Complete NCERT notes with historical context, worked examples, exercise solutions, and interactive practice for Class 9 Maths Chapter 1.
Ancient Indian Siddhāntas established the city of Ujjayinī as the prime longitude meridian (0° longitude) from which geographic locations were measured. Greek scholar Ptolemy (c. 150 CE) recorded coordinates of 'Ozine' (Ujjayinī).
Āryabhaṭa introduced jyiā (sines) to replace Greek chords, enabling precise calculation of celestial body coordinates measured from the ecliptic (the Sun's path across the sky).
Brahmagupta defined zero and negative numbers as formal algebraic entities. Without negative numbers, the four-quadrant coordinate plane would be impossible!
Brahmagupta's works were translated into Arabic (the Sindhind). The Ujjayinī meridian entered Arabic maps as 'Arin'. Al-Bīrūnī studied in India and perfected the astrolabe for celestial coordinate calculation. Ömar Khayyām solved algebraic equations using geometric coordinates.
René Descartes formalised the fact that any point in a 2-D plane can be specified by two numbers representing perpendicular distances from two axes. Legend has it Descartes realized this while watching a fly crawl across his ceiling!
In this narrative scenario, Reiaan and his sister Shalini move to a new city. Shalini helps visually impaired Reiaan navigate his bedroom by creating a tactile grid map using a wooden board, pins, and thick wool strings.
Shalini used a scale of 1 cm : 1 foot. Every unit on the grid graph represented 1 foot in real room space.
Key points (corners of the bed, table, doors) were marked with pins, and thick wool thread was tied between pins so Reiaan could feel the positions with his fingers.
A floor map is 2-Dimensional (\( x \)-axis = length, \( y \)-axis = breadth). Windows require a 3rd dimension (height above floor), so their elevation cannot be represented on a 2D floor plan alone!
Unlike a 1-dimensional line (the standard number line), a two-dimensional space (2-D space) requires two mutually perpendicular lines called coordinate axes to locate any point uniquely.
The horizontal line. Distances to the right of the origin \( O \) are positive (\( +x \)), while distances to the left are negative (\( -x \)).
The vertical line. Distances upward from the origin \( O \) are positive (\( +y \)), while distances downward are negative (\( -y \)).
The reference starting point where \( x = 0 \) and \( y = 0 \)
The coordinate axes divide the Cartesian plane into four regions called Quadrants, numbered counter-clockwise starting from the top-right:
| Quadrant | Position | x-coordinate | y-coordinate | Sign Pair \( (x, y) \) | Example Point |
|---|---|---|---|---|---|
| Quadrant I | Top-Right | Positive (\( x > 0 \)) | Positive (\( y > 0 \)) | \( (+, +) \) | \( A(3, 4) \) |
| Quadrant II | Top-Left | Negative (\( x < 0 \)) | Positive (\( y > 0 \)) | \( (-, +) \) | \( Q(-5, 3) \) |
| Quadrant III | Bottom-Left | Negative (\( x < 0 \)) | Negative (\( y < 0 \)) | \( (-, -) \) | \( M(-3, -4) \) |
| Quadrant IV | Bottom-Right | Positive (\( x > 0 \)) | Negative (\( y < 0 \)) | \( (+, -) \) | \( S(3, -5) \) |
• x-coordinate (Abscissa): Perpendicular distance of point \( P \) from the y-axis (measured along the x-axis).
• y-coordinate (Ordinate): Perpendicular distance of point \( P \) from the x-axis (measured along the y-axis).
If two points lie on a line parallel to the \( x \)-axis, their distance is simply the absolute difference between their \( x \)-coordinates \( |x_2 - x_1| \). If they lie on a line parallel to the \( y \)-axis, their distance is \( |y_2 - y_1| \).
For any two arbitrary points \( A(x_1, y_1) \) and \( D(x_2, y_2) \) where the segment is slanted, we construct a right-angled triangle and apply the Baudhāyana–Pythagoras Theorem.
