NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 01 Orienting Yourself: The Use of Coordinates

Exercise Set 1.1

Question (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?
Answer: The room door sits on the x-axis, which means its distance from the x-axis equals 0 units. Looking at the figure, D₁ = (8, 0) and R₁ = (11.5, 0). This tells us the door starts 8 units away from the y-axis.
In simple words: The door begins 8 units from the left wall and touches the horizontal line (x-axis).

Exam Tip: Always identify which axis a point touches to determine its distance from that axis - points on the x-axis have distance 0 from it.

Question (ii). What are the coordinates of D₁?
Answer: The coordinates of D₁ are (8, 0).
In simple words: Point D₁ sits 8 units to the right and on the horizontal line.

Exam Tip: When reading coordinates from a figure, check the x-value (horizontal position) and y-value (vertical position) carefully.

Question (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 she/he be able to do so easily?
Answer: The coordinates show D₁ = (8, 0) and R₁ = (11.5, 0). So the door's width = distance between these points = 11.5 - 8 = 3.5 units.

If the unit represents feet, then 3.5 ft is about 42 inches, which is quite comfortable for a room door. Standard residential doors are usually 30 - 36 inches wide, so this is actually a bit wider than the usual size, which is good.

For wheelchair accessibility: A wheelchair typically needs at least 32 inches (around 2.7 ft) of clear width. Since 3.5 ft is greater than 2.7 ft, the door has enough space. So yes, this door width is comfortable and a person using a wheelchair should be able to enter without trouble.
In simple words: The door is 3.5 units wide - wide enough for someone in a wheelchair to go through easily.

Exam Tip: Always show distance calculations clearly and relate abstract units to real-world measurements for practical context.

Question (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?
Answer: The coordinates of B₁ = (0, 1.5) and B₂ = (0, 4). So the bathroom door's width = distance between these two points = 4 - 1.5 = 2.5 units. Since 2.5 is less than 3.5, the bathroom door is narrower than the room door.
In simple words: The bathroom door measures 2.5 units - smaller than the room door's 3.5 units.

Exam Tip: Compare distances directly - a smaller difference between coordinates means a smaller distance.

Exercise Set 1.2

Question 1(i). Place Reiaan's rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7). Where will the fourth foot of the table be?
Answer: The three given points form three corners of a rectangle: A = (8, 9), B = (11, 9), C = (11, 7). To complete the rectangle, the fourth point must have the same x-coordinate as point A, which is 8, and the same y-coordinate as point C, which is 7. So the fourth foot is at (8, 7).
In simple words: The missing corner has the same x-number as the first point and the same y-number as the third point.

Exam Tip: In a rectangle, opposite corners share x-coordinates and y-coordinates with their opposite sides.

Question 1(ii). Is this a good spot for the table?
Answer: Yes, this is a good spot because the table fits neatly inside the room without blocking any doors or pathways. Additionally, it is positioned near a wall, which is useful for studying.
In simple words: The table is in a good place - it does not get in the way and is close to a wall.

Exam Tip: When evaluating placement, consider practical factors like blocking pathways, proximity to walls, and overall functionality.

Question 1(iii). What is the width of the table? The length? Can you make out the height of the table?
Answer: Width: The distance between (8, 9) and (11, 9) is 11 - 8 = 3 units. Length: The distance between (11, 9) and (11, 7) is 9 - 7 = 2 units. Height: The height cannot be worked out from the given diagram since the figure shows only a top view in 2D, not the vertical dimension.
In simple words: The table is 3 units wide and 2 units long. We cannot see how tall it is from the top view.

Exam Tip: Recognize the difference between 2D views (which show only length and width) and 3D information (which would include height).

Question 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?
Answer: From the figure, B₁ = (0, 1.5) and B₂ = (0, 4). The bathroom door's width is 4 - 1.5 = 2.5 units. If the door is hinged at B₁ and opens into the bedroom, it sweeps an arc with radius 2.5 units from B₁. The wardrobe starts at W₁ = (3, 0) and W₄ = (3, 2). The nearest point of the wardrobe from B₁(0, 1.5) is around x = 3, which is farther than the door width of 2.5 units. So the bathroom door will not hit the wardrobe.

Suggestions if the door becomes wider: If the door becomes much wider, it might get close to or hit the wardrobe. In that case, the door could be made to swing inward into the bathroom instead, the wardrobe could be moved slightly to the right, or the door width could be kept limited for comfortable movement.
In simple words: The bathroom door does not hit the wardrobe now. If the door gets wider, we would need to move the wardrobe or change how the door opens.

Exam Tip: Visualize the arc created by a swinging door from its hinge point to check for collisions with nearby objects.

