Question 1

Referring to Fig. 1.3, answer the following questions:

(i) If  D₁R₁ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

(ii) What are the coordinates of  D₁?

(iii) If  R₁ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so 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?

Accepted Answer

Solution:

(i) The door D₁R₁ is shown in Fig. 1.3. The distance of the door from the left wall (y-axis) is the x-coordinate of the door points. Since D₁ lies on the y-axis (x=0), the door is at 0 units from the y-axis (left wall). The distance from the x-axis is the y-coordinate of D₁.

(ii) The coordinates of D₁ can be read from the figure. Since it lies on the y-axis, its x-coordinate is 0. The y-coordinate is the vertical distance from the x-axis. From the figure, D₁ is at (0, 2.5).

(iii) R₁ is given as (11.5, 0). The width of the door is the distance between D₁ and R₁ along the x-axis, which is 11.5 - 0 = 11.5 units. This seems quite wide for a door (usually doors are about 3 feet or 0.9 meters wide). So, 11.5 units (assuming units in feet or meters) is very wide and comfortable. A person in a wheelchair will easily be able to enter through such a wide door.

(iv) Bathroom door ends are B₁(0, 1.5) and B₂(0, 4). The width is the distance between these points along the y-axis: 4 - 1.5 = 2.5 units. Comparing with the room door width (11.5 units), the bathroom door is narrower. — Step-by-step:
- Distance from y-axis is x-coordinate.
- Distance from x-axis is y-coordinate.
- Door width is difference in x-coordinates of D₁ and R₁.
- Bathroom door width is difference in y-coordinates of B₁ and B₂.
- Compare widths to conclude which is wider.

Question 2

Think and Reflect

1. 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?

Accepted Answer

Answers will vary depending on local standards and observations.

1. Standard widths for room doors typically range from about 2.5 feet (0.76 m) to 3 feet (0.91 m) or more. Wider doors are preferred for accessibility.

2. This depends on the school. Some schools have wider doors to accommodate wheelchairs, while others may not. Observations and measurements can confirm suitability. — These are reflective questions encouraging students to observe and think about real-world applications of door widths and accessibility.

Question 3

Copy Fig. 1.4 and mark S and Q in your diagram. Mark any point P in Quadrant I and any point R in Quadrant III, and write down their coordinates.

Accepted Answer

Solution: - Mark point S at (3, -5) in Quadrant IV (x positive, y negative). - Mark point Q at (-5, 3) in Quadrant II (x negative, y positive). - Choose any point P in Quadrant I, for example (2, 3) where both coordinates are positive. - Choose any point R in Quadrant III, for example (-4, -2) where both coordinates are negative. Write down the coordinates as above. — Step-by-step: - Quadrant I: x > 0, y > 0 - Quadrant II: x < 0, y > 0 - Quadrant III: x < 0, y < 0 - Quadrant IV: x > 0, y < 0 - Plot points accordingly and note coordinates.

Question 4

Think and Reflect

1. 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?

Accepted Answer

Answers:

1. The x-coordinate of any point on the y-axis is 0.

2. Yes, the y-coordinate of any point on the x-axis is 0.

3. Point Q (y, x) coincides with point P (x, y) only if x = y. Otherwise, they are different points.

4. The claim is true. If x ≠ y, then (x, y) ≠ (y, x). They are equal only when x = y. — Step-by-step:
- Points on y-axis have x=0.
- Points on x-axis have y=0.
- Equality of ordered pairs requires equality of both coordinates.
- Hence, (x, y) = (y, x) iff x = y.

Question 5

On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (-7, 0) to (13, 0) on the x-axis and from (0, -15) to (0, 12) on the y-axis. (Use the scale 1cm = 1 unit.) Using Fig. 1.5, answer the given questions.

1. Place 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?

2. If 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?

3. Look at Reiaan's bathroom.

(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 in Reiaan's bathroom? Write the coordinates of the four corners.

(iii) Mark off a 3ft × 2ft space for the washbasin and a 2ft × 3ft space for the toilet. Write the coordinates of the corners of these spaces.

4. Other rooms in the house:

(i) Reiaan's room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.

(ii) Place a rectangular 5ft × 3ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Accepted Answer

1. (i) The three feet of the rectangular table are at (8,9), (11,9), and (11,7). Since the table is rectangular, the fourth foot will be at (8,7) to complete the rectangle.

(ii) To check if this is a good spot, we need to see if the table fits without obstruction. If the area is free and the table does not block pathways or other furniture, it is a good spot. From the figure, if the point (8,7) is free, then yes.

(iii) Width of the table = distance between (11,9) and (11,7) = |9 - 7| = 2 units.
Length of the table = distance between (8,9) and (11,9) = |11 - 8| = 3 units.
Height cannot be determined from the 2D coordinates.

2. If the bathroom door has a hinge at B1 and opens into the bedroom, check if the door swing overlaps with the wardrobe area. If yes, the door will hit the wardrobe.
If the door is made wider, it may hit the wardrobe more. Suggest changing the door to open outward or sliding door to avoid hitting.

