Question 1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
The x-axis and y-axis intersect at the origin.
Coordinates of the origin are:
(0, 0)
Answer:
x-coordinate = 0
y-coordinate = 0
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Question 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?
A line parallel to the y-axis has the same x-coordinate at every point.
Since W has x-coordinate = -5, point H will also have x-coordinate = -5.
Therefore:
Coordinates of H can be written as:
(-5, y)
where y can be any real number.
If y is positive, H lies in Quadrant II.
If y is negative, H lies in Quadrant III.
Answer:
Coordinates of H = (-5, y)
H can lie in Quadrant II or Quadrant III.
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Question 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
Points:
R = (3, 0)
A = (0, -2)
M = (-5, -2)
P = (-5, 2)
AM is horizontal because y-coordinate is same.
MP is vertical because x-coordinate is same.
A horizontal line and a vertical line are perpendicular.
Answer:
AM and MP are perpendicular to each other.
(ii) One side of RAMP that is parallel to one of the axes
AM has constant y-coordinate.
Therefore, AM is parallel to the x-axis.
MP has constant x-coordinate.
Therefore, MP is parallel to the y-axis.
Answer:
AM is parallel to the x-axis.
MP is parallel to the y-axis.
(iii) Two points that are mirror images of each other in one axis. Which axis will this be?
Points:
M = (-5, -2)
P = (-5, 2)
The x-coordinates are same and y-coordinates are opposite.
Therefore, they are mirror images in the x-axis.
Answer:
M and P are mirror images of each other in the x-axis.
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Question 4. Plot the point Z (5, -6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.
Point Z = (5, -6)
Take:
I = (5, 0)
N = (0, 0)
Then triangle IZN is right-angled.
Length IZ:
= 6 units
Length IN:
= 5 units
Length ZN:
Using Pythagoras theorem:
ZN² = 5² + 6²
ZN² = 25 + 36
ZN² = 61
ZN = √61 units
Answer:
IZ = 6 units
IN = 5 units
ZN = √61 units
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Question 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?
Without negative numbers, points could only lie in the first quadrant and on the positive axes.
We would not be able to represent points in Quadrants II, III and IV.
Therefore, all points on the plane could not be located.
Answer:
No, such a system would not allow us to locate all points on a 2-D plane.
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Question 6. Are the points M (-3, -4), A (0, 0) and G (6, 8) on the same straight line? Check using distance formula.
Distance formula:
Distance between two points:
d = √[(x2 – x1)² + (y2 – y1)²]
Step 1: Find MA
M = (-3, -4)
A = (0, 0)
MA = √[(0 + 3)² + (0 + 4)²]
= √[3² + 4²]
= √[9 + 16]
= √25
= 5 units
Step 2: Find AG
A = (0, 0)
G = (6, 8)
AG = √[(6 – 0)² + (8 – 0)²]
= √[6² + 8²]
= √[36 + 64]
= √100
= 10 units
Step 3: Find MG
M = (-3, -4)
G = (6, 8)
MG = √[(6 + 3)² + (8 + 4)²]
= √[9² + 12²]
= √[81 + 144]
= √225
= 15 units
Step 4: Check condition
MA + AG = 5 + 10 = 15
MG = 15
Since:
MA + AG = MG
Therefore, the three points are on the same straight line.
Answer:
Yes, the points M, A and G are collinear.
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Question 7. Check using distance formula if the points R (-5, -1), B (-2, -5) and C (4, -12) are on the same straight line.
Step 1: Find RB
R = (-5, -1)
B = (-2, -5)
RB = √[(-2 + 5)² + (-5 + 1)²]
= √[3² + (-4)²]
= √[9 + 16]
= √25
= 5 units
Step 2: Find BC
B = (-2, -5)
C = (4, -12)
BC = √[(4 + 2)² + (-12 + 5)²]
= √[6² + (-7)²]
= √[36 + 49]
= √85 units
Step 3: Find RC
R = (-5, -1)
C = (4, -12)
RC = √[(4 + 5)² + (-12 + 1)²]
= √[9² + (-11)²]
= √[81 + 121]
= √202 units
Step 4: Check condition
RB + BC ≠ RC
Therefore, the points are not on the same straight line.
Answer:
No, the points R, B and C are not collinear.
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Question 8. Using the origin as one vertex, plot the vertices of:
(i) A right-angled isosceles triangle
One possible set of vertices:
(0, 0), (4, 0), (0, 4)
This forms a right-angled isosceles triangle.