Question 3(i). Look at Reiaan's bathroom. What are the coordinates of the four corners O, F, R, and P of the bathroom?
Answer: From the figure, the four corners of the bathroom are: O = (0, 0), F = (0, 9), R = (-6, 9), P = (-6, 0).
In simple words: The bathroom corners are at these four points forming a rectangle.

Exam Tip: Label all four vertices clearly and verify they form the correct shape by checking that opposite corners are aligned.

Question 3(ii). What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the four corners.
Answer: From the figure: S = (-6, 6), H = (-3, 6), W = (-2, 9), R = (-6, 9). Since one pair of opposite sides is parallel, SHWR is a trapezium. Coordinates of the corners are S = (-6, 6), H = (-3, 6), W = (-2, 9), R = (-6, 9).
In simple words: The shower area is a trapezium - a four-sided shape where only one pair of opposite sides are parallel.

Exam Tip: Check which sides are parallel by comparing if they have the same slope or by verifying that y-coordinates match in pairs.

Question 3(iii). Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.
Answer: Washbasin space (3 ft × 2 ft): Consider a rectangle near the bottom-left corner of the bathroom. Coordinates of corners: (-6, 0.5), (-5, 0.5), (-5, 2), (-6, 2).

Toilet space (2 ft × 3 ft): Take a rectangle above the washbasin. Coordinates of corners: (-6, 3), (-4.5, 3), (-4.5, 4), (-6, 4).
In simple words: The washbasin fits in the lower-left area and the toilet goes just above it.

Exam Tip: When placing fixtures in a room, ensure they do not overlap and fit within the bathroom boundaries.

Question 4(i). Other rooms in the house: Reiaan's room door leads from the dining room which has 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.
Answer: From the figure, coordinates of P = (-6, 0) and coordinates of A = (12, 0). The length PA = 12 - (-6) = 18 ft, which matches the given length. If the dining room is 15 ft wide and lies below PA, then its upper side is PA and it extends downward 15 units. So the coordinates of the four corners are: P = (-6, 0), A = (12, 0), Q = (12, -15), S = (-6, -15).
In simple words: The dining room is a rectangle that sits below the line connecting P and A, going down 15 units.

Exam Tip: Always verify that calculated dimensions match the given measurements before finalizing coordinates.

Question 4(ii). Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.
Answer: The dining room extends from x = -6 to x = 12 and from y = 0 to y = -15. The centre of the dining room: x-coordinate of centre = (-6 + 12) / 2 = 3, y-coordinate of centre = (0 + (-15)) / 2 = -7.5.

Now place a 5 ft × 3 ft table at the centre, with length = 5 units along the x-axis and width = 3 units along the y-axis. Half-length = 2.5, Half-width = 1.5. So the coordinates of the corners (feet) of the table are: (3 - 2.5, -7.5 - 1.5) = (0.5, -9), (3 + 2.5, -7.5 - 1.5) = (5.5, -9), (3 + 2.5, -7.5 + 1.5) = (5.5, -6), (3 - 2.5, -7.5 + 1.5) = (0.5, -6).
In simple words: The table sits in the middle of the dining room with corners at these four points.

Exam Tip: To centre a rectangle, find the midpoint of the room and subtract/add half the table's dimensions from that centre point.

End-of-Chapter Exercises

Question 1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Answer: The x-axis and y-axis meet at the origin. At this point, the x-coordinate is 0 and the y-coordinate is 0. So the point of intersection is (0, 0).
In simple words: The two lines cross at the point where both numbers are zero.

Exam Tip: The origin is always (0, 0) - this is the only point that lies on both axes simultaneously.

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?
Answer: If point W has x-coordinate -5, then any point on a line through W that is parallel to the y-axis will also have x-coordinate -5. So the coordinates of H will be of the form H = (-5, y), where y can be any real number.

Depending on the value of y: If y > 0, then H lies in Quadrant II. If y < 0, then H lies in Quadrant III. If y = 0, then H lies on the x-axis. So H can lie in Quadrant II, Quadrant III, or on the x-axis.
In simple words: Point H has the same left-right position as W but can move up or down. It will be in Quadrant II if it goes up, Quadrant III if it goes down, or on the x-axis line if y = 0.

Exam Tip: A vertical line has constant x-coordinate; the y-value determines which quadrant or axis the point falls on.

Question 3(i). 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: Two sides of RAMP that are perpendicular to each other.
Answer: Let us observe: AM joins A(0, -2) to M(-5, -2), making it horizontal. MP joins M(-5, -2) to P(-5, 2), making it vertical. A horizontal line and a vertical line are perpendicular to each other. So the two perpendicular sides are AM and MP.
In simple words: AM goes left-right and MP goes up-down, so they meet at a right angle.

