Orienting Yourself: The Use of Coordinates — Notes

Introduction

A coordinate system is a numerical method used to describe the exact physical location of points or objects in space. It provides a structured framework, much like the grid lines on a map or graph paper, that allows us to pinpoint locations.

The concept of using grids for spatial organization has ancient roots in Bhārat.

• The Sindhu-Sarasvatī Civilisation (thousands of years ago) is noted for its highly organized urban planning, featuring streets laid out with precision in North-South and East-West directions at uniform intervals. This functioned as a practical coordinate system, enabling merchants to locate places by counting units of distance from a central point.
• Baudhāyana (c. 800 BCE) used East-West and North-South lines in his geometric constructions, which led to the development of the Baudhāyana–Pythagoras Theorem, a foundational element for coordinate geometry.

Later, coordinates became crucial for navigation.

• Ujjayinī was recognized in ancient Indian texts (Siddhāntas, 4th century BCE) as a reference meridian (a line of longitude) from which other locations were measured.
• Ptolemy (c. 150 BCE), a Greek mathematician, compiled extensive lists of latitudes and longitudes for thousands of locations, including Ujjayinī (referred to as ‘Ozine’).
• Āryabhaṭa (c. 499 CE) introduced sines for calculating coordinates more easily, particularly for celestial bodies. He mapped the sky using Celestial Coordinates, measuring distances from the ecliptic.
• Brahmagupta (c. 628 CE) formalized the use of zero and negative numbers as algebraic concepts. His work was essential for the development of the four-quadrant Cartesian plane, which includes negative values.
• Brahmagupta's work, translated as the Sindhind, influenced Arabic geography. The Ujjayinī meridian was known as 'Arin' and used as a zero-longitude reference on early Arabic maps, which also incorporated negative numbers.
• Al-Bīrūnī (c. 1000 CE), an Arab scholar, studied Indian texts and used Indian trigonometry to calculate coordinates of cities across Asia. He also improved the astrolabe, a device for determining location using stars.
• Ömar Khayyām (c. 1100 CE) used Indian mathematics and algebra to solve algebraic problems geometrically by interpreting them using coordinates in a plane.

The modern concept of coordinate geometry was largely formalized in Europe.

• Following the work of Fermat, René Descartes (1637 CE) established that any point in a two-dimensional plane could be precisely defined by two numbers. These numbers represent the point's distances from two perpendicular axes.
• This system, known as the Cartesian Coordinate System, allows for the precise description of points and geometric shapes using algebra and equations. It effectively bridges the fields of geometry and algebra.

In this chapter, we will study this coordinate system, learn to locate objects with accuracy, and understand how coordinates help us visualize algebraic equations as geometric shapes and vice versa.

Remember

Coordinate systems are fundamental tools for mapping, navigation, and representing geometric and algebraic relationships numerically. 🎯 Exam-smart: Questions may involve understanding the historical development of coordinate systems or the purpose of using numerical systems to describe location, especially in contexts like maps or planning.

Settling In

This section introduces the idea of using a system to locate objects and understand spatial relationships, particularly in the context of a new environment for Reiaan and Shalini. It highlights how a coordinate system can help someone navigate and understand their surroundings, even if they have sensory challenges like Reiaan, who has a visual impairment.

The example of Reiaan's room uses a rectangular grid to map out the floor. Pins mark key points (like corners of furniture or doorways), and threads connect these pins to help Reiaan physically feel the layout. This is a practical, tangible way to understand spatial positioning.

The core idea here is that position can be described and understood using a structured system. Even though the full Cartesian coordinate system (with x and y axes) is detailed in the next section, this introduces the why and how of needing such a system.

• Scale: A scale of 1 cm : 1 foot is used, meaning 1 centimetre on the map represents 1 foot in reality. This is crucial for accurately representing distances and sizes in the physical space.
• Limitations: The text points out that the map of the floor cannot show the position of windows. This is because windows are typically on walls, which are vertical, and the map only represents the horizontal floor. This hints at the need for a system that can describe positions in more than one dimension.

Remember

A map or a grid is a simplified representation of a real-world space, and it's important to understand what it can and cannot show. The use of tactile elements (pins and threads) helps make abstract spatial information accessible.

Exam-smart

This section lays the groundwork for understanding how diagrams and representations are used in mathematics and real life. Questions might ask you to interpret a simple map or grid, or explain why certain features might be missing from a 2D representation of a 3D space.

The 2-d Cartesian Coordinate System

The 2-dimensional Cartesian coordinate system is a method for locating any point on a flat surface, called a plane, using two perpendicular lines. Think of it like giving directions on a map where you need both a street name and a house number.

Key Components:

• Coordinate Axes: These are the two perpendicular lines.
• The x-axis is the horizontal line.
• The y-axis is the vertical line.
• Origin (O): This is the point where the x-axis and y-axis intersect. Its coordinates are always (0, 0).
• Coordinates: A pair of numbers written as (x, y) that specify the location of a point.
• The x-coordinate (also called the abscissa) is the distance of the point from the y-axis, measured along the x-axis.
• The y-coordinate (also called the ordinate) is the distance of the point from the x-axis, measured along the y-axis.

