Orienting Yourself: The Use of Coordinates
1. What is a Coordinate System?
A coordinate system is like a map grid. It uses numbers to tell the exact position of a point or object. Imagine a treasure map with horizontal and vertical lines — coordinates are the instructions that say "go 3 steps right and 2 steps up."
Imp: A coordinate system changes shapes and locations into numbers. This helps us solve geometry problems using algebra.
The idea of grid-based thinking is very old. Thousands of years ago, cities of the Sindhu-Sarasvati Civilisation were built with streets in perfect North–South and East–West directions, about 10 metres apart. A merchant could find any shop by counting steps from the city centre. This was a real-life coordinate system.
Later, mathematicians like Baudhayana used East–West and North–South lines for geometry. Aryabhata mapped the sky using celestial coordinates. Brahmagupta gave us zero and negative numbers, which made the four-quadrant plane possible. Finally, Rene Descartes showed that every point on a flat surface can be described using just two numbers. This linked algebra and geometry forever.
2. The 2D Cartesian Coordinate System
In earlier grades, you used a number line, which is one-dimensional. To describe a point on a flat surface like a floor or a sheet of paper, we need two number lines that cross each other at right angles.
- The horizontal line is called the x-axis.
- The vertical line is called the y-axis.
- The point where they meet is called the origin, written as O (0, 0).
Distances to the right of the origin or upwards from it are positive. Distances to the left or downwards are negative.
Fig 1.2: Structure of the coordinate plane
Imp: A point P(x, 0) always lies on the x-axis. A point P(0, y) always lies on the y-axis.
Points on the Axes
- If x is positive, the point lies to the right of the origin.
- If x is negative, the point lies to the left of the origin.
- If y is positive, the point lies above the origin.
- If y is negative, the point lies below the origin.
3. Reading a Room Map with Coordinates
Reiaan and his sister Shalini use a grid to map their new room. The scale is 1 cm = 1 foot. The corner points of the room are marked so Reiaan, who cannot see, can feel where things are.
Fig 1.3: Reiaan's room on a coordinate grid
Exercise Set 1.1 — Solutions
(i) If 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?
The left wall is the y-axis (x = 0). Point D₁ is at (11, 0), so the door starts 11 feet from the left wall. The entire door lies on the x-axis, so its distance from the x-axis is 0 feet.
(ii) What are the coordinates of D₁?
From the grid, D₁ is located at (11, 0).
(iii) If R₁ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?
Width = Distance between D₁(11, 0) and R₁(11.5, 0) = 0.5 feet (or 6 inches).
This is not comfortable at all. A standard room door is about 2.5 to 3 feet wide. A wheelchair needs at least 2.5 feet (about 32 inches) to pass through. Therefore, a person in a wheelchair cannot enter easily.
This is not comfortable at all. A standard room door is about 2.5 to 3 feet wide. A wheelchair needs at least 2.5 feet (about 32 inches) to pass through. Therefore, a person in a wheelchair cannot enter easily.
(iv) 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?
Bathroom door width = |4 − 1.5| = 2.5 feet.
Room door width = 0.5 feet.
The bathroom door is wider than the room door.
Room door width = 0.5 feet.
The bathroom door is wider than the room door.
4. The Four Quadrants
The two axes divide the plane into four parts called quadrants. They are numbered counter-clockwise starting from the top-right.
Fig 1.4: The four quadrants of the coordinate plane
- Quadrant I (top-right): x is positive, y is positive → (+, +)
- Quadrant II (top-left): x is negative, y is positive → (−, +)
- Quadrant III (bottom-left): x is negative, y is negative → (−, −)
- Quadrant IV (bottom-right): x is positive, y is negative → (+, −)
Imp: The x-coordinate tells how far the point is from the y-axis. The y-coordinate tells how far the point is from the x-axis.
Think and Reflect — Answers
1. What is the x-coordinate of a point on the y-axis?
Zero. Every point on the y-axis has the form (0, y).
