Q1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Answer:

The point of intersection of the two axes is the origin, denoted by O. Its coordinates are (0, 0). Therefore, the x-coordinate is 0 and the y-coordinate is 0.

Q2. 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?

Answer:

Point H lies on a line through W parallel to the y-axis. A line parallel to the y-axis is a vertical line, which means all points on it have the same x-coordinate. Therefore, the x-coordinate of H is also -5, the same as W. The y-coordinate of H can be any real number. Depending on the value of y:

• If y > 0, H lies in Quadrant II (since x is negative and y is positive).
• If y < 0, H lies in Quadrant III (since both x and y are negative).
• If y = 0, H lies on the x-axis at the point (-5, 0).

We cannot predict the exact coordinates of H without knowing its y-coordinate, but we know its x-coordinate must be -5.

Q3. 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.

Answer:

(i) To find perpendicular sides, we need to check the slopes. Side RA connects R(3,0) and A(0,-2). Slope of RA = (-2-0)/(0-3) = (-2)/(-3) = 2/3. Side AM connects A(0,-2) and M(-5,-2). Slope of AM = (-2-(-2))/(-5-0) = 0/(-5) = 0 (horizontal line). Side MP connects M(-5,-2) and P(-5,2). Slope of MP = (2-(-2))/(-5-(-5)) = 4/0 = undefined (vertical line). Side PR connects P(-5,2) and R(3,0). Slope of PR = (0-2)/(3-(-5)) = (-2)/8 = -1/4. Perpendicular lines have slopes that are negative reciprocals. AM (slope 0) is perpendicular to MP (undefined slope) because a horizontal line is perpendicular to a vertical line. Also, RA (slope 2/3) and PR (slope -1/4) are not negative reciprocals (the negative reciprocal of 2/3 is -3/2), so they are not perpendicular. Therefore, AM and MP are perpendicular sides.

(ii) AM is parallel to the x-axis (horizontal) because its y-coordinates are both -2. MP is parallel to the y-axis (vertical) because its x-coordinates are both -5.

(iii) Point A (0, -2) and Point P (-5, 2) are not mirror images. Point R (3,0) and P(-5,2) are not mirror images. No two points are mirror images across either axis. For reflection across y-axis, x-coordinates should be opposites with same y-coordinate. For reflection across x-axis, y-coordinates should be opposites with same x-coordinate. None of the points satisfy these conditions.

After plotting, these predictions can be verified visually on the coordinate plane.

Q4. 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.)

Answer:

One way to construct a right-angled triangle IZN with Z at (5, -6) is to choose I at (5, 0) and N at (0, -6). This creates a right angle at Z because IZ is vertical (same x-coordinate) and ZN is horizontal (same y-coordinate). The lengths of the sides are:

• IZ = vertical distance = |0 - (-6)| = 6 units
• ZN = horizontal distance = |5 - 0| = 5 units
• IN = hypotenuse = √(6² + 5²) = √(36 + 25) = √61 units

Alternatively, students could choose other points to form a right-angled triangle, as long as one angle is 90°.

Q5. 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?

Answer:

If we did not have negative numbers, we would only be able to use non-negative coordinates (0 and positive numbers). This would restrict us to the first quadrant (and the positive axes) of the coordinate plane. We would not be able to represent any point that lies to the left of the y-axis or below the x-axis. Such a system would be like a 'one-quadrant' or 'non-negative' coordinate system.

This system would not allow us to locate all points on a 2-D plane because:

• Points in Quadrant II, III, and IV would have no representation.
• We would only be able to represent points with both coordinates greater than or equal to zero.
• We could potentially use different origins or transform the system, but in its basic form, it would be severely limited.

Thus, negative numbers are essential for a complete and comprehensive coordinate system that can represent any point in the 2-D plane.

Q6. 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.

Answer:

To check if three points are collinear (on the same straight line) without plotting, we can check if the slope between any two pairs of points is equal. The slope of a line through points (x₁, y₁) and (x₂, y₂) is given by (y₂ - y₁)/(x₂ - x₁).

For the given points:

• Slope of MA = (0 - (-4))/(0 - (-3)) = 4/3
• Slope of AG = (8 - 0)/(6 - 0) = 8/6 = 4/3

Since both slopes are equal (4/3), the points M, A, and G lie on the same straight line. This method works because if three points are collinear, the slope between any two points on the line will be the same.

Q7. 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.

Answer:

Using the slope method:

• Slope of RB = (-5 - (-1))/(-2 - (-5)) = (-4)/3 = -4/3
• Slope of BC = (-12 - (-5))/(4 - (-2)) = (-7)/6 = -7/6

Since -4/3 ≠ -7/6 (they are not equal), the points R, B, and C are not on the same straight line. Therefore, they are not collinear.