How Coordinates Work:

• Positive values: Distances to the right of the origin along the x-axis and distances above the origin along the y-axis are considered positive.
• Negative values: Distances to the left of the origin along the x-axis and distances below the origin along the y-axis are considered negative.
• Points on Axes:
• A point on the x-axis will always have a y-coordinate of 0, written as (x, 0).
• A point on the y-axis will always have an x-coordinate of 0, written as (0, y).
• Notation: While the full coordinates are written as a pair of numbers, like Point P = (x, y), it's common practice on graphs to simply write P(x, y).

Example 1: Locating Points on Axes

Given: Points A, B, and C plotted on a coordinate plane. Point A is 3 units to the right of the origin on the x-axis. Point B is 5 units below the origin on the y-axis. Point C is 2 units to the left of the origin on the x-axis.

To find: The coordinates of points A, B, and C.

Solution:

• Point A is on the x-axis and 3 units to the right. Its y-coordinate is 0. So, A = (3, 0).
• Point B is on the y-axis and 5 units down. Its x-coordinate is 0. Down means negative. So, B = (0, -5).
• Point C is on the x-axis and 2 units to the left. Its y-coordinate is 0. Left means negative. So, C = (-2, 0).

Answer: A = (3, 0), B = (0, -5), C = (-2, 0)

Points Not on the Axes and Quadrants

When points are not on either the x-axis or the y-axis, they fall into one of four regions called quadrants. The coordinate axes divide the entire plane into these four quadrants.

• The plane formed by the x and y axes is called the Cartesian plane, coordinate plane, or xy-plane.

Quadrants:

How to determine the quadrant: Look at the signs of the x and y coordinates.

• x-coordinate (Abscissa): Represents the horizontal distance from the y-axis.
• y-coordinate (Ordinate): Represents the vertical distance from the x-axis.

Example 2: Identifying Quadrants and Coordinates

Given: Points P, Q, R, and S plotted on a coordinate plane.

• P is 5 units to the right and 4 units up.
• Q is 6 units to the left and 3 units up.
• R is 2 units to the left and 5 units down.
• S is 8 units to the right and 2 units down.

To find: The coordinates of P, Q, R, and S, and the quadrant they lie in.

Solution:

• Point P: 5 units right (positive x), 4 units up (positive y). Coordinates: (5, 4). This is in Quadrant I.
• Point Q: 6 units left (negative x), 3 units up (positive y). Coordinates: (-6, 3). This is in Quadrant II.
• Point R: 2 units left (negative x), 5 units down (negative y). Coordinates: (-2, -5). This is in Quadrant III.
• Point S: 8 units right (positive x), 2 units down (negative y). Coordinates: (8, -2). This is in Quadrant IV.

Answer: P(5, 4) in Quadrant I, Q(-6, 3) in Quadrant II, R(-2, -5) in Quadrant III, S(8, -2) in Quadrant IV.

Example 3: Interpreting a Diagram (Similar to Exercise 1.1)

Given: Figure 1.3 showing Reiaan's room with corners marked OABC, and a door DR. O is the origin (0,0). Assume the scale is 1 unit = 1 foot.

To find: (i) Distance of the door from the left wall (y-axis) and from the x-axis. (ii) Coordinates of D. (iii) Width of the door if R is at (11.5, 0).

Solution: (i) The left wall is the y-axis. The distance of the door from the y-axis is its x-coordinate. From the diagram, the door (DR) is marked at x = 11.5. So, it is 11.5 units from the y-axis. The distance from the x-axis is its y-coordinate. The door is along the line y=3. So, it is 3 units from the x-axis.

(ii) Point D is on the y-axis (meaning its x-coordinate is 0). It is located at a height of 3 units from the x-axis. So, the coordinates of D are (0, 3).

(iii) R is given as (11.5, 0). This point represents the right edge of the door on the floor (x-axis). D is (0, 3), which is on the y-axis. The question likely implies a rectangular door opening. Assuming the diagram implies D is at (0,3) and R is at (11.5, 3) to define the door's opening. If R is at (11.5, 0), and D is at (0,3), the door's position needs clarification. Let's assume R is the point on the floor directly below the right edge of the door, so R=(11.5, 0). The diagram shows the door extends from the y-axis. If the door's left edge is at D(0,3), and its right edge is at a point with x=11.5, then the width would be 11.5 - 0 = 11.5 units. However, the diagram shows DR as a segment representing the door opening. If D is (0,3) and we assume the right edge of the door is at x=11.5 on the same floor level, then the width is 11.5 feet. If R (11.5, 0) is a corner of the room and the door's opening width is between two points, say $D_{left}$ and $D_{right}$ on the y-axis (like a cabinet door), the width would be their y-difference. Re-interpreting based on standard room layout: If D is the point (0,3) on the y-axis (left edge of door), and R is the point (11.5, 3) on the line y=3 (right edge of door on the same level), then the width is $11.5-0=11.5$ units. If R is (11.5, 0) on the floor, and the door opening extends from (0,3) to (11.5, 3), then the width is 11.5. Let's assume R is indeed (11.5, 3) for a standard door opening. The width is 11.5 feet. This is a very wide door, comfortable for wheelchairs.