Answer:
(0, 0), (4, 0), (0, 4)
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV
One possible set of vertices:
(0, 0), (-3, -4), (3, -4)
This forms an isosceles triangle.
Answer:
(0, 0), (-3, -4), (3, -4)
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Question 9. The following table shows the coordinates of points S, M and T. Check whether M is the midpoint of ST using distance formula.
(i)
S = (-3, 0)
M = (0, 0)
T = (3, 0)
Find SM
SM = √[(0 + 3)² + (0 – 0)²]
= √[3²]
= 3 units
Find MT
MT = √[(3 – 0)² + (0 – 0)²]
= √[3²]
= 3 units
Since:
SM = MT
Therefore, M is the midpoint.
Answer:
Yes, M is the midpoint of ST.
(ii)
S = (2, 3)
M = (3, 4)
T = (4, 5)
Find SM
SM = √[(3 – 2)² + (4 – 3)²]
= √[1² + 1²]
= √2 units
Find MT
MT = √[(4 – 3)² + (5 – 4)²]
= √[1² + 1²]
= √2 units
Since:
SM = MT
Therefore, M is the midpoint.
Answer:
Yes, M is the midpoint of ST.
(iii)
S = (0, 0)
M = (0, 5)
T = (0, -10)
Find SM
SM = √[(0 – 0)² + (5 – 0)²]
= √25
= 5 units
Find MT
MT = √[(0 – 0)² + (-10 – 5)²]
= √225
= 15 units
Since:
SM ≠ MT
Therefore, M is not the midpoint.
Answer:
No, M is not the midpoint of ST.
(iv)
S = (-8, 7)
M = (0, -2)
T = (6, -3)
Find SM
SM = √[(0 + 8)² + (-2 – 7)²]
= √[8² + (-9)²]
= √[64 + 81]
= √145 units
Find MT
MT = √[(6 – 0)² + (-3 + 2)²]
= √[6² + (-1)²]
= √[36 + 1]
= √37 units
Since:
SM ≠ MT
Therefore, M is not the midpoint.
Answer:
No, M is not the midpoint of ST.
Students can also check other exercise solutions of Chapter 1 “Orienting Yourself: The Use of Coordinates” for better understanding and exam preparation. Visit Exercise Set 1.1 Solutions and Exercise Set 1.2 Solutions to practice more coordinate geometry questions with easy step-by-step explanations.
Question 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).
Midpoint formula:
M = ((x1 + x2)/2 , (y1 + y2)/2)
Given:
M = (-7, 1)
A = (3, -4)
B = (x, y)
Using x-coordinate:
(-7) = (3 + x)/2
-14 = 3 + x
x = -17
Using y-coordinate:
1 = (-4 + y)/2
2 = -4 + y
y = 6
Therefore:
B = (-17, 6)
Answer:
Coordinates of B are (-17, 6).
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Question 11. Let P and Q be points of trisection of AB, with P closer to A and Q closer to B. Find the coordinates of P and Q for A (4, 7) and B (16, -2).
Given:
A = (4, 7)
B = (16, -2)
Difference in x-coordinates:
16 – 4 = 12
Difference in y-coordinates:
-2 – 7 = -9
Since P and Q trisect the segment:
Step in x-direction:
12/3 = 4
Step in y-direction:
-9/3 = -3
Coordinates of P
P is one-third distance from A.
P = (4 + 4 , 7 – 3)
P = (8, 4)
Coordinates of Q
Q is two-third distance from A.
Q = (4 + 8 , 7 – 6)
Q = (12, 1)
Answer:
P = (8, 4)
Q = (12, 1)
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Question 12.
(i) Given the points A (1, -8), B (-4, 7) and C (-7, -4), show that they lie on a circle K whose centre is the origin O (0, 0). What is the radius of the circle?
Distance formula:
d = √[(x2 – x1)² + (y2 – y1)²]
Distance OA
OA = √[(1 – 0)² + (-8 – 0)²]
= √[1 + 64]
= √65
Distance OB
OB = √[(-4 – 0)² + (7 – 0)²]
= √[16 + 49]
= √65
Distance OC
OC = √[(-7 – 0)² + (-4 – 0)²]
= √[49 + 16]
= √65
Since:
OA = OB = OC
All three points lie on the same circle centered at the origin.
Radius = √65 units
Answer:
The points lie on the circle K.
Radius of the circle = √65 units.
(ii) Given the points D (-5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.
Radius of circle K = √65
Distance OD
OD = √[(-5)² + 6²]
= √[25 + 36]
= √61
Since:
√61 < √65
Point D lies inside the circle.
