Chapter 1 - Orienting Yourself: The Use of Coordinates
Exercise Set 1.1
Q(i) If D R 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?
Solution: The room door lies on the x-axis, so its distance from the x-axis = 0. The coordinates are given as D = (8, 0) and R = (11.5, 0).
Q(ii) What are the coordinates of D1?
Solution: The coordinates of D1 are (8, 0).
Q(iii) If R 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 she/he be able to do so easily?
Solution:
Coordinates of D = (8, 0) and R = (11.5, 0).
Width of the door = distance between D and R = 11.5 - 8 = 3.5 units.
Since 3.5 ft > 2.7 ft (the minimum clear width for a wheelchair), it is a comfortable width.
Width of the door = distance between D and R = 11.5 - 8 = 3.5 units.
Since 3.5 ft > 2.7 ft (the minimum clear width for a wheelchair), it is a comfortable width.
Q(iv) If (0, 1.5) and (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?
Solution:
The coordinates of the ends are (0, 1.5) and (0, 4).
Width of the bathroom door = distance between them = 4 - 1.5 = 2.5 units.
Since 2.5 < 3.5, the bathroom door is narrower than the room door.
Width of the bathroom door = distance between them = 4 - 1.5 = 2.5 units.
Since 2.5 < 3.5, the bathroom door is narrower than the room door.
Exercise Set 1.2
Q1. Evaluate the table placement at (8,9), (11,9), and (11,7).
Solution:
Coordinates: A = (8, 9), B = (11, 9), C = (11, 7).
Distance between (8, 9) and (11, 9) = 11 - 8 = 3.
Distance between (11, 9) and (11, 7) = 9 - 7 = 2.
Distance between (8, 9) and (11, 9) = 11 - 8 = 3.
Distance between (11, 9) and (11, 7) = 9 - 7 = 2.
Q2. If the bathroom door has a hinge at B and opens into the bedroom, will it hit the wardrobe?
Solution:
B1 = (0, 1.5) and B2 = (0, 4).
Door width = 4 - 1.5 = 2.5.
Wardrobe coordinates: W1 = (3, 0), W2 = (3, 2).
The nearest point of the wardrobe from (0, 1.5) is 3 units away. Since 3 > 2.5 (the width of the door), the bathroom door will not hit the wardrobe.
Door width = 4 - 1.5 = 2.5.
Wardrobe coordinates: W1 = (3, 0), W2 = (3, 2).
The nearest point of the wardrobe from (0, 1.5) is 3 units away. Since 3 > 2.5 (the width of the door), the bathroom door will not hit the wardrobe.
Q3. What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the 3ft × 2ft space for the washbasin and a 2ft × 3ft space for the toilet.
Solution:
Shower area coordinates: S = (-6, 5), H = (-3, 5), W = (-2, 9), R = (-6, 9).
Washbasin space (3ft × 2ft): (-6, 0), (-3, 0), (-3, 2), and (-6, 2).
Toilet space (2ft × 3ft): (-6, 2), (-4, 2), (-4, 5), and (-6, 5).
Shower area coordinates: S = (-6, 5), H = (-3, 5), W = (-2, 9), R = (-6, 9).
Washbasin space (3ft × 2ft): (-6, 0), (-3, 0), (-3, 2), and (-6, 2).
Toilet space (2ft × 3ft): (-6, 2), (-4, 2), (-4, 5), and (-6, 5).
Q4. Reiaan's room door leads from the dining room which has a length of 18 ft and width of 15 ft. Place a 5ft × 3ft table at the center and mark the coordinates of its feet.
Solution:
Room corners: P = (-6, 0), A = (12, 0). Length = 12 - (-6) = 18 ft.
Other corners: Q = (12, 15), S = (-6, 15).
Center coordinates:
x-coordinate of center = (-6 + 12) / 2 = 3.
y-coordinate of center = (0 + 15) / 2 = 7.5.
Taking table length = 5 (half-length = 2.5) and width = 3 (half-width = 1.5).
Feet coordinates:
Room corners: P = (-6, 0), A = (12, 0). Length = 12 - (-6) = 18 ft.
Other corners: Q = (12, 15), S = (-6, 15).
Center coordinates:
x-coordinate of center = (-6 + 12) / 2 = 3.
y-coordinate of center = (0 + 15) / 2 = 7.5.
Taking table length = 5 (half-length = 2.5) and width = 3 (half-width = 1.5).