• (Self-correction based on diagram): The diagram shows 'DR' as a segment. If D is (0, 3) and R is the point on the floor at x=11.5, it's possible R is meant to be (11.5, 3) to define a rectangular opening. The question states R is (11.5, 0). This seems to place R on the floor. The diagram suggests D is on the y-axis. If the door spans from x=0 (y-axis) to x=11.5, then width is 11.5. This is a common approach in floor plans.

Answer: (i) The door is 11.5 units from the left wall (y-axis) and 3 units from the x-axis. (ii) The coordinates of D are (0, 3). (iii) If R is (11.5, 0) and D is (0, 3), and assuming the door opening is a rectangle from x=0 to x=11.5 at y=3, the width is 11.5 feet. This is very wide and comfortable for a wheelchair.

Example 4: Interpreting Points on Axes (Similar to Exercise 1.1)

Given: Figure 1.2, which shows points on the x and y axes.

• Point B is on the x-axis, 4.5 units to the right of O.
• Point G is on the y-axis, 4.5 units down from O.
• Point H is on the y-axis, 4 units up from O.

To find: The coordinates of points B, G, and H.

Solution:

• Point B: On the x-axis means its y-coordinate is 0. 4.5 units to the right means its x-coordinate is +4.5. So, B = (4.5, 0).
• Point G: On the y-axis means its x-coordinate is 0. 4.5 units down means its y-coordinate is -4.5. So, G = (0, -4.5).
• Point H: On the y-axis means its x-coordinate is 0. 4 units up means its y-coordinate is +4. So, H = (0, 4).

Answer: B = (4.5, 0), G = (0, -4.5), H = (0, 4).

Remember

Points on the x-axis have the form (x, 0), and points on the y-axis have the form (0, y). The origin is always (0, 0).

Exam-smart

Questions might give you a diagram with points plotted and ask for their coordinates, or give coordinates and ask you to plot them. Be careful with signs (positive/negative) and which axis the point lies on. For word problems, visualize the scenario on a coordinate plane.

Exercises 1.1

Example 5: Room Layout Interpretation

Given: Figure 1.3 shows Reiaan’s room corners OABC. O is the origin. Assume the scale is 1 unit = 1 foot.

To find: (i) If DR represents the door, how far is it from the left wall (y-axis) and the x-axis? (ii) Coordinates of D. (iii) Width of the door if R is (11.5, 0).

Solution: (i) From the diagram, the door DR is positioned such that its left edge D is on the y-axis, and its opening extends horizontally. The distance from the y-axis is the x-coordinate. The diagram suggests the door opening spans from x=0 to x=11.5. So, the distance from the y-axis is 11.5 units. The door is shown at a height of 3 units from the floor (x-axis). So, the distance from the x-axis is 3 units.

(ii) Point D is on the y-axis (so its x-coordinate is 0) and at a height of 3 units from the x-axis. Therefore, the coordinates of D are (0, 3).

(iii) If R is (11.5, 0) and D is (0, 3), and we assume the door is a rectangle with width along the line y=3 and extending from x=0 to x=11.5, then the width is $11.5-0=11.5$ feet. If R is a point on the floor at (11.5, 0), and D is (0,3), this implies a diagonal measurement or a different interpretation of 'R'. Based on typical floor plans and the visual, the door spans from the y-axis (x=0) to x=11.5. So, the width is 11.5 feet. This is a very wide door.

Answer: (i) The door is 11.5 units from the left wall (y-axis) and 3 units from the x-axis. (ii) The coordinates of D are (0, 3). (iii) The width of the door is 11.5 feet. This is a comfortable width for a wheelchair.

Example 6: Comparing Door Widths

Given: Bathroom door ends are at B (0, 1.5) and B (0, 4). This implies two points on the y-axis representing the door frame.

To find: Is the bathroom door narrower or wider than the room door (which we found to be 11.5 feet wide in the previous example)?

Solution: The bathroom door has ends at y-coordinates 1.5 and 4, while its x-coordinate is 0 (it's on the y-axis). This means the door opens along the y-axis. The width (or height, if it's a sliding door) is the difference between the y-coordinates: $4-1.5=2.5$ units. If 1 unit = 1 foot, the bathroom door is 2.5 feet wide.

Comparing this to the room door (11.5 feet), the bathroom door (2.5 feet) is much narrower.

Answer: The bathroom door is narrower than the room door.

Remember

When interpreting diagrams for real-world scenarios, assume standard conventions unless specified otherwise. For room layouts, distances from axes often represent measurements like distance from a wall or height.

Exam-smart

Questions involving room layouts will expect you to translate textual descriptions or diagrammatic representations into coordinate points and distances. Pay close attention to which axis represents which dimension (e.g., length vs. width vs. height from floor).

Exercises 1.2

Example 7: Placing a Table

Given: A rectangular study table with three feet at P(8, 9), Q(11, 9), and R(11, 7).

To find: (i) Where will the fourth foot of the table be? (ii) What is the width and length of the table?

Solution: Let the fourth foot be at S(x, y). The points P(8, 9) and Q(11, 9) share the same y-coordinate (9). This means the line segment PQ is horizontal. The distance PQ is $\mathrm{\mid }11-8\mathrm{\mid }=3$ units. The points Q(11, 9) and R(11, 7) share the same x-coordinate (11). This means the line segment QR is vertical. The distance QR is $\mathrm{\mid }9-7\mathrm{\mid }=2$ units.