Distance OE
OE = √[(0)² + 9²]
= √81
= 9
Since:
9 > √65
Point E lies outside the circle.
Answer:
D lies inside the circle.
E lies outside the circle.
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Question 13. The midpoints of the sides of triangle ABC are 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.
Let:
A = (x1, y1)
B = (x2, y2)
C = (x3, y3)
Given:
D = midpoint of BC = (5, 1)
E = midpoint of CA = (6, 5)
F = midpoint of AB = (0, 3)
Using midpoint relations:
A = E + F – D
B = D + F – E
C = D + E – F
Coordinates of A
A = (6 + 0 – 5 , 5 + 3 – 1)
A = (1, 7)
Coordinates of B
B = (5 + 0 – 6 , 1 + 3 – 5)
B = (-1, -1)
Coordinates of C
C = (5 + 6 – 0 , 1 + 5 – 3)
C = (11, 3)
Answer:
A = (1, 7)
B = (-1, -1)
C = (11, 3)
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Question 14. A city has two main roads crossing at the centre. Streets are parallel and 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 roads by single lines.
Draw:
- 10 parallel vertical lines
- 10 parallel horizontal lines
- Distance between consecutive lines = 1 cm
Answer:
A square grid of streets should be drawn with spacing 1 cm between consecutive roads.
(ii) Find:
(a) How many street intersections can be referred to as (4, 3)?
There is only one unique intersection formed by:
4th N-S street
and
3rd E-W street
Answer:
1 intersection
(b) How many street intersections can be referred to as (3, 4)?
There is only one unique intersection formed by:
3rd N-S street
and
4th E-W street
Answer:
1 intersection
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Question 15. A computer graphics program displays images on a rectangular screen whose origin is at the bottom-left corner.
Screen dimensions:
Width = 800 pixels
Height = 600 pixels
Circle A:
Centre = (100, 150)
Radius = 80 pixels
Circle B:
Centre = (250, 230)
Radius = 100 pixels
(i) Whether any part of either circle lies outside the screen
Circle A
Left edge:
100 – 80 = 20
Right edge:
100 + 80 = 180
Bottom edge:
150 – 80 = 70
Top edge:
150 + 80 = 230
All values lie within screen dimensions.
Therefore, Circle A lies completely inside the screen.
Circle B
Left edge:
250 – 100 = 150
Right edge:
250 + 100 = 350
Bottom edge:
230 – 100 = 130
Top edge:
230 + 100 = 330
All values lie within screen dimensions.
Therefore, Circle B also lies completely inside the screen.
Answer:
No part of either circle lies outside the screen.
(ii) Whether the two circles intersect each other
Distance between centres:
d = √[(250 – 100)² + (230 – 150)²]
= √[150² + 80²]
= √[22500 + 6400]
= √28900
= 170
Sum of radii:
80 + 100 = 180
Since:
170 < 180
The circles intersect each other.
Answer:
Yes, the two circles intersect each other.
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Question 16. Plot the points A (2, 1), B (-1, 2), C (-2, -1), and D (1, -2). Is ABCD a square? What is the area of this square?
Step 1: Find side lengths
AB
AB = √[(-1 – 2)² + (2 – 1)²]
= √[(-3)² + 1²]
= √[9 + 1]
= √10
BC
BC = √[(-2 + 1)² + (-1 – 2)²]
= √[(-1)² + (-3)²]
= √[1 + 9]
= √10
CD
CD = √[(1 + 2)² + (-2 + 1)²]
= √[3² + (-1)²]
= √10
DA
DA = √[(2 – 1)² + (1 + 2)²]
= √[1² + 3²]
= √10
All sides are equal.
Step 2: Check diagonals
AC
AC = √[(-2 – 2)² + (-1 – 1)²]
= √[(-4)² + (-2)²]
= √[16 + 4]
= √20
BD
BD = √[(1 + 1)² + (-2 – 2)²]
= √[2² + (-4)²]
= √[4 + 16]
= √20
Diagonals are equal.
Therefore, ABCD is a square.
Step 3: Find area
Side = √10
Area = side²
= (√10)²
= 10 square units
Answer:
ABCD is a square.
Area = 10 square units.
Students can also check all other exercise solutions of the chapter “Introduction to Linear Polynomial” for complete preparation. Visit Exercise set 1.1, Exercise 2.1 Solutions, Exercise 2.2 Solutions, Exercise 2.3 Solutions, Exercise 2.4 Solutions, Exercise 2.5 Solutions, and Exercise 2.6 Solutions to understand every concept step by step and score better in exams.