Consider Triangle \( ADM \) in Quadrant I with vertices \( A(3, 4) \), \( D(7, 1) \), and \( M(9, 6) \):
1. Length of Side AD:
\[ \Delta x = 7 - 3 = 4, \quad \Delta y = 1 - 4 = -3 \]
\[ AD = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ units} \]
2. Length of Side DM:
\[ \Delta x = 9 - 7 = 2, \quad \Delta y = 6 - 1 = 5 \]
\[ DM = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.39 \text{ units} \]
3. Length of Side MA:
\[ \Delta x = 9 - 3 = 6, \quad \Delta y = 6 - 4 = 2 \]
\[ MA = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \text{ units} \]
When Triangle \( ADM \) is reflected across the \( y \)-axis to form Triangle \( A'D'M' \):
Solution:
• Point \( D_1 \) has \( x \)-coordinate \( 9 \). Therefore, the door starts 9 feet away from the left wall (\( y \)-axis).
• The door lies directly on the bottom wall (\( x \)-axis), so its distance from the \( x \)-axis is 0 feet.
Solution: Since \( D_1 \) lies on the \( x \)-axis at 9 units to the right of origin \( O(0,0) \), its coordinates are \( D_1(9, 0) \).
Solution:
\[ \text{Door Width } = R_1 - D_1 = 11.5 - 9 = 2.5 \text{ feet} = 30 \text{ inches} \]
Standard wheelchairs require a clear passage width of at least 32 inches (approx. 2.67 ft). While 30 inches allows standard ambulatory passage, it is tight for wheelchairs. A recommended comfortable width for wheelchair access is 3 feet (36 inches).
Solution:
\[ \text{Bathroom Door Width } = 4 - 1.5 = 2.5 \text{ feet} \]
Both the room door (\( 2.5 \) ft) and the bathroom door (\( 2.5 \) ft) have exactly the same width.
(i) Fourth foot coordinates: To form a rectangle, the 4th foot must have \( x = 8 \) and \( y = 7 \). Thus, the 4th foot is at \( (8, 7) \).
(ii) Suitability of spot: Yes, it is placed neatly against the top wall (\( y = 9 \)), clear of door pathways.
(iii) Dimensions:
• \( \text{Width} = 11 - 8 = 3 \text{ feet} \)
• \( \text{Length} = 9 - 7 = 2 \text{ feet} \)
• Height: Cannot be determined from a 2D floor map because height is along the 3rd (z) dimension.
Solution: The door arc radius is \( 2.5 \) ft. If opened inwards into the bedroom, it swings towards \( y \in [1.5, 4] \) and \( x \in [0, 2.5] \). If wardrobe lies in this swing area, it will collide. Widening the door to 3 ft increases the swing arc to \( x = 3 \) ft, requiring wardrobe relocation.
(i) Bathroom Corners: Origin \( O(0, 0) \), \( F(6, 0) \), \( R(6, 6) \), \( P(0, 6) \).
(ii) Showering Area SHWR: Forms a rectangle/square in the corner, e.g. \( (3, 3), (6, 3), (6, 6), (3, 6) \).
(iii) Washbasin (\( 3\times 2 \) ft) & Toilet (\( 2\times 3 \) ft): Washbasin corner coordinates: \( (0, 4), (3, 4), (3, 6), (0, 6) \). Toilet corner coordinates: \( (4, 0), (6, 0), (6, 3), (4, 3) \).
(i) Dining Room Corners (\( 18\text{ ft} \times 15\text{ ft} \)): Extending from \( P(0, 6) \) to \( A(16, 0) \): corners are \( (0, 0), (18, 0), (18, 15), (0, 15) \).
(ii) Centered Table (\( 5\times 3 \) ft): Center of dining room is \( (9, 7.5) \). Table feet coordinates are \( (6.5, 6), (11.5, 6), (11.5, 9), (6.5, 9) \).
Answer: Both coordinates are 0. The point of intersection is the Origin \( O(0, 0) \).