For a rectangle, opposite sides must be parallel and equal in length.

• PQ is horizontal. Therefore, SR must also be horizontal, meaning S and R must have the same y-coordinate. So, $y=7$.
• QR is vertical. Therefore, PS must also be vertical, meaning P and S must have the same x-coordinate. So, $x=8$.

Thus, the fourth foot S is at (8, 7).

The sides of the rectangle are PQ (length) and QR (width). Length (horizontal side PQ) = $\mathrm{\mid }11-8\mathrm{\mid }=3$ units. Width (vertical side QR) = $\mathrm{\mid }9-7\mathrm{\mid }=2$ units.

If 1 unit = 1 foot: (i) The fourth foot will be at (8, 7). (ii) The width of the table is 2 feet and the length is 3 feet. (The height cannot be determined from these 2D coordinates).

Answer: (i) The fourth foot of the table will be at (8, 7). (ii) The width of the table is 2 units, and the length is 3 units.

Example 8: Reading Coordinates from a Diagram (Similar to Exercise 1.2, Q3)

Given: Figure 1.5, showing Reiaan's bathroom.

To find: Coordinates of the four corners O, F, R, and P of the bathroom.

Solution:

• Point O is the origin: O = (0, 0).
• Point F is on the x-axis, 6 units to the right of O. So, F = (6, 0).
• Point P is on the y-axis, 4 units up from O. So, P = (0, 4).
• Point R has an x-coordinate corresponding to the distance from the y-axis (which is 6, same as F) and a y-coordinate corresponding to the distance from the x-axis (which is 4, same as P). So, R = (6, 4).

Answer: The coordinates of the four corners are O(0, 0), F(6, 0), R(6, 4), and P(0, 4).

Example 9: Calculating Distance on a Coordinate Plane (Conceptual)

Given: A coordinate plane with axes and points marked.

To find: If a point is at (5, 7), what is its distance from the y-axis and the x-axis?

Solution: The x-coordinate (5) tells us the horizontal distance from the y-axis. So, the distance from the y-axis is 5 units. The y-coordinate (7) tells us the vertical distance from the x-axis. So, the distance from the x-axis is 7 units.

Answer: The point (5, 7) is 5 units from the y-axis and 7 units from the x-axis.

Remember

The x-coordinate is the distance from the y-axis, and the y-coordinate is the distance from the x-axis. Always check the signs for direction (positive/negative).

Exam-smart

You might be given a real-world scenario (like a room, a table, a map) and asked to represent parts of it using coordinates. You'll need to identify the origin and the scale, and then assign coordinates to points based on their position relative to the axes. For objects with dimensions, you'll use coordinate differences to find length or width.

Distance Between Two Points in the 2-D Plane

We can easily find the distance between two points if they lie on the same horizontal or vertical line (i.e., are parallel to the x-axis or y-axis). However, when the line segment connecting two points is not parallel to either axis, we can use the Baudhāyana–Pythagoras theorem to find the distance. This theorem relates the sides of a right-angled triangle.

Consider two points, $A(x_1,y_1)$ and $D(x_2,y_2)$, in the Cartesian plane. We can form a right-angled triangle by drawing a horizontal line through one point and a vertical line through the other. Let's draw a horizontal line through $A$ and a vertical line through $D$. These lines intersect at a point, let's call it $C$.

The coordinates of point $C$ will be $(x_2,y_1)$.

Now, we have a right-angled triangle $ADC$, with the right angle at $C$. The length of the side $AC$ is the difference in the x-coordinates: $\mathrm{\mid }{x}_2-x_1\mathrm{\mid }$. The length of the side $CD$ is the difference in the y-coordinates: $\mathrm{\mid }{y}_2-y_1\mathrm{\mid }$.

According to the Baudhāyana–Pythagoras theorem, in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In triangle $ADC$, the hypotenuse is $AD$, which is the distance between points $A$ and $D$.

So, we have:

To find the distance $AD$, we take the square root of both sides:

This is the distance formula. It calculates the straight-line distance between any two points $(x_1,y_1)$ and $(x_2,y_2)$ in a 2-D plane.

Key points:

• The order of the points $(x_1,y_1)$ and $(x_2,y_2)$ does not matter because the differences are squared, making the result always positive. For example, $(x_2-x_1{)}^2=(x_1-x_2{)}^2$.
• The distance is always a non-negative value.

Example 1

Given: Points $A(3,4)$ and $D(7,1)$. To find: The distance between $A$ and $D$. Solution:

Answer: The distance between points $A$ and $D$ is 5 units.

Example 2

Given: Points $A^{\mathrm{\prime }}(\text{–}3,4)$ and $D^{\mathrm{\prime }}(\text{–}7,1)$. To find: The distance between $A^{\mathrm{\prime }}$ and $D^{\mathrm{\prime }}$. Solution:

Answer: The distance between points $A^{\mathrm{\prime }}$ and $D^{\mathrm{\prime }}$ is 5 units. This demonstrates that reflections preserve distances.