3. (i) Coordinates of bathroom corners O, F, R, and P can be read from the figure (Fig. 1.5). For example, O(0,0), F(0,6), R(6,6), P(6,0) (assuming from figure).

(ii) The showering area SHWR is rectangular or square; coordinates of its corners can be read from the figure.

(iii) Mark a 3ft × 2ft space for washbasin and 2ft × 3ft for toilet; write coordinates accordingly based on the figure.

4. (i) Dining room corners can be marked with length 18 ft and width 15 ft extending from P to A; coordinates depend on figure.

(ii) Place a 5ft × 3ft dining table at the centre of the dining room; calculate centre coordinates and then coordinates of table feet accordingly. — Step-by-step:
1. (i) Since three feet are at (8,9), (11,9), (11,7), the fourth foot must be at (8,7) to form a rectangle.
(ii) Check if (8,7) is free space; if yes, good spot.
(iii) Width = vertical distance = 2 units; Length = horizontal distance = 3 units.
Height cannot be determined from 2D coordinates.

2. Door hinge at B1 opening into bedroom may hit wardrobe if door swing overlaps wardrobe area. Wider door increases chance of hitting. Suggest door opening outward or sliding door.

3. (i) Read coordinates from figure.
(ii) Identify shape of SHWR and write corner coordinates.
(iii) Mark spaces for washbasin and toilet and write corner coordinates.

4. (i) Sketch dining room with given dimensions and mark corners.
(ii) Find centre of dining room and place table; write feet coordinates accordingly.

Question 6

Think and Reflect

1. 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?

Accepted Answer

1. Distance covered along x-axis = x-coordinate of D - x-coordinate of A = 7 - 3 = 4 units.
Distance covered along y-axis = y-coordinate of A - y-coordinate of D = 4 - 1 = 3 units.

2. Yes, these distances form the legs of a right triangle with AD as the hypotenuse. Using Pythagoras theorem:
AD = sqrt((4)^2 + (3)^2) = sqrt(16 + 9) = sqrt(25) = 5 units. — Step 1: Calculate horizontal distance: 7 - 3 = 4 units.
Step 2: Calculate vertical distance: 4 - 1 = 3 units.
Step 3: Apply Pythagoras theorem: AD = sqrt(4^2 + 3^2) = sqrt(16 + 9) = 5 units.

Question 7

Think and Reflect

1. 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)?

Accepted Answer

1. With reflection in the y-axis:
- The lengths of the sides of the triangle remain the same (distance preserved).
- The coordinates of points change sign in the x-coordinate (x becomes -x), but y-coordinates remain the same.

2. If ΔADM is reflected in the x-axis:
- The y-coordinates would change sign (y becomes -y), x-coordinates remain the same.
- Lengths of sides remain unchanged.
- So, the observations about lengths being preserved remain the same, but the coordinate changes differ. — Reflection in y-axis changes x-coordinates to their negatives, y-coordinates remain same; distances remain unchanged.
Reflection in x-axis changes y-coordinates to their negatives, x-coordinates remain same; distances remain unchanged.
Thus, lengths are preserved in both reflections, but coordinate changes differ.

Question 8

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

Accepted Answer

The point of intersection of the two axes is called the origin. The x-coordinate of the origin is 0 and the y-coordinate of the origin is 0. Therefore, the coordinates of the point of intersection are (0, 0). — The x-axis and y-axis intersect at the origin. By definition, the origin has coordinates (0, 0) because it is zero units away from both axes.

Question 9

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?

Accepted Answer

Since point W has x-coordinate -5, any point H on the line through W parallel to the y-axis will have the same x-coordinate -5 but can have any y-coordinate. So, the coordinates of H will be (-5, y), where y is any real number. The line x = -5 lies in Quadrants II and III. Therefore, point H can lie in Quadrant II if y > 0, or Quadrant III if y < 0. If y = 0, H lies on the x-axis. — Lines parallel to the y-axis have constant x-coordinate. Since W has x = -5, the line through W parallel to y-axis is x = -5. Points on this line have coordinates (-5, y). The sign of y determines the quadrant: positive y means Quadrant II, negative y means Quadrant III.

Question 10

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.

Accepted Answer

(i) Two sides of RAMP that are perpendicular to each other are RA and AM.
- RA joins R(3,0) to A(0,-2). Slope of RA = (-2 - 0)/(0 - 3) = -2 / -3 = 2/3.
- AM joins A(0,-2) to M(-5,-2). Slope of AM = (-2 - (-2))/(-5 - 0) = 0 / -5 = 0.
Since slope of AM is 0 (horizontal line) and slope of RA is 2/3 (non-zero), these two sides are perpendicular if their slopes multiply to -1. But 0 * (2/3) = 0 ≠ -1, so check other sides.
Check sides AM and MP:
- MP joins M(-5,-2) to P(-5,2). Slope = (2 - (-2))/(-5 - (-5)) = 4/0 = undefined (vertical line).
- AM is horizontal (slope 0), MP is vertical (undefined slope), so AM ⟂ MP.
Therefore, sides AM and MP are perpendicular.