2. Is there a similar rule for a point on the x-axis?
Yes. Every point on the x-axis has the form (x, 0). Its y-coordinate is always zero.
3. Does Q(y, x) ever coincide with P(x, y)? Justify.
Only when x = y. For example, (3, 3) is the same point when swapped. But (2, 5) is not the same as (5, 2).
4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this true?
Yes, it is true. Two ordered pairs are equal only when their first parts are equal and their second parts are equal.
5. Reiaan's Room — Extended Map
The bathroom is attached to the left side of the bedroom. Its corners use negative x-values because they lie to the left of the y-axis.
Fig 1.5: Bedroom and bathroom on one coordinate grid
Exercise Set 1.2 — Solutions
1. Place Reiaan's rectangular study table with three feet at (8, 9), (11, 9) and (11, 7).
(i) Where will the fourth foot be?
(ii) Is this a good spot?
(iii) What is the width and length? Can you find the height?
(i) Where will the fourth foot be?
(ii) Is this a good spot?
(iii) What is the width and length? Can you find the height?
(i) Since opposite sides of a rectangle are equal, the fourth foot is at (8, 7).
(ii) It is an okay spot but tight. It sits in the empty corner near the bed, leaving little walking space.
(iii) Width = 11 − 8 = 3 feet. Length = 9 − 7 = 2 feet. Height cannot be found from a floor map because the map only shows length and width, not height.
(ii) It is an okay spot but tight. It sits in the empty corner near the bed, leaving little walking space.
(iii) Width = 11 − 8 = 3 feet. Length = 9 − 7 = 2 feet. Height cannot be found from a floor map because the map only shows length and width, not height.
2. If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the wardrobe? What changes would you suggest if the door is made wider?
The bathroom door runs from B₁(0, 1.5) to B₂(0, 4), so its length is 2.5 feet. When it swings open 90° into the bedroom around B₁, the outer edge B₂ moves to about (2.5, 1.5). The wardrobe starts at x = 3. Since 2.5 < 3, the door does not hit the wardrobe.
If the door is made wider, it might strike the wardrobe. In that case, move the wardrobe further right or use a sliding door.
If the door is made wider, it might strike the wardrobe. In that case, move the wardrobe further right or use a sliding door.
3. Look at Reiaan's bathroom.
(i) What are the coordinates of the four corners O, F, R, and P?
(ii) What is the shape of the showering area SHWR? Write the coordinates of the four corners.
(iii) Mark off a 1 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates.
(i) What are the coordinates of the four corners O, F, R, and P?
(ii) What is the shape of the showering area SHWR? Write the coordinates of the four corners.
(iii) Mark off a 1 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates.
(i) O (0, 0), F (0, 9), R (-6, 9), P (-6, 0).
(ii) The showering area SHWR is a rectangle. Its corners are S (-6, 5), H (-3, 5), W (-3, 9), R (-6, 9).
(iii) One possible arrangement:
Washbasin (1 ft × 2 ft): (-5, 0), (-3, 0), (-3, 1), (-5, 1)
Toilet (2 ft × 3 ft): (-6, 0), (-4, 0), (-4, 3), (-6, 3)
(Answers may vary based on placement, but dimensions must match.)
(ii) The showering area SHWR is a rectangle. Its corners are S (-6, 5), H (-3, 5), W (-3, 9), R (-6, 9).
(iii) One possible arrangement:
Washbasin (1 ft × 2 ft): (-5, 0), (-3, 0), (-3, 1), (-5, 1)
Toilet (2 ft × 3 ft): (-6, 0), (-4, 0), (-4, 3), (-6, 3)
(Answers may vary based on placement, but dimensions must match.)
4. Other rooms in the house:
(i) The dining room has length 18 ft and width 15 ft. Its length runs from P to 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. Write the coordinates of the feet.
(i) The dining room has length 18 ft and width 15 ft. Its length runs from P to 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. Write the coordinates of the feet.