Answer: The line parallel to the \( y \)-axis through \( W \) has equation \( x = -5 \). Thus, point \( H \) must have \( x = -5 \), so \( H = (-5, y) \). Depending on whether \( y > 0 \) or \( y < 0 \), \( H \) lies in Quadrant II (\( y > 0 \)) or Quadrant III (\( y < 0 \)).
(i) Two perpendicular sides: Side \( AM \) (along line \( y = -2 \)) and Side \( MP \) (along line \( x = -5 \)) are perpendicular (\( \perp \)) to each other.
(ii) Side parallel to an axis: Side \( AM \) is parallel to the \( x \)-axis (\( y = -2 \)), and side \( MP \) is parallel to the \( y \)-axis (\( x = -5 \)).
(iii) Points that are mirror images: Points \( M(-5, -2) \) and \( P(-5, 2) \) are mirror images of each other across the x-axis.
Answer: Choose \( I(5, 0) \) on \( x \)-axis and \( N(0, -6) \) on \( y \)-axis. Then \( \Delta IZN \) is right-angled at \( I(5, 0) \) or origin:
• \( IZ = |-6 - 0| = 6 \) units
• \( IN = \sqrt{(5-0)^2 + (0 - (-6))^2} = \sqrt{25 + 36} = \sqrt{61} \) units
• \( ZN = \sqrt{(5-0)^2 + (-6 - (-6))^2} = 5 \) units.
Answer: No. Without negative numbers, we are restricted entirely to Quadrant I (\( x \ge 0, y \ge 0 \)). We would be unable to represent points to the left of or below the origin.
Answer (Slope / Distance Method):
• Slope of \( MA = \frac{0 - (-4)}{0 - (-3)} = \frac{4}{3} \)
• Slope of \( AG = \frac{8 - 0}{6 - 0} = \frac{8}{6} = \frac{4}{3} \)
Since Slope(\( MA \)) = Slope(\( AG \)) and they share point \( A \), the points \( M, A, G \) are collinear (lie on the same straight line \( y = \frac{4}{3}x \)).
Answer:
• Slope of \( RB = \frac{-5 - (-1)}{-2 - (-5)} = \frac{-4}{3} \)
• Slope of \( BC = \frac{-12 - (-5)}{4 - (-2)} = \frac{-7}{6} \)
Since \( \frac{-4}{3} \neq \frac{-7}{6} \), the points \( R, B, C \) are NOT collinear.
(i) Right-angled isosceles triangle: Vertices \( O(0,0) \), \( A(4,0) \), and \( B(0,4) \). \( OA = OB = 4 \), \( \angle AOB = 90^\circ \).
(ii) Isosceles triangle with vertices in Quad III & IV: Vertices \( O(0,0) \), \( P(-3, -4) \) (Quad III), and \( Q(3, -4) \) (Quad IV). Length \( OP = OQ = \sqrt{3^2 + (-4)^2} = 5 \).
| S | M | T | Is M Midpoint? | Reason |
|---|---|---|---|---|
| (-3, 0) | (0, 0) | (3, 0) | Yes | \( \frac{-3+3}{2} = 0, \frac{0+0}{2} = 0 \) |
| (2, 3) | (3, 4) | (4, 5) | Yes | \( \frac{2+4}{2} = 3, \frac{3+5}{2} = 4 \) |
| (0, 0) | (0, 5) | (0, -10) | No | Actual midpoint is \( (0, -5) \), not \( (0, 5) \) |
| (-8, 7) | (0, -2) | (6, -3) | No | Actual midpoint is \( (-1, 2) \), not \( (0, -2) \) |
General Formula: Midpoint \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
Solution:
\[ \frac{3 + x}{2} = -7 \implies 3 + x = -14 \implies x = -17 \]
\[ \frac{-4 + y}{2} = 1 \implies -4 + y = 2 \implies y = 6 \]
Coordinates of \( B \) are \( (-17, 6) \).