Remember

The distance formula is derived directly from the Pythagorean theorem. Always ensure you are squaring the differences in the x and y coordinates separately before adding them.

Exam-smart

You might be given the coordinates of two points and asked to find the distance directly. Alternatively, you might be given a diagram with points plotted, or a word problem describing locations, and you'll need to identify the coordinates first before applying the distance formula. Some problems might ask you to show that three points are collinear (lie on the same line) by checking if the sum of the distances between two pairs of points equals the distance between the remaining pair.

Exercises (End of Chapter)

1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

• Solution: The point of intersection of the x-axis and the y-axis is the origin. Its coordinates are $(0,0)$.

2. Point 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?

• Solution: If a point is on a line through $W(\text{–}5,y_W)$ parallel to the y-axis, its x-coordinate must be the same as $W$, which is –5. So, the coordinates of $H$ will be $(\text{–}5,y_H)$, where $y_H$ can be any real number.
• If $y_H>0$, $H$ lies in Quadrant II.
• If $y_H<0$, $H$ lies in Quadrant III.
• If $y_H=0$, $H$ lies on the negative x-axis.

3. Consider 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.

• Predictions:

• (i) Side $AM$ is vertical (constant x = -5 for M, x=0 for A, not vertical). Side $MP$ is vertical (constant x = -5). Side $RA$ is not vertical or horizontal. Let's check slopes. Slope of $RA=\frac{-2-0}{0-3}=\frac{-2}{-3}=\frac{2}{3}$. Slope of $AM=\frac{-2-0}{0-3}=\frac{-2}{-3}=\frac{2}{3}$. Points R(3,0), A(0,-2), M(-5,-2), P(-5,2).

• Side $AM$: Slope is $\frac{-2-0}{0-3}=\frac{-2}{-3}=\frac{2}{3}$.

• Side $MP$: $M(-5,-2)$ and $P(-5,2)$. This is a vertical line segment as the x-coordinates are the same. Its length is $\mathrm{\mid }2-(-2)\mathrm{\mid }=4$.

• Side $PR$: $P(-5,2)$ and $R(3,0)$. Slope is $\frac{0-2}{3-(-5)}=\frac{-2}{8}=-\frac{1}{4}$.

• Side $RA$: $R(3,0)$ and $A(0,-2)$. Slope is $\frac{-2-0}{0-3}=\frac{-2}{-3}=\frac{2}{3}$.

• Side $RM$: $R(3,0)$ and $M(-5,-2)$. Slope is $\frac{-2-0}{-5-3}=\frac{-2}{-8}=\frac{1}{4}$.

• Side $AP$: $A(0,-2)$ and $P(-5,2)$. Slope is $\frac{2-(-2)}{-5-0}=\frac{4}{-5}=-\frac{4}{5}$.

• Perpendicular sides: If the product of slopes is -1, they are perpendicular.

• Slope $MP$ is undefined (vertical line). A horizontal line would be perpendicular. No side is horizontal.

• Slope $RA=2\mathrm{/}3$, Slope $AM=2\mathrm{/}3$. These are parallel.

• Slope $PR=-1\mathrm{/}4$, Slope $RM=1\mathrm{/}4$. Not perpendicular.

• Slope $AP=-4\mathrm{/}5$.

• Let's re-examine the points and the figure they form. R(3,0), A(0,-2), M(-5,-2), P(-5,2).

• $MP$ is a vertical line segment (x = -5). $AM$ is a line segment. $RA$ is a line segment. $PR$ is a line segment.

• $MP$ is parallel to the y-axis.

• The points are $R(3,0)$, $A(0,-2)$, $M(-5,-2)$, $P(-5,2)$.

• Side $MP$ is vertical. Side $MA$ is not vertical. Side $RA$ is not vertical. Side $RP$ is not vertical.

• Consider the segments connecting these points. $RM$ has slope $1\mathrm{/}4$. $RA$ has slope $2\mathrm{/}3$. $RP$ has slope $-1\mathrm{/}4$. $AM$ has slope $2\mathrm{/}3$. $AP$ has slope $-4\mathrm{/}5$. $MP$ is vertical.

• There seems to be no pair of perpendicular sides among the sides of the quadrilateral RAMP when joined in that order.

• Let's consider the segments and their slopes:

• $RA$: Slope = $2\mathrm{/}3$

• $AM$: Slope = $2\mathrm{/}3$

• $MP$: Undefined (vertical)

• $PR$: Slope = $-1\mathrm{/}4$

• The problem states "If they are joined in the same order". This implies R to A, A to M, M to P, P to R.

• Wait, the reference diagram for this problem shows A, M, P, R forming a quadrilateral.

• Let's recheck the slopes for perpendicularity:

• $AM$ slope: $\frac{-2-0}{0-3}=\frac{-2}{-3}=\frac{2}{3}$

• $MP$ slope: Undefined (vertical line segment)

• $PR$ slope: $\frac{0-2}{3-(-5)}=\frac{-2}{8}=-\frac{1}{4}$

• $RA$ slope: $\frac{-2-0}{0-3}=\frac{-2}{-3}=\frac{2}{3}$

• The question asks for "Two sides of RAMP that are perpendicular". None of the sides of RAMP appear perpendicular. However, if we consider segments that can be formed from these points.