(i) P is (-6, 0) and A is (12, 0). The length PA = 18 ft. The dining room extends 15 ft downward (negative y direction). Its corners are:
P (-6, 0), A (12, 0), (12, -15), (-6, -15).
(ii) The centre of the room is at (3, -7.5). A 5 ft × 3 ft table placed centrally can have its feet at:
(0.5, -6), (5.5, -6), (5.5, -9), (0.5, -9) (if length 5 ft runs along x and width 3 ft along y).
P (-6, 0), A (12, 0), (12, -15), (-6, -15).
(ii) The centre of the room is at (3, -7.5). A 5 ft × 3 ft table placed centrally can have its feet at:
(0.5, -6), (5.5, -6), (5.5, -9), (0.5, -9) (if length 5 ft runs along x and width 3 ft along y).
6. Distance Between Any Two Points
When two points form a line segment parallel to an axis, the distance is just the difference of the matching coordinates.
- Distance between (x₁, y) and (x₂, y) = |x₂ − x₁|
- Distance between (x, y₁) and (x, y₂) = |y₂ − y₁|
But what if the segment is slanted? We use the Baudhayana–Pythagoras Theorem.
Fig 1.8: Distance formula from the Baudhayana–Pythagoras Theorem
Distance between A(x₁, y₁) and D(x₂, y₂) = √[(x₂ − x₁)² + (y₂ − y₁)²]
Imp: It does not matter if (x₂ − x₁) or (y₂ − y₁) is negative. Squaring removes the negative sign, so distance is always positive.
Reflection and Distance
When a shape is reflected in the y-axis, the x-coordinate changes its sign while the y-coordinate stays the same. Reflection does not change the lengths of sides.
Fig 1.9: Reflection in the y-axis keeps side lengths the same
7. End-of-Chapter Exercises — Solutions
1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
The axes meet at the origin, whose coordinates are (0, 0). So both the x-coordinate and y-coordinate are 0.
2. Point W has x-coordinate equal to -5. Point H is on the line through W parallel to the y-axis. Which quadrants can H lie in?
A line through W parallel to the y-axis is the vertical line x = -5. Any point on this line with a positive y lies in Quadrant II. Any point with a negative y lies in Quadrant III. If y = 0, H is on the axis, not inside any quadrant.
3. Consider the points R(3, 0), A(0, -2), M(-5, -2) and P(-5, 2).
(i) Predict two sides of RAMP that are perpendicular.
(ii) Predict one side parallel to an axis.
(iii) Predict two points that are mirror images in one axis. Which axis?
(i) Predict two sides of RAMP that are perpendicular.
(ii) Predict one side parallel to an axis.
(iii) Predict two points that are mirror images in one axis. Which axis?
(i) AM and MP are perpendicular because AM is horizontal (y = -2) and MP is vertical (x = -5).
(ii) AM is parallel to the x-axis, and MP is parallel to the y-axis.
(iii) M(-5, -2) and P(-5, 2) are mirror images in the x-axis. When reflected in the x-axis, the x-value stays the same and the y-value changes sign.
(ii) AM is parallel to the x-axis, and MP is parallel to the y-axis.
(iii) M(-5, -2) and P(-5, 2) are mirror images in the x-axis. When reflected in the x-axis, the x-value stays the same and the y-value changes sign.
4. Plot Z(5, -6). Construct a right-angled triangle IZN and find the lengths of the three sides.
One simple choice: Let I = (5, 0) and N = (11, -6).
Then:
IZ = |0 − (-6)| = 6 units (vertical)
ZN = |11 − 5| = 6 units (horizontal)
IN = √[(11−5)² + (0−(−6))²] = √[36 + 36] = √72 = 6√2 units.
Note: Your triangle may be different, but the method remains the same.
Then:
IZ = |0 − (-6)| = 6 units (vertical)
ZN = |11 − 5| = 6 units (horizontal)
IN = √[(11−5)² + (0−(−6))²] = √[36 + 36] = √72 = 6√2 units.