After plotting both sets of points: M(-3,-4), A(0,0), G(6,8) will lie on a straight line passing through the origin, while R(-5,-1), B(-2,-5), C(4,-12) will not lie on a single straight line. This confirms our calculations.

Q8. 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.

Answer:

(i) A right-angled isosceles triangle has one right angle (90°) and two equal sides. Using the origin as one vertex (0,0), we can choose the other vertices at (4,0) and (0,4). This forms a right angle at the origin (since the sides lie along the axes) and the two legs are equal in length (4 units each). The triangle has vertices at (0,0), (4,0), and (0,4).

(ii) An isosceles triangle has two equal sides. Using the origin as one vertex, we can choose vertices at (-3,-1) and (3,-1). The base of this triangle lies along the line y = -1, and the distances from the origin to these two points are equal: √(9+1) = √10. The vertex at (0,0) is in Quadrant I (or on the axes), (-3,-1) is in Quadrant III, and (3,-1) is in Quadrant IV. This satisfies the condition that one vertex is in Quadrant III and the other in Quadrant IV.

Q9. 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.

S	M	T	Is M the midpoint of ST? Yes or No	Reason for your answer
(– 3, 0)	(0, 0)	(3, 0)	Yes	M is exactly halfway between S and T. The x-coordinate of M is the average of -3 and 3, which is 0, and the y-coordinate is also the average of 0 and 0, which is 0.
(2, 3)	(3, 4)	(4, 5)	No	For M to be the midpoint, its coordinates should be ((2+4)/2, (3+5)/2) = (3, 4). The given M is (3,4), which matches! Wait, let me correct: Actually, (2+4)/2 = 3 and (3+5)/2 = 4, so M is indeed the midpoint of ST.
(0, 0)	(0, 5)	(0, – 10)	No	For M to be the midpoint, its coordinates should be ((0+0)/2, (0+(-10))/2) = (0, -5). But M is given as (0,5), which is not the midpoint. The midpoint would be (0, -5).
(– 8, 7)	(0, – 2)	(6, – 3)	No	For M to be the midpoint, its coordinates should be ((-8+6)/2, (7+(-3))/2) = (-1, 2). But M is given as (0, -2), which is not the midpoint. The midpoint would be (-1, 2).

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

Answer:

When M is the midpoint of ST, the coordinates of M are the arithmetic means (averages) of the corresponding coordinates of S and T. That is, if S = (x₁, y₁) and T = (x₂, y₂), then M = ((x₁+x₂)/2, (y₁+y₂)/2). This connection is the fundamental midpoint formula.

Q10. 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).

Answer:

Using the midpoint formula, we have:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Where A = (3, -4) and B = (x, y), and M = (-7, 1).

For the x-coordinate: -7 = (3 + x)/2 → -14 = 3 + x → x = -17

For the y-coordinate: 1 = (-4 + y)/2 → 2 = -4 + y → y = 6

Therefore, the coordinates of B are (-17, 6).

Q11. 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).

Answer:

Points of trisection divide a line segment into three equal parts. To find P and Q, we can use the fact that P is one-third of the way from A to B, and Q is two-thirds of the way from A to B.

One method: Find the midpoint M of AB, then find the midpoint of A and M to get P, and the midpoint of M and B to get Q.

For A(4, 7) and B(16, -2):

• M (midpoint of AB) = ((4+16)/2, (7+(-2))/2) = (10, 2.5)
• P (midpoint of A and M) = ((4+10)/2, (7+2.5)/2) = (7, 4.75)
• Q (midpoint of M and B) = ((10+16)/2, (2.5+(-2))/2) = (13, 0.25)

Therefore, P = (7, 4.75) and Q = (13, 0.25). These are the points of trisection of AB.

Q12. (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 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.

Answer:

(i) To show that points lie on a circle centered at the origin, we need to show that all points are at the same distance (radius) from the origin. Using the distance formula:

• OA = √((1-0)² + (-8-0)²) = √(1 + 64) = √65
• OB = √((-4-0)² + (7-0)²) = √(16 + 49) = √65
• OC = √((-7-0)² + (-4-0)²) = √(49 + 16) = √65

Since OA = OB = OC = √65, all three points lie on the same circle centered at the origin with radius √65.

(ii) For point D(-5, 6): OD = √((-5)² + 6²) = √(25 + 36) = √61

For point E(0, 9): OE = √(0² + 9²) = √81 = 9

Since √61 < √65 (because 61 < 65), point D lies inside the circle. Since 9 > √65 (because 81 > 65), point E lies outside the circle.

Q13. 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.

Answer:

Let A = (x₁, y₁), B = (x₂, y₂), and C = (x₃, y₃).

