+7}{2}\right)=(6,5)=E\) ✓; midpoint of AB \(=\left(\frac{1-1}{2},\frac{7-1}{2}\right)=(0,3)=F\) ✓.
14A city has two main roads crossing at the centre, along N–S and E–W directions, with all other streets parallel and 200 m apart (10 streets each direction). An intersection is named (a, b) where a is the N–S street number and b is the E–W street number. Using this convention, find: (a) how many street intersections can be referred to as (4, 3). (b) how many street intersections can be referred to as (3, 4).
(4,3) and (3,4) are two different intersections — order matters, just like point coordinates
Each ordered pair (a, b) names one specific N–S street and one specific E–W street; exactly one point marks where that particular pair of streets crosses. So each label (a, b) refers to exactly one intersection.
(a) (4, 3): exactly 1 intersection — where the 4th N–S street meets the 3rd E–W street.
(b) (3, 4): exactly 1 intersection — where the 3rd N–S street meets the 4th E–W street. This is a different physical location from (4, 3), just as (x, y) ≠ (y, x) in general — swapping the order changes which streets are being named.
15A computer graphics screen is 800 × 600 pixels, origin at the bottom-left corner. Circle A: radius 80, centre (100, 150). Circle B: radius 100, centre (250, 230). Determine: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.
The 800×600 screen with circle A (centre 100,150, radius 80) and circle B (centre 250,230, radius 100) — both fully inside the screen and overlapping each other.
(i) For circle A (centre (100,150), radius 80): leftmost x = \(100-80=20\), rightmost x = \(100+80=180\), bottom y = \(150-80=70\), top y = \(150+80=230\). All of these lie within \(0 \le x \le 800\) and \(0 \le y \le 600\), so circle A is entirely on-screen.
For circle B (centre (250,230), radius 100): leftmost x = \(250-100=150\), rightmost x = \(250+100=350\), bottom y = \(230-100=130\), top y = \(230+100=330\). These also all lie within the screen, so circle B is entirely on-screen too.
(ii) Distance between the centres: \(\sqrt{(250-100)^2+(230-150)^2}=\sqrt{150^2+80^2}=\sqrt{22500+6400}=\sqrt{28900}=170\).
Sum of radii \(=80+100=180\); difference of radii \(=100-80=20\). Since \(20 < 170 < 180\), the distance between centres is less than the sum of the radii but more than their difference — so the two circles intersect (overlap partially, crossing at two points).
16Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?
Square ABCD: A(2,1), B(-1,2), C(-2,-1), D(1,-2), with equal diagonals AC and BD shown.
Compute all four sides using the distance formula:
\(AB=\sqrt{(2-(-1))^2+(1-2)^2}=\sqrt{9+1}=\sqrt{10}\)
\(BC=\sqrt{(-1-(-2))^2+(2-(-1))^2}=\sqrt{1+9}=\sqrt{10}\)
\(CD=\sqrt{(-2-1)^2+(-1-(-2))^2}=\sqrt{9+1}=\sqrt{10}\)
\(DA=\sqrt{(1-2)^2+(-2-1)^2}=\sqrt{1+9}=\sqrt{10}\)
All four sides are equal (\(\sqrt{10}\)), so ABCD is at least a rhombus. Now check the diagonals:
\(AC=\sqrt{(2-(-2))^2+(1-(-1))^2}=\sqrt{16+4}=\sqrt{20}\)
\(BD=\sqrt{(-1-1)^2+(2-(-2))^2}=\sqrt{4+16}=\sqrt{20}\)
The diagonals are also equal. A rhombus (all sides equal) whose diagonals are also equal must be a square — so ABCD is indeed a square.
Area \(=\text{side}^2=(\sqrt{10})^2=10\) square units. (Check: for a square, area \(=\frac{1}{2}d^2 = \frac{1}{2}(\sqrt{20})^2=\frac{1}{2}(20)=10\), which matches.)
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