Note: Your triangle may be different, but the method remains the same.
5. What would a coordinate system be like if we did not have negative numbers? Would it allow us to locate all points on a 2-D plane?
Without negative numbers, we could only use positive values. We would have only Quadrant I and the positive parts of the axes. All points to the left of the y-axis or below the x-axis could not be described. So we cannot locate every point on the full plane.
6. Are the points M(-3, -4), A(0, 0) and G(6, 8) on the same straight line? Suggest a method to check without plotting.
Yes, they are collinear.
Method: Find the slope between two pairs of points. If the slopes are equal, the points lie on the same straight line.
Slope of AM = (-4 − 0) / (-3 − 0) = 4/3
Slope of AG = (8 − 0) / (6 − 0) = 8/6 = 4/3
Since both slopes are equal and they share point A, all three points lie on one straight line.
Method: Find the slope between two pairs of points. If the slopes are equal, the points lie on the same straight line.
Slope of AM = (-4 − 0) / (-3 − 0) = 4/3
Slope of AG = (8 − 0) / (6 − 0) = 8/6 = 4/3
Since both slopes are equal and they share point A, all three points lie on one straight line.
7. Use your method from Problem 6 to check if R(-5, -1), B(-2, -5) and C(4, -12) are on the same straight line.
Slope of RB = (-5 − (-1)) / (-2 − (-5)) = -4/3
Slope of BC = (-12 − (-5)) / (4 − (-2)) = -7/6
Since -4/3 ≠ -7/6, the points are not collinear.
Slope of BC = (-12 − (-5)) / (4 − (-2)) = -7/6
Since -4/3 ≠ -7/6, the points are not collinear.
8. Using the origin as one vertex, plot:
(i) A right-angled isosceles triangle.
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
(i) A right-angled isosceles triangle.
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
(i) Let the vertices be O(0, 0), A(4, 0), B(0, 4).
OA = 4, OB = 4, and AB = √32. Since OA = OB and the angle at O is 90°, it is a right-angled isosceles triangle.
(ii) Let the vertices be O(0, 0), P(-2, -2) (Quadrant III) and Q(2, -2) (Quadrant IV).
OP = √8, OQ = √8, and PQ = 4. Since OP = OQ, it is isosceles with the base in Quadrants III and IV.
OA = 4, OB = 4, and AB = √32. Since OA = OB and the angle at O is 90°, it is a right-angled isosceles triangle.
(ii) Let the vertices be O(0, 0), P(-2, -2) (Quadrant III) and Q(2, -2) (Quadrant IV).
OP = √8, OQ = √8, and PQ = 4. Since OP = OQ, it is isosceles with the base in Quadrants III and IV.
9. The table shows points S, M and T. Is M the midpoint of ST? Justify. Can you find a connection between the coordinates?
| S | M | T | Is M the midpoint? | Reason |
|---|---|---|---|---|
| (-3, 0) | (0, 0) | (3, 0) | Yes | (-3+3)/2 = 0, (0+0)/2 = 0 |
| (2, 3) | (3, 4) | (4, 5) | Yes | (2+4)/2 = 3, (3+5)/2 = 4 |
| (0, 0) | (0, 5) | (0, -10) | No | (0 + (-10))/2 = -5 ≠ 5 |
| (-8, 7) | (0, -2) | (6, -3) | No | (-8+6)/2 = -1 ≠ 0, (7+(-3))/2 = 2 ≠ -2 |
M = ( (Sx + Tx)/2 , (Sy + Ty)/2 )
10. Use the connection to find B(x, y) if M(-7, 1) is the midpoint of A(3, -4) and B.
Using the midpoint formula:
-7 = (3 + x) / 2 → -14 = 3 + x → x = -17
1 = (-4 + y) / 2 → 2 = -4 + y → y = 6
So B = (-17, 6).
-7 = (3 + x) / 2 → -14 = 3 + x → x = -17
1 = (-4 + y) / 2 → 2 = -4 + y → y = 6
So B = (-17, 6).