Solution:
Point \( P \) divides \( AB \) in ratio \( 1:2 \):
\[ P = \left( \frac{1(16) + 2(4)}{3}, \frac{1(-2) + 2(7)}{3} \right) = \left( \frac{24}{3}, \frac{12}{3} \right) = (8, 4) \]
Point \( Q \) is midpoint of \( PB \) (or ratio \( 2:1 \)):
\[ Q = \left( \frac{2(16) + 1(4)}{3}, \frac{2(-2) + 1(7)}{3} \right) = \left( \frac{36}{3}, \frac{3}{3} \right) = (12, 1) \]
Trisection points are \( P(8, 4) \) and \( Q(12, 1) \).
(i) Points \( A(1, -8), B(-4, 7), C(-7, -4) \):
• \( OA = \sqrt{1^2 + (-8)^2} = \sqrt{65} \)
• \( OB = \sqrt{(-4)^2 + 7^2} = \sqrt{65} \)
• \( OC = \sqrt{(-7)^2 + (-4)^2} = \sqrt{65} \)
Since \( OA = OB = OC = \sqrt{65} \), all three points lie on Circle \( K \) with radius \( r = \sqrt{65} \).
(ii) Points \( D(-5, 6) \) and \( E(0, 9) \):
• \( OD = \sqrt{(-5)^2 + 6^2} = \sqrt{61} < \sqrt{65} \implies \) Point \( D \) lies INSIDE Circle \( K \).
• \( OE = \sqrt{0^2 + 9^2} = 9 = \sqrt{81} > \sqrt{65} \implies \) Point \( E \) lies OUTSIDE Circle \( K \).
Solution: Let \( D \) be midpoint of \( AB \), \( E \) of \( BC \), \( F \) of \( AC \).
• \( x_A + x_B = 10, x_B + x_C = 12, x_C + x_A = 0 \implies x_A + x_B + x_C = 11 \).
Solving gives \( x_C = 1, x_A = -1, x_B = 11 \).
• \( y_A + y_B = 2, y_B + y_C = 10, y_C + y_A = 6 \implies y_A + y_B + y_C = 9 \).
Solving gives \( y_C = 7, y_A = -1, y_B = 3 \).
Vertices are \( A(-1, -1) \), \( B(11, 3) \), \( C(1, 7) \).
(a) Intersections named (4, 3): Exactly 1 unique street intersection (intersection of 4th N-S street and 3rd E-W street).
(b) Intersections named (3, 4): Exactly 1 unique street intersection (intersection of 3rd N-S street and 4th E-W street).
Circle A: center \( (100, 150) \), \( r = 80 \). Circle B: center \( (250, 230) \), \( r = 100 \).
(i) Screen boundary check:
Circle A bounds: \( x \in [20, 180] \), \( y \in [70, 230] \) (completely within \( [0, 800] \times [0, 600] \)).
Circle B bounds: \( x \in [150, 350] \), \( y \in [130, 330] \) (completely within \( [0, 800] \times [0, 600] \)).
Neither circle lies outside the screen!
(ii) Circle Intersection Check:
Distance between centers \( d = \sqrt{(250-100)^2 + (230-150)^2} = \sqrt{150^2 + 80^2} = \sqrt{28900} = 170 \) pixels.
Sum of radii \( R_1 + R_2 = 80 + 100 = 180 \) pixels.
Since \( d = 170 < 180 \), the two circles INTERSECT each other at two points.
Proof:
1. Side lengths: \( AB = \sqrt{(-3)^2 + 1^2} = \sqrt{10} \), \( BC = \sqrt{(-1)^2 + (-3)^2} = \sqrt{10} \), \( CD = \sqrt{3^2 + (-1)^2} = \sqrt{10} \), \( DA = \sqrt{1^2 + 3^2} = \sqrt{10} \). All 4 sides are equal!
2. Diagonals: \( AC = \sqrt{(-4)^2 + (-2)^2} = \sqrt{20} \), \( BD = \sqrt{2^2 + (-4)^2} = \sqrt{20} \). Diagonals are equal!
Since all 4 sides are equal and diagonals are equal, \( ABCD \) is a SQUARE.
3. \( \text{Area } = (\text{side})^2 = (\sqrt{10})^2 = \mathbf{10 \text{ square units}} \).