• Let's consider the possibility that points R, A, M, P are vertices.

• Side $MP$ is vertical. For perpendicularity, we need a horizontal side. None exists.

• Let's reconsider the "Think and Reflect" question in the main text. Point A (3, 4) to D (7, 1). Distance along x-axis is 4. Distance along y-axis is 3.

• The example points are R(3, 0), A(0, – 2), M(– 5, – 2), P(– 5, 2).

• Plotting these: R is on positive x-axis. A is on negative y-axis. M and P have the same x-coordinate (-5), so MP is a vertical segment. MA connects (0,-2) and (-5,-2), which is a horizontal segment.

• So, the sides are:

• $RA$: slope $2\mathrm{/}3$

• $AM$: horizontal segment (y = -2). Length $\mathrm{\mid }0-(-5)\mathrm{\mid }=5$.

• $MP$: vertical segment (x = -5). Length $\mathrm{\mid }2-(-2)\mathrm{\mid }=4$.

• $PR$: slope $-1\mathrm{/}4$.

• (i) Two sides that are perpendicular to each other: $AM$ (horizontal) and $MP$ (vertical).

• (ii) One side parallel to an axis: $AM$ is parallel to the x-axis. $MP$ is parallel to the y-axis.

• (iii) Two points that are mirror images: $A(0,-2)$ and $M(-5,-2)$. If we consider mirroring across the y-axis, $A(0,-2)$ becomes $(0,-2)$ (points on y-axis don't change x). $M(-5,-2)$ would become $(5,-2)$. These are not mirror images.

• Let's consider mirroring across the x-axis. $A(0,-2)$ becomes $(0,2)$. $M(-5,-2)$ becomes $(-5,2)$. This is point $P$. So, $M$ and $P$ are mirror images of each other in the x-axis.

• Verification by plotting:

• Plot R(3,0), A(0,-2), M(-5,-2), P(-5,2).

• Join R to A, A to M, M to P, P to R.

• $AM$ is a horizontal line segment. $MP$ is a vertical line segment. They are perpendicular.

• $AM$ is parallel to the x-axis. $MP$ is parallel to the y-axis.

• Point $M(-5,-2)$ reflected in the x-axis is $(-5,2)$, which is point $P$.

4. Plot 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.)

• Solution: To construct a right-angled triangle IZN with Z(5, -6), we can choose I and N such that one side is horizontal and another is vertical.
• Let $I$ be $(5,0)$. This makes $ZI$ a vertical line segment along $x=5$. Length of $ZI=\mathrm{\mid }0-(-6)\mathrm{\mid }=6$.
• Let $N$ be $(0,-6)$. This makes $ZN$ a horizontal line segment along $y=-6$. Length of $ZN=\mathrm{\mid }5-0\mathrm{\mid }=5$.
• The triangle $IZN$ has vertices $I(5,0)$, $Z(5,-6)$, and $N(0,-6)$. The right angle is at $Z$.
• Length of $ZI=6$ units.
• Length of $ZN=5$ units.
• Length of hypotenuse $IN$ using the distance formula:

• Answer: The lengths of the three sides of triangle $IZN$ can be 6 units, 5 units, and $\sqrt{61}$ units.

5. What 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?

• Solution: If we did not have negative numbers, our coordinate system would only cover the first quadrant. We would not be able to represent points that lie in the second, third, or fourth quadrants, nor points on the negative x-axis or negative y-axis. Therefore, this system would not allow us to locate all the points on a 2-D 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 this without plotting and joining the points.

• Method: Points are collinear if the slope between any two pairs of points is the same.
• Solution:
• Calculate the slope between $M(-3,-4)$ and $A(0,0)$:

• Calculate the slope between $A(0,0)$ and $G(6,8)$:

• Since $m_{MA}=m_{AG}$, the points $M$, $A$, and $G$ 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.

• Solution:
• Calculate the slope between $R(-5,-1)$ and $B(-2,-5)$:

• Calculate the slope between $B(-2,-5)$ and $C(4,-12)$:

• Since $m_{RB}\mathrm{\ne }{m}_{BC}$, the points $R$, $B$, and $C$ are not 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.

• Solution:
• (i) A right-angled isosceles triangle with the origin $O(0,0)$ as one vertex can have vertices $O(0,0)$, $A(3,0)$, and $B(0,3)$. The sides $OA$ and $OB$ are equal in length (3 units each) and are perpendicular (one along x-axis, one along y-axis).
• (ii) An isosceles triangle with the origin $O(0,0)$ as one vertex. Let the other two vertices be $P$ and $Q$.
• We need one vertex in Quadrant III (both coordinates negative) and one in Quadrant IV (x positive, y negative).
• Let $P=(-2,-3)$ (Quadrant III).
• Let $Q=(4,-3)$ (Quadrant IV).
• To make it isosceles with $O$ as a vertex, we need two sides to be equal.
• If $OP=OQ$:

$OP\mathrm{\ne }OQ$.

• If $OP=PQ$:

$OP\mathrm{\ne }PQ$.