11. P and Q are points of trisection of AB, with P closer to A and Q closer to B. Find P and Q for A(4, 7) and B(16, -2).
P divides AB in the ratio 1 : 2.
P = ( (2×4 + 1×16)/3 , (2×7 + 1×(-2))/3 ) = (24/3, 12/3) = (8, 4).
Q divides AB in the ratio 2 : 1.
Q = ( (1×4 + 2×16)/3 , (1×7 + 2×(-2))/3 ) = (36/3, 3/3) = (12, 1).
P = ( (2×4 + 1×16)/3 , (2×7 + 1×(-2))/3 ) = (24/3, 12/3) = (8, 4).
Q divides AB in the ratio 2 : 1.
Q = ( (1×4 + 2×16)/3 , (1×7 + 2×(-2))/3 ) = (36/3, 3/3) = (12, 1).
12. (i) Show that A(1, -8), B(-4, 7) and C(-7, -4) lie on a circle K with centre O(0, 0). Find the radius.
(ii) Check whether D(-5, 6) and E(0, 9) lie inside, on, or outside the circle.
(ii) Check whether D(-5, 6) and E(0, 9) lie inside, on, or outside the circle.
(i) Distance from origin:
OA = √[(1)² + (-8)²] = √65
OB = √[(-4)² + (7)²] = √65
OC = √[(-7)² + (-4)²] = √65
All three distances equal √65, so they lie on the same circle. Radius = √65 units.
(ii) OD = √[(-5)² + (6)²] = √61. Since √61 < √65, point D is inside the circle.
OE = √[(0)² + (9)²] = √81 = 9. Since 9 > √65 (≈ 8.06), point E is outside the circle.
OA = √[(1)² + (-8)²] = √65
OB = √[(-4)² + (7)²] = √65
OC = √[(-7)² + (-4)²] = √65
All three distances equal √65, so they lie on the same circle. Radius = √65 units.
(ii) OD = √[(-5)² + (6)²] = √61. Since √61 < √65, point D is inside the circle.
OE = √[(0)² + (9)²] = √81 = 9. Since 9 > √65 (≈ 8.06), point E is outside the circle.
13. The midpoints of the sides of triangle ABC are D(5, 1), E(6, 5) and F(0, 3). Find the coordinates of A, B and C.
Let D, E, F be midpoints of BC, AC, AB respectively.
Using the midpoint connection in reverse:
x₁ + x₂ = 2×Fx = 0, y₁ + y₂ = 2×Fy = 6
x₁ + x₃ = 2×Ex = 12, y₁ + y₃ = 2×Ey = 10
x₂ + x₃ = 2×Dx = 10, y₂ + y₃ = 2×Dy = 2
Adding all x-equations: 2(x₁+x₂+x₃) = 22 → x₁+x₂+x₃ = 11
x₃ = 11 − 0 = 11, x₂ = 11 − 12 = -1, x₁ = 11 − 10 = 1
Adding all y-equations: 2(y₁+y₂+y₃) = 18 → y₁+y₂+y₃ = 9
y₃ = 9 − 6 = 3, y₂ = 9 − 10 = -1, y₁ = 9 − 2 = 7
A (1, 7), B (-1, -1), C (11, 3).
Using the midpoint connection in reverse:
x₁ + x₂ = 2×Fx = 0, y₁ + y₂ = 2×Fy = 6
x₁ + x₃ = 2×Ex = 12, y₁ + y₃ = 2×Ey = 10
x₂ + x₃ = 2×Dx = 10, y₂ + y₃ = 2×Dy = 2
Adding all x-equations: 2(x₁+x₂+x₃) = 22 → x₁+x₂+x₃ = 11
x₃ = 11 − 0 = 11, x₂ = 11 − 12 = -1, x₁ = 11 − 10 = 1
Adding all y-equations: 2(y₁+y₂+y₃) = 18 → y₁+y₂+y₃ = 9
y₃ = 9 − 6 = 3, y₂ = 9 − 10 = -1, y₁ = 9 − 2 = 7
A (1, 7), B (-1, -1), C (11, 3).