Exam Tip: Horizontal and vertical lines are always perpendicular - look for constant y-values (horizontal) paired with constant x-values (vertical).

Question 3(ii). One side of RAMP that is parallel to one of the axes.
Answer: AM is parallel to the x-axis because both points A and M have the same y-coordinate (-2). MP is parallel to the y-axis because both points M and P have the same x-coordinate (-5). So one side parallel to an axis is AM, which is parallel to the x-axis, or MP, which is parallel to the y-axis.
In simple words: AM runs left-right like the x-axis. MP runs up-down like the y-axis.

Exam Tip: Lines with the same y-value are parallel to the x-axis; lines with the same x-value are parallel to the y-axis.

Question 3(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: Comparing M(-5, -2) and P(-5, 2): They have the same x-coordinate (-5). Their y-coordinates are equal in size but opposite in sign (-2 and 2). So they are mirror images of each other in the x-axis. The points M and P are mirror images of each other. The axis is the x-axis.
In simple words: Points M and P are the same distance from the x-axis but on opposite sides of it.

Exam Tip: Mirror images across the x-axis have the same x-coordinate but opposite y-coordinates.

Question 4. Plot point Z (5, -6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.
Answer: Point Z is (5, -6). To form a right-angled triangle easily, take I = (5, 0) on the x-axis and N = (0, -6) on the y-axis. Let us verify: IZ is vertical (same x-value of 5) and ZN is horizontal (same y-value of -6). So triangle IZN is right-angled at Z.

Coordinates of the points: I = (5, 0), Z = (5, -6), N = (0, -6).

Now find the lengths of the sides:
1. IZ: Distance between (5, 0) and (5, -6) is 0 - (-6) = 6 units.
2. ZN: Distance between (5, -6) and (0, -6) is 5 - 0 = 5 units.
3. IN: Using distance formula: IN = \(\sqrt{(5 - 0)^2 + (0 - (-6))^2}\) = \(\sqrt{5^2 + 6^2}\) = \(\sqrt{25 + 36}\) = \(\sqrt{61}\) units.

One possible right-angled triangle is formed by I = (5, 0), Z = (5, -6), N = (0, -6). The lengths of the sides are: IZ = 6 units, ZN = 5 units, IN = \(\sqrt{61}\) units.
In simple words: The triangle has sides of length 6, 5, and about 7.8 units.

Exam Tip: When constructing a right-angled triangle, align one vertex with the x-axis and one with the y-axis through your main point to ensure a 90-degree angle.

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?
Answer: If negative numbers did not exist, coordinates could only be zero or positive. This means on the x-axis we could mark only points to the right of the origin, and on the y-axis only points above the origin.

We could locate points only in Quadrant I, on the positive part of the x-axis, on the positive part of the y-axis, and at the origin. We would not be able to reach any points in Quadrant II, Quadrant III, or Quadrant IV, and we could not mark any points on the negative parts of the axes. So such a system would not let us locate all the points on a 2-D plane.
In simple words: Without negative numbers, we could only use the top-right part of the plane, not the whole plane.

Exam Tip: Negative coordinates are essential for mapping all four quadrants of the coordinate plane.

Question 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.
Answer: To check if points M (-3, -4), A (0, 0), and G (6, 8) lie on the same straight line, use the distance formula: d = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

MA = \(\sqrt{(0 + 3)^2 + (0 + 4)^2}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5

AG = \(\sqrt{(6 - 0)^2 + (8 - 0)^2}\) = \(\sqrt{36 + 64}\) = \(\sqrt{100}\) = 10

MG = \(\sqrt{(6 + 3)^2 + (8 + 4)^2}\) = \(\sqrt{81 + 144}\) = \(\sqrt{225}\) = 15

Now checking: MA + AG = 5 + 10 = 15 = MG. Since the sum of two distances equals the third, the points M, A and G lie on the same straight line.
In simple words: When you add the distances from M to A and from A to G, you get the distance from M to G. This means the three points are collinear.

Exam Tip: Three points are collinear if and only if the sum of two smaller distances equals the largest distance.