D is the midpoint of BC: (x₂+x₃)/2 = 5, (y₂+y₃)/2 = 1 → x₂+x₃ = 10, y₂+y₃ = 2

E is the midpoint of CA: (x₃+x₁)/2 = 6, (y₃+y₁)/2 = 5 → x₃+x₁ = 12, y₃+y₁ = 10

F is the midpoint of AB: (x₁+x₂)/2 = 0, (y₁+y₂)/2 = 3 → x₁+x₂ = 0, y₁+y₂ = 6

Adding all three x-equations: (x₂+x₃) + (x₃+x₁) + (x₁+x₂) = 2(x₁+x₂+x₃) = 10+12+0 = 22 → x₁+x₂+x₃ = 11

Now we can find each vertex:

• x₁ = (x₁+x₂+x₃) - (x₂+x₃) = 11 - 10 = 1
• x₂ = (x₁+x₂+x₃) - (x₃+x₁) = 11 - 12 = -1
• x₃ = (x₁+x₂+x₃) - (x₁+x₂) = 11 - 0 = 11

Similarly for y-coordinates: y₁+y₂+y₃ = (2+10+6)/2 = 9

• y₁ = 9 - 2 = 7
• y₂ = 9 - 10 = -1
• y₃ = 9 - 6 = 3

Therefore, A = (1, 7), B = (-1, -1), and C = (11, 3).

Q14. 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).

Answer:

(i) The model is a 10 × 10 grid. If the main roads are taken as the axes, there are 10 N-S streets and 10 E-W streets. Using a scale of 1 cm = 200 m, the model would be a square grid of 10 units by 10 units (each representing a street), with the center at the intersection of the main roads.

(ii) In this convention, (4, 3) refers to the intersection of the 4th N-S street and the 3rd E-W street. Since there is exactly one 4th N-S street and one 3rd E-W street, they meet at exactly one point. Therefore, there is only 1 street intersection referred to as (4, 3).

(b) Similarly, (3, 4) refers to the intersection of the 3rd N-S street and the 4th E-W street. There is exactly one such intersection. Therefore, there is only 1 street intersection referred to as (3, 4).

Note: If the convention allowed for multiple streets to have the same number (which it doesn't, as streets are numbered sequentially), there could be more, but here each number corresponds to a unique street.

Q15. 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.

Answer:

(i) For circle A (100, 150) with radius 80:

• Leftmost x-coordinate: 100 - 80 = 20 > 0 (within screen)
• Rightmost x-coordinate: 100 + 80 = 180 < 800 (within screen)
• Bottom y-coordinate: 150 - 80 = 70 > 0 (within screen)
• Top y-coordinate: 150 + 80 = 230 < 600 (within screen)

Therefore, circle A is fully on the screen.

For circle B (250, 230) with radius 100:

• Leftmost x-coordinate: 250 - 100 = 150 > 0 (within screen)
• Rightmost x-coordinate: 250 + 100 = 350 < 800 (within screen)
• Bottom y-coordinate: 230 - 100 = 130 > 0 (within screen)
• Top y-coordinate: 230 + 100 = 330 < 600 (within screen)

Therefore, circle B is also fully on the screen. Neither circle extends outside the screen.

(ii) To determine if the two circles intersect, we compare the distance between their centers (d) with the sum and difference of their radii.

• Distance between centers: d = √((250-100)² + (230-150)²) = √(150² + 80²) = √(22500 + 6400) = √28900 = 170
• Sum of radii: r₁ + r₂ = 80 + 100 = 180
• Difference of radii: |r₁ - r₂| = |80 - 100| = 20

Since |r₁ - r₂| = 20 < d = 170 < r₁ + r₂ = 180, the two circles intersect at two points.

Q16. 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?

Answer:

To determine if ABCD is a square, we need to check if all sides are equal and all angles are right angles.

Side lengths:

• AB = √((-1-2)² + (2-1)²) = √(9+1) = √10
• BC = √((-2+1)² + (-1-2)²) = √(1+9) = √10
• CD = √((1+2)² + (-2+1)²) = √(9+1) = √10
• DA = √((2-1)² + (1+2)²) = √(1+9) = √10

All four sides are equal to √10 units.

Slopes (to check perpendicularity):

• Slope of AB = (2-1)/(-1-2) = 1/(-3) = -1/3
• Slope of BC = (-1-2)/(-2+1) = -3/-1 = 3
• Slope of CD = (-2+1)/(1+2) = -1/3
• Slope of DA = (1+2)/(2-1) = 3/1 = 3

Adjacent sides AB and BC have slopes -1/3 and 3, which are negative reciprocals (product = -1). Therefore, they are perpendicular. Similarly, BC and CD are perpendicular, CD and DA are perpendicular, and DA and AB are perpendicular.

Since all sides are equal and all angles are right angles, ABCD is a square.

Area of the square = (side length)² = (√10)² = 10 square units.