• If $OQ=PQ$: $OQ=5$ and $PQ=6$. $OQ\mathrm{\ne }PQ$.
• Let's try different points for $P$ and $Q$. We need $OP=OQ$ or $OP=PQ$ or $OQ=PQ$.
• For $OP=OQ$, we need $\sqrt{x_P^2+y_P^2}=\sqrt{x_Q^2+y_Q^2}$.
• Let $P(-4,-3)$ (Quadrant III). $OP=\sqrt{(-4{)}^2+(-3{)}^2}=\sqrt{16+9}=\sqrt{25}=5$.
• We need $Q(x_Q,y_Q)$ such that $x_Q>0$, $y_Q<0$, and $OQ=5$.
• So, $x_Q^2+y_Q^2=25$. For example, $Q(3,-4)$ (Quadrant IV). $OQ=\sqrt{3^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5$.
• Vertices: $O(0,0)$, $P(-4,-3)$ (Quadrant III), $Q(3,-4)$ (Quadrant IV). This triangle $OPQ$ is isosceles with $OP=OQ$.

9. The following 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. When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

• Connection when M is the midpoint: If $M(x_M,y_M)$ is the midpoint of segment $ST$ with $S(x_S,y_S)$ and $T(x_T,y_T)$, then the coordinates of $M$ are given by the midpoint formula:

This means that the x-coordinate of the midpoint is the average of the x-coordinates of the endpoints, and similarly for the y-coordinate.

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).

• Given: $M(-7,1)$ is the midpoint of $A(3,-4)$ and $B(x,y)$.
• To find: Coordinates of $B(x,y)$.
• Solution:
• Using the midpoint formula for the x-coordinate:

• Using the midpoint formula for the y-coordinate:

• Answer: The coordinates of point $B$ are $(-17,6)$.

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).

• Method:
• Trisection means dividing a line segment into three equal parts. If $P$ and $Q$ trisect $AB$ with $P$ closer to $A$, then $AP=PQ=QB$.
• Point $P$ divides $AB$ in the ratio $1:2$.
• Point $Q$ divides $AB$ in the ratio $2:1$.
• We can find the coordinates of $P$ by considering it as the midpoint of $A$ and $Q$, or by using the section formula for ratio $1:2$.
• Alternatively, we can find $P$ first using the section formula for ratio $1:2$. Then, $Q$ is the midpoint of $P$ and $B$.
• Using section formula for P: Let $A(x_1,y_1)=(4,7)$ and $B(x_2,y_2)=(16,-2)$. $P$ divides $AB$ in the ratio $1:2$. The section formula for a point dividing a segment joining $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ is:

For $P$, $m=1,n=2$:

So, $P=(8,4)$.

• Using midpoint formula for Q: $Q$ divides $AB$ in the ratio $2:1$. For $Q$, $m=2,n=1$:

So, $Q=(12,1)$.

• Alternative Method (using midpoint):

1. Find $P$ using the section formula as above: $P(8,4)$.
2. $Q$ is the midpoint of $P(8,4)$ and $B(16,-2)$.

So, $Q=(12,1)$.

• Answer: The coordinates of $P$ are $(8,4)$ and the coordinates of $Q$ are $(12,1)$.

12. (i) Given the points A (1, – 8), B (– 4, 7) and C (–7, – 4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?

• Solution: A point lies on a circle centered at the origin if its distance from the origin is constant (which is the radius). We calculate the distance of each point from $O(0,0)$ using the distance formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. Here, $(x_1,y_1)=(0,0)$.
• Distance $OA$:

• Distance $OB$:

• Distance $OC$:

• Since $OA=OB=OC=\sqrt{65}$, points $A$, $B$, and $C$ are equidistant from the origin. Therefore, they lie on a circle $K$ with center $O(0,0)$.
• Answer: The radius of circle $K$ is $\sqrt{65}$ units.

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.

• Solution: Let $A(x_A,y_A)$, $B(x_B,y_B)$, and $C(x_C,y_C)$ be the vertices of triangle $ABC$. Let $D(5,1)$ be the midpoint of $BC$. Let $E(6,5)$ be the midpoint of $AC$. Let $F(0,3)$ be the midpoint of $AB$. Using the midpoint formula:
• For $D(5,1)$ (midpoint of $BC$):

• For $E(6,5)$ (midpoint of $AC$):

• For $F(0,3)$ (midpoint of $AB$):

Now we solve these systems of linear equations:

For x-coordinates:

1. $x_B+x_C=10$
2. $x_A+x_C=12$
3. $x_A+x_B=0\Rightarrow x_B=-x_A$

Substitute (3) into (1): $(-x_A)+x_C=10\Rightarrow x_C-x_A=10$. We now have two equations with $x_A$ and $x_C$:

Adding these two equations: $2x_C=22\Rightarrow x_C=11$. Substitute $x_C=11$ into $x_A+x_C=12$: $x_A+11=12\Rightarrow x_A=1$. Substitute $x_A=1$ into $x_A+x_B=0$: $1+x_B=0\Rightarrow x_B=-1$. So, $x_A=1,x_B=-1,x_C=11$.