Question 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.
Answer: To check if points R (-5, -1), B (-2, -5) and C (4, -12) lie on the same straight line, use the distance formula: d = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

RB = \(\sqrt{(-2 + 5)^2 + (-5 + 1)^2}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5

BC = \(\sqrt{(4 + 2)^2 + (-12 + 5)^2}\) = \(\sqrt{36 + 49}\) = \(\sqrt{85}\)

RC = \(\sqrt{(4 + 5)^2 + (-12 + 1)^2}\) = \(\sqrt{81 + 121}\) = \(\sqrt{202}\)

Now check: RB + BC = 5 + \(\sqrt{85}\) ≠ \(\sqrt{202}\). Since the sum of two distances is not equal to the third, the points R, B and C do not lie on the same straight line.
In simple words: When you add the first two distances, they do not equal the third distance, so these points are not in a straight line.

Exam Tip: If the sum of two distances does not equal the third, the points form a triangle, not a line.

Question 8(i). Using the origin as one vertex, plot the vertices of: A right-angled isosceles triangle.
Answer: One possible set of vertices of triangle OAB is: O = (0, 0), A = (4, 0), B = (0, 4). This works because: OA = 4 units, OB = 4 units, and OA is perpendicular to OB. So triangle OAB is a right-angled isosceles triangle.
In simple words: The triangle has two equal sides that meet at a right angle at the origin.

Exam Tip: A right-angled isosceles triangle has two equal sides meeting at 90 degrees.

Question 8(ii). An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
Answer: One possible set of vertices is: O = (0, 0), P = (-3, -4), Q = (3, -4). Explanation: P lies in Quadrant III, Q lies in Quadrant IV. OP = \(\sqrt{(-3)^2 + (-4)^2}\) = \(\sqrt{9 + 16}\) = 5 units, and OQ = \(\sqrt{3^2 + (-4)^2}\) = \(\sqrt{9 + 16}\) = 5 units. So triangle OPQ is an isosceles triangle.
In simple words: The two sides from the origin to the other vertices are equal in length, making it isosceles.

Exam Tip: An isosceles triangle has two sides of equal length - ensure distances from one vertex to the other two are the same.

Question 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.
Answer:

Exam Tip: Use the midpoint formula: Midpoint = \(((x_1 + x_2)/2, (y_1 + y_2)/2)\) to verify answers.

Question 9. When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?
Answer: When M serves as the mid-point of segment ST, the distances SM and MT are always equal. This means M sits exactly halfway along the line segment. To verify this, you can use the distance formula to calculate how far M is from S and how far M is from T - if these two distances match, then M is truly the mid-point. Looking at examples: when S is at (-3, 0), M at (0, 0), and T at (3, 0), both SM and MT equal 3 units. Similarly, for S(2, 3), M(3, 4), and T(4, 5), we get SM = MT = √2. However, when the distances are not equal (like SM = 5 and MT = 15), M is not the mid-point.
In simple words: If M is the mid-point of ST, then M is the same distance from S as it is from T. You can check this by measuring both distances.

Exam Tip: Always use the distance formula to verify the mid-point - if SM equals MT, you have confirmed M is the mid-point. Showing your distance calculations step-by-step is what examiners look for.

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).
Answer: Since M is the mid-point of segment AB, point M must be positioned at equal distance from both A and B. Working with the x-coordinate: the gap from A to M is -7 - 3 = -10. For M to be truly the mid-point, the gap from M to B must also equal -10. Therefore, x = -7 - 10 = -17. For the y-coordinate: the gap from A to M is 1 - (-4) = 5. The gap from M to B must also equal 5. Therefore, y = 1 + 5 = 6. We can verify this using the midpoint formula: (3 + x)/2 = -7 gives 3 + x = -14, so x = -17, and (-4 + y)/2 = 1 gives -4 + y = 2, so y = 6. The coordinates of B are (-17, 6).
In simple words: Find how far A is from M in both x and y. Then move that same distance from M to find B. You can also use the midpoint formula rearranged to solve for the unknown coordinates.

Exam Tip: Show both methods - the symmetry method and the midpoint formula method - to demonstrate complete understanding. Always verify your final answer by checking that the calculated mid-point matches the given point M.

Question 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).
Answer: To trisect AB means to divide the segment into three equal parts using two points, P and Q. The key insight is that P acts as the mid-point between A and some point, while Q acts as the mid-point between P and B. Setting up equations: P is the mid-point of A and Q, so x₁ = (4 + x₂)/2 and y₁ = (7 + y₂)/2. Also, Q is the mid-point of P and B, so x₂ = (x₁ + 16)/2 and y₂ = (y₁ - 2)/2. Solving the x-coordinates: substitute the first equation into the third to get x₂ = ((4 + x₂)/2 + 16)/2, which simplifies to 4x₂ = x₂ + 36, giving x₂ = 12. Then x₁ = (4 + 12)/2 = 8. Solving the y-coordinates similarly: y₂ = ((7 + y₂)/2 - 2)/2, which simplifies to 4y₂ = y₂ + 3, giving y₂ = 1. Then y₁ = (7 + 1)/2 = 4. Therefore, P = (8, 4) and Q = (12, 1).
In simple words: P is halfway between A and Q. Q is halfway between P and B. Use the mid-point formula twice, treating the unknown points as the targets. Substitute and solve the equations to find both sets of coordinates.