Feet coordinates:
- (3 - 2.5, 7.5 - 1.5) = (0.5, 6)
- (3 + 2.5, 7.5 - 1.5) = (5.5, 6)
- (3 + 2.5, 7.5 + 1.5) = (5.5, 9)
- (3 - 2.5, 7.5 + 1.5) = (0.5, 9)
End Of Chapter Exercises
1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Solution: The x-axis and y-axis intersect at the origin where x-coordinate = 0 and y-coordinate = 0. So, the point of intersection is (0, 0).
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.
Solution: Use the distance formula d = √[(x2 - x1)2 + (y2 - y1)2].
MA = √[(0 + 3)2 + (0 + 4)2] = √(9 + 16) = √25 = 5.
AG = √[(6 - 0)2 + (8 - 0)2] = √(36 + 64) = √100 = 10.
MG = √[(6 + 3)2 + (8 + 4)2] = √(81 + 144) = √225 = 15.
Since MA + AG = MG (5 + 10 = 15), the points M, A, and G lie on the same straight line.
MA = √[(0 + 3)2 + (0 + 4)2] = √(9 + 16) = √25 = 5.
AG = √[(6 - 0)2 + (8 - 0)2] = √(36 + 64) = √100 = 10.
MG = √[(6 + 3)2 + (8 + 4)2] = √(81 + 144) = √225 = 15.
Since MA + AG = MG (5 + 10 = 15), the points M, A, and G lie on the same straight line.
7. Check if R(-5, -1), B(-2, -5) and C(4, -12) are on the same straight line.
Solution: Use the distance formula.
RB = √[(-2 + 5)2 + (-5 + 1)2] = √(9 + 16) = √25 = 5.
BC = √[(4 + 2)2 + (-12 + 5)2] = √(36 + 49) = √85.
RC = √[(4 + 5)2 + (-12 + 1)2] = √(81 + 121) = √202.
Since RB + BC ≠ RC (5 + √85 ≠ √202), the points R, B, and C do not lie on the same straight line.
RB = √[(-2 + 5)2 + (-5 + 1)2] = √(9 + 16) = √25 = 5.
BC = √[(4 + 2)2 + (-12 + 5)2] = √(36 + 49) = √85.
RC = √[(4 + 5)2 + (-12 + 1)2] = √(81 + 121) = √202.
Since RB + BC ≠ RC (5 + √85 ≠ √202), the points R, B, and C do not lie on the same straight line.
10. If (-7, 1) is the midpoint of A(3, -4) and B(x, y), find x and y.
Solution: Using the midpoint relation M = [(x1 + x2) / 2, (y1 + y2) / 2].
(-7, 1) = [(3 + x) / 2, (-4 + y) / 2].
-7 = (3 + x) / 2 ⇒ -14 = 3 + x ⇒ x = -17.
1 = (-4 + y) / 2 ⇒ 2 = -4 + y ⇒ y = 6.
(-7, 1) = [(3 + x) / 2, (-4 + y) / 2].
-7 = (3 + x) / 2 ⇒ -14 = 3 + x ⇒ x = -17.
1 = (-4 + y) / 2 ⇒ 2 = -4 + y ⇒ y = 6.
11. Find coordinates of points P and Q that trisect the line segment A(4, 7) and B(16, -2).
Solution: Let P(x1, y1) and Q(x2, y2) trisect the segment.
Since P is the midpoint of AQ:
x1 = (4 + x2) / 2 ...(1)
y1 = (7 + y2) / 2 ...(2)
Since Q is the midpoint of PB:
x2 = (x1 + 16) / 2 ...(3)
y2 = (y1 - 2) / 2 ...(4)
Substitute (1) into (3):
x2 = [ (4 + x2)/2 + 16 ] / 2
4x2 = 4 + x2 + 32 ⇒ 3x2 = 36 ⇒ x2 = 12.
From (1): x1 = (4 + 12) / 2 = 8.
Substitute (2) into (4):
y2 = [ (7 + y2)/2 - 2 ] / 2
4y2 = 7 + y2 - 4 ⇒ 3y2 = 3 ⇒ y2 = 1.
From (2): y1 = (7 + 1) / 2 = 4.
Hence, P = (8, 4) and Q = (12, 1).