For y-coordinates:

1. $y_B+y_C=2$
2. $y_A+y_C=10$
3. $y_A+y_B=6\Rightarrow y_B=6-y_A$

Substitute (3) into (1): $(6-y_A)+y_C=2\Rightarrow y_C-y_A=2-6=-4$. We now have two equations with $y_A$ and $y_C$:

Adding these two equations: $2y_C=6\Rightarrow y_C=3$. Substitute $y_C=3$ into $y_A+y_C=10$: $y_A+3=10\Rightarrow y_A=7$. Substitute $y_A=7$ into $y_A+y_B=6$: $7+y_B=6\Rightarrow y_B=-1$. So, $y_A=7,y_B=-1,y_C=3$.

• Answer: The coordinates of the vertices are $A(1,7)$, $B(-1,-1)$, and $C(11,3)$.

14. A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction. (i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines. (ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). 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).

• Solution:
• (i) This part requires drawing a grid. Imagine 10 vertical lines and 10 horizontal lines, equally spaced.
• (ii) The convention $(n,m)$ means the $n$-th street running N–S and the $m$-th street running E–W.
• (a) An intersection referred to as $(4,3)$ means the 4th street in the N–S direction and the 3rd street in the E–W direction. There is only one such specific intersection.
• (b) An intersection referred to as $(3,4)$ means the 3rd street in the N–S direction and the 4th street in the E–W direction. There is only one such specific intersection.

15. A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.

• Solution:

• Screen boundaries: $x$ from 0 to 800, $y$ from 0 to 600.

• Circle 1 (Center A(100, 150), Radius $r_A=80$):

• Leftmost point: $100-80=20$ (within $x\ge 0$)

• Rightmost point: $100+80=180$ (within $x\le 800$)

• Bottommost point: $150-80=70$ (within $y\ge 0$)

• Topmost point: $150+80=230$ (within $y\le 600$)

• No part of Circle 1 lies outside the screen.

• Circle 2 (Center B(250, 230), Radius $r_B=100$):

• Leftmost point: $250-100=150$ (within $x\ge 0$)

• Rightmost point: $250+100=350$ (within $x\le 800$)

• Bottommost point: $230-100=130$ (within $y\ge 0$)

• Topmost point: $230+100=330$ (within $y\le 600$)

• No part of Circle 2 lies outside the screen.

• (i) Answer: No part of either circle lies outside the screen.

• (ii) To check if two circles intersect, we compare the distance between their centers ($d$) with the sum and difference of their radii ($r_A+r_B$ and $\mathrm{\mid }{r}_A-r_B\mathrm{\mid }$).

• Distance between centers $A(100,150)$ and $B(250,230)$:

• Sum of radii: $r_A+r_B=80+100=180$.

• Difference of radii: $\mathrm{\mid }{r}_A-r_B\mathrm{\mid }=\mathrm{\mid }80-100\mathrm{\mid }=\mathrm{\mid }-20\mathrm{\mid }=20$.

• The condition for intersection is $\mathrm{\mid }{r}_A-r_B\mathrm{\mid }\le d\le r_A+r_B$.

• Here, $20\le 170\le 180$. This condition is satisfied.

• (ii) Answer: Yes, the two circles intersect each other.

16. Plot 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?

• Solution:

• Plotting: Plot these four points on the Cartesian plane.

• Check side lengths:

• $AB=\sqrt{(-1-2{)}^2+(2-1{)}^2}=\sqrt{(-3{)}^2+1^2}=\sqrt{9+1}=\sqrt{10}$

• $BC=\sqrt{(-2-(-1){)}^2+(-1-2{)}^2}=\sqrt{(-1{)}^2+(-3{)}^2}=\sqrt{1+9}=\sqrt{10}$

• $CD=\sqrt{(1-(-2){)}^2+(-2-(-1){)}^2}=\sqrt{(3{)}^2+(-1{)}^2}=\sqrt{9+1}=\sqrt{10}$

• $DA=\sqrt{(2-1{)}^2+(1-(-2){)}^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}$ All sides are equal. This indicates it could be a rhombus or a square.

• Check diagonals:

• $AC=\sqrt{(-2-2{)}^2+(-1-1{)}^2}=\sqrt{(-4{)}^2+(-2{)}^2}=\sqrt{16+4}=\sqrt{20}$

• $BD=\sqrt{(1-(-1){)}^2+(-2-2{)}^2}=\sqrt{(2{)}^2+(-4{)}^2}=\sqrt{4+16}=\sqrt{20}$ The diagonals are equal.

• Check slopes for perpendicularity:

• Slope of $AB=\frac{2-1}{-1-2}=\frac{1}{-3}=-\frac{1}{3}$

• Slope of $BC=\frac{-1-2}{-2-(-1)}=\frac{-3}{-1}=3$ The product of slopes $AB$ and $BC$ is $(-\frac{1}{3})\times 3=-1$. This means $AB\perp BC$. So, there is a right angle at $B$.

• Explanation: Since all four sides are equal in length ($\sqrt{10}$ units) and the diagonals are equal ($\sqrt{20}$ units), OR since all sides are equal and one angle is a right angle (e.g., $\mathrm{\angle }ABC$), the quadrilateral $ABCD$ is a square.

• Area: The area of a square is side $\times$ side.

Alternatively, the area of a rhombus (and thus a square) is $\frac{1}{2}\times d_1\times d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.

• Answer: Yes, $ABCD$ is a square. The area of the square is 10 square units.