(ii) One side parallel to one of the axes:
- AM is horizontal (parallel to x-axis).
- MP is vertical (parallel to y-axis).
So, AM or MP can be chosen.

(iii) Two points that are mirror images of each other in one axis:
- Points R(3,0) and P(-5,2) are not mirror images.
- Points A(0,-2) and M(-5,-2) share the same y-coordinate (-2), so they are mirror images about the y-axis if their x-coordinates are negatives of each other. But 0 and -5 are not negatives.
- Points R(3,0) and A(0,-2) are not mirror images.
- Points P(-5,2) and M(-5,-2) share the same x-coordinate (-5), so they are mirror images about the x-axis if their y-coordinates are negatives. 2 and -2 are negatives.
Therefore, points P and M are mirror images about the x-axis.

Plotting the points on the Cartesian plane will verify these predictions. — Calculate slopes of sides to check perpendicularity. Horizontal and vertical lines are perpendicular. Points with same x or y coordinate and opposite other coordinate are mirror images about the respective axis.

Question 11

Plot point Z (5, -6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.

Accepted Answer

Since this is a construction and measurement question, the answer may vary. However, one way is:
- Plot Z at (5, -6).
- Choose point I on the x-axis with the same x-coordinate as Z, so I = (5, 0).
- Choose point N on the y-axis with the same y-coordinate as Z, so N = (0, -6).
Triangle IZN is right-angled at point I or N.
Lengths:
- IZ = distance between I(5,0) and Z(5,-6) = |0 - (-6)| = 6 units.
- ZN = distance between Z(5,-6) and N(0,-6) = |5 - 0| = 5 units.
- IN = distance between I(5,0) and N(0,-6) = sqrt((5-0)^2 + (0 - (-6))^2) = sqrt(25 + 36) = sqrt(61) units.
Thus, the sides are 5 units, 6 units, and sqrt(61) units. — By choosing points I and N on the axes with coordinates matching Z's x and y respectively, the triangle IZN formed is right-angled. Use distance formula to find lengths.

Question 12

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?

Accepted Answer

If we did not have negative numbers, the coordinate system would be limited to only the first quadrant where both x and y are non-negative. This means the coordinate axes would only represent positive values and zero. Such a system would not allow us to locate points in the other three quadrants (II, III, and IV) of the 2-D plane. Therefore, it would not be possible to represent all points on the plane using only non-negative coordinates. — Negative numbers are essential to represent points lying to the left of the y-axis (negative x) and below the x-axis (negative y). Without negatives, the coordinate system is restricted to the first quadrant.

Question 13

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.

Accepted Answer

To check if three points are collinear without plotting, we can check if the slopes of the line segments joining pairs of points are equal.
Calculate slope of MA:
= (0 - (-4)) / (0 - (-3)) = 4 / 3.
Calculate slope of AG:
= (8 - 0) / (6 - 0) = 8 / 6 = 4 / 3.
Since slope MA = slope AG = 4/3, points M, A, and G lie on the same straight line.
Method: Calculate slopes of two segments sharing a common point; if equal, points are collinear. — Equal slopes between consecutive points indicate collinearity. This method avoids plotting.

Question 14

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.

Accepted Answer

Calculate slope of RB:
= (-5 - (-1)) / (-2 - (-5)) = (-4) / (3) = -4/3.
Calculate slope of BC:
= (-12 - (-5)) / (4 - (-2)) = (-7) / (6) = -7/6.
Since slope RB ≠ slope BC, the points R, B, and C are not collinear.
Plotting the points will confirm this result. — Different slopes between consecutive segments indicate points are not on the same line.

Question 15

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.

Accepted Answer

(i) Right-angled isosceles triangle with origin as one vertex:
- Choose points A(0,0), B(2,0), and C(0,2).
- Triangle ABC has sides AB = 2 units, AC = 2 units, and BC = sqrt(2^2 + 2^2) = sqrt(8) = 2√2 units.
- Sides AB and AC are equal, and angle at A is 90°, so triangle is right-angled isosceles.

(ii) Isosceles triangle with one vertex in Quadrant III and another in Quadrant IV:
- Choose points A(0,0), B(-3,-4) in Quadrant III, and C(4,-4) in Quadrant IV.
- Sides AB and AC are equal in length:
  AB = sqrt((-3 - 0)^2 + (-4 - 0)^2) = 5 units.
  AC = sqrt((4 - 0)^2 + (-4 - 0)^2) = sqrt(16 + 16) = sqrt(32) ≈ 5.66 units.
- To make it isosceles, adjust points accordingly, for example B(-4,-3) and C(4,-3) both in Quadrants III and IV respectively, with equal distances from A.
- Plotting these points will show the required triangle. — Select points carefully to satisfy the conditions of right angle and equal sides for isosceles triangles in specified quadrants.

Orienting Yourself: The Use of

Orienting Yourself: The Use of — Study Notes

1.1 INTRODUCTION

1.2 SETTLING IN

1.3 THE 2-D CARTESIAN COORDINATE SYSTEM

Practice Questions — Orienting Yourself: The Use of

All 8 Chapters in Ganita Manjari (English)