14. A city has two main roads crossing at the centre. Other streets are parallel, 200 m apart, with 10 streets in each direction.
(i) Using 1 cm = 200 m, draw a model.
(ii) Using the convention (N-S street, E-W street), find:
(a) how many intersections can be called (4, 3).
(b) how many intersections can be called (3, 4).
(i) Using 1 cm = 200 m, draw a model.
(ii) Using the convention (N-S street, E-W street), find:
(a) how many intersections can be called (4, 3).
(b) how many intersections can be called (3, 4).
(i) Draw 11 parallel lines for each direction (including the central road), 1 cm apart, intersecting to form a grid.
(ii) Under the given convention, each ordered pair names exactly one crossing.
(a) One intersection: 4th N-S street crossing 3rd E-W street.
(b) One intersection: 3rd N-S street crossing 4th E-W street.
This shows that (4, 3) and (3, 4) are different points. Order matters in coordinates.
(ii) Under the given convention, each ordered pair names exactly one crossing.
(a) One intersection: 4th N-S street crossing 3rd E-W street.
(b) One intersection: 3rd N-S street crossing 4th E-W street.
This shows that (4, 3) and (3, 4) are different points. Order matters in coordinates.
15. A screen is 800 × 600 pixels with origin at the bottom-left. Circle A has centre (100, 150) and radius 80. Circle B has centre (250, 230) and radius 100.
(i) Does any part of either circle lie outside the screen?
(ii) Do the two circles intersect?
(i) Does any part of either circle lie outside the screen?
(ii) Do the two circles intersect?
(i) Circle A: Left = 100−80 = 20, Right = 180, Bottom = 150−80 = 70, Top = 230. All values are between 0 and 800 (or 0 and 600). Fully inside.
Circle B: Left = 150, Right = 350, Bottom = 130, Top = 330. All values are within limits. Fully inside.
So no part lies outside.
(ii) Distance between centres AB = √[(250−100)² + (230−150)²] = √[22500 + 6400] = 170 pixels.
Sum of radii = 80 + 100 = 180 pixels.
Since 170 < 180, the circles do intersect each other.
Circle B: Left = 150, Right = 350, Bottom = 130, Top = 330. All values are within limits. Fully inside.
So no part lies outside.
(ii) Distance between centres AB = √[(250−100)² + (230−150)²] = √[22500 + 6400] = 170 pixels.
Sum of radii = 80 + 100 = 180 pixels.
Since 170 < 180, the circles do intersect each other.
16. Plot A(2, 1), B(-1, 2), C(-2, -1) and D(1, -2). Is ABCD a square? Explain. What is the area?
Calculate all side lengths:
AB = √[(2+1)² + (1−2)²] = √10
BC = √[(-1+2)² + (2+1)²] = √10
CD = √[(-2−1)² + (-1+2)²] = √10
DA = √[(1−2)² + (-2−1)²] = √10
All four sides are equal.
Calculate diagonals:
AC = √[(2+2)² + (1+1)²] = √20
BD = √[(-1−1)² + (2+2)²] = √20
Since diagonals are also equal, ABCD is a square.
Area = (side)² = 10 square units.
(Or use diagonal formula: Area = d²/2 = 20/2 = 10.)
AB = √[(2+1)² + (1−2)²] = √10
BC = √[(-1+2)² + (2+1)²] = √10
CD = √[(-2−1)² + (-1+2)²] = √10
DA = √[(1−2)² + (-2−1)²] = √10
All four sides are equal.
Calculate diagonals:
AC = √[(2+2)² + (1+1)²] = √20
BD = √[(-1−1)² + (2+2)²] = √20
Since diagonals are also equal, ABCD is a square.
Area = (side)² = 10 square units.
(Or use diagonal formula: Area = d²/2 = 20/2 = 10.)