Since P is the midpoint of AQ:
x1 = (4 + x2) / 2 ...(1)
y1 = (7 + y2) / 2 ...(2)
Since Q is the midpoint of PB:
x2 = (x1 + 16) / 2 ...(3)
y2 = (y1 - 2) / 2 ...(4)
Substitute (1) into (3):
x2 = [ (4 + x2)/2 + 16 ] / 2
4x2 = 4 + x2 + 32 ⇒ 3x2 = 36 ⇒ x2 = 12.
From (1): x1 = (4 + 12) / 2 = 8.
Substitute (2) into (4):
y2 = [ (7 + y2)/2 - 2 ] / 2
4y2 = 7 + y2 - 4 ⇒ 3y2 = 3 ⇒ y2 = 1.
From (2): y1 = (7 + 1) / 2 = 4.
Hence, P = (8, 4) and Q = (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?
(ii) Given points D(-5, 6) and E(0, 9), check whether D and E lie within, on, or outside the circle K.
(ii) Given points D(-5, 6) and E(0, 9), check whether D and E lie within, on, or outside the circle K.
Solution:
(i) Calculate distance from origin O(0, 0):
OA = √[(1)2 + (-8)2] = √(1 + 64) = √65.
OB = √[(-4)2 + (7)2] = √(16 + 49) = √65.
OC = √[(-7)2 + (-4)2] = √(49 + 16) = √65.
Since OA = OB = OC = √65, all three points are at the same distance from the origin, lying on circle K with radius = √65.
(ii) Check distances for D and E:
OD = √[(-5)2 + (6)2] = √(25 + 36) = √61. Since √61 < √65, D lies inside the circle.
OE = √[(0)2 + (9)2] = √81 = 9. Since 9 > √65, E lies outside the circle.
(i) Calculate distance from origin O(0, 0):
OA = √[(1)2 + (-8)2] = √(1 + 64) = √65.
OB = √[(-4)2 + (7)2] = √(16 + 49) = √65.
OC = √[(-7)2 + (-4)2] = √(49 + 16) = √65.
Since OA = OB = OC = √65, all three points are at the same distance from the origin, lying on circle K with radius = √65.
(ii) Check distances for D and E:
OD = √[(-5)2 + (6)2] = √(25 + 36) = √61. Since √61 < √65, D lies inside the circle.
OE = √[(0)2 + (9)2] = √81 = 9. Since 9 > √65, E lies outside the circle.
16. Plot points A(2, 1), B(-1, 2), C(-2, -1), and D(1, -2) in the coordinate plane. Is ABCD a square? Calculate its area.
Solution: Calculate the length of all sides and diagonals.
Sides:
AB = √[(-1 - 2)2 + (2 - 1)2] = √((-3)2 + 12) = √(9 + 1) = √10.
BC = √[(-2 + 1)2 + (-1 - 2)2] = √((-1)2 + (-3)2) = √(1 + 9) = √10.
CD = √[(1 + 2)2 + (-2 + 1)2] = √(32 + (-1)2) = √(9 + 1) = √10.
DA = √[(2 - 1)2 + (1 + 2)2] = √(12 + 32) = √(1 + 9) = √10.
Diagonals:
AC = √[(-2 - 2)2 + (-1 - 1)2] = √((-4)2 + (-2)2) = √(16 + 4) = √20.
BD = √[(1 + 1)2 + (-2 - 2)2] = √(22 + (-4)2) = √(4 + 16) = √20.
Since all sides are equal (√10) and diagonals are equal (√20), the figure is a square.
Area of the square = Side2 = (√10)2 = 10 square units.
Sides:
AB = √[(-1 - 2)2 + (2 - 1)2] = √((-3)2 + 12) = √(9 + 1) = √10.
BC = √[(-2 + 1)2 + (-1 - 2)2] = √((-1)2 + (-3)2) = √(1 + 9) = √10.
CD = √[(1 + 2)2 + (-2 + 1)2] = √(32 + (-1)2) = √(9 + 1) = √10.
DA = √[(2 - 1)2 + (1 + 2)2] = √(12 + 32) = √(1 + 9) = √10.
Diagonals:
AC = √[(-2 - 2)2 + (-1 - 1)2] = √((-4)2 + (-2)2) = √(16 + 4) = √20.
BD = √[(1 + 1)2 + (-2 - 2)2] = √(22 + (-4)2) = √(4 + 16) = √20.
Since all sides are equal (√10) and diagonals are equal (√20), the figure is a square.
Area of the square = Side2 = (√10)2 = 10 square units.
0 Comments