Exam Tip: Label your equations clearly (1), (2), (3), (4) and show each substitution step. Examiners reward the method as much as the final answer - clear working demonstrates you understand the relationship between mid-points and trisection.

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 center is the origin O (0, 0). What is the radius of circle K?
Answer: To prove all three points sit on the same circle centered at the origin, calculate the distance from O to each point using the distance formula. For point A: distance OA = √(1² + (-8)²) = √(1 + 64) = √65. For point B: distance OB = √((-4)² + 7²) = √(16 + 49) = √65. For point C: distance OC = √((-7)² + (-4)²) = √(49 + 16) = √65. Since all three distances are equal to √65, all three points are positioned at the same distance from the origin. This confirms they all lie on a circle centered at O(0, 0) with a radius of √65 units.
In simple words: Measure the distance from the center O to each point A, B, and C. If all three distances are the same, the points lie on a circle. That equal distance is the radius.

Exam Tip: Always calculate and display all three distances separately before concluding they are equal. This step-by-step approach shows the examiner your reasoning, not just your final answer.

Question 12(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: Using the circle from Question 12(i) with center O(0, 0) and radius √65 (approximately 8.06), calculate the distance from O to each new point. For point D: distance OD = √((-5)² + 6²) = √(25 + 36) = √61 (approximately 7.81). Comparing √61 ≈ 7.81 with the radius √65 ≈ 8.06, we see that √61 is less than √65. This means D is closer to the center than the circle's boundary, so D lies inside the circle. For point E: distance OE = √(0² + 9²) = 9. Comparing 9 with the radius √65 ≈ 8.06, we see that 9 is greater than √65. This means E is farther from the center than the circle's boundary, so E lies outside the circle. Therefore: D lies inside the circle, and E lies outside the circle.
In simple words: Find the distance from the center O to each point. If that distance is less than the radius, the point is inside. If equal to the radius, the point is on the circle. If greater than the radius, the point is outside.

Exam Tip: Show the numerical comparison clearly (e.g., "√61 ≈ 7.81 < √65 ≈ 8.06"). This direct comparison is what earns marks - avoid just stating a conclusion without showing the calculation.

Question 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.
Answer: Let the vertices of triangle ABC be A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃). Since D is the mid-point of BC, using the mid-point formula: x₂ + x₃ = 10 (equation 1) and y₂ + y₃ = 2 (equation 2). Since E is the mid-point of CA: x₃ + x₁ = 12 (equation 3) and y₃ + y₁ = 10 (equation 4). Since F is the mid-point of AB: x₁ + x₂ = 0 (equation 5) and y₁ + y₂ = 6 (equation 6). From equation 5: x₂ = -x₁. Substituting into equation 1 gives -x₁ + x₃ = 10, so x₃ = 10 + x₁. Substituting this into equation 3: (10 + x₁) + x₁ = 12, which gives 2x₁ = 2, so x₁ = 1. Then x₂ = -1 and x₃ = 11. From equation 6: y₂ = 6 - y₁. Substituting into equation 2 gives (6 - y₁) + y₃ = 2, so y₃ = y₁ - 4. Substituting this into equation 4: (y₁ - 4) + y₁ = 10, which gives 2y₁ = 14, so y₁ = 7. Then y₂ = -1 and y₃ = 3. Therefore, the coordinates are A(1, 7), B(-1, -1), and C(11, 3).
In simple words: Write the mid-point formula for each pair of sides. This gives you six equations with six unknowns. Solve them by substitution, finding x-coordinates first, then y-coordinates separately.

Exam Tip: Label all six equations clearly and show each substitution step. Examiners appreciate seeing the algebraic reasoning - solving in a logical order (x-coordinates, then y-coordinates) demonstrates solid mathematical method.

Question 14(i). 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. Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.
Answer: To build a scaled model of the city, apply the given scale where 1 cm on paper equals 200 m in reality. Since streets are spaced 200 m apart, each separation becomes 1 cm on the model. The city layout calls for 10 N-S streets (drawn as vertical lines) and 10 E-W streets (drawn as horizontal lines). Arrange these so that consecutive lines are 1 cm apart in both directions. The result is a square grid pattern with 10 × 10 = 100 intersection points, each representing where an N-S street meets an E-W street.
In simple words: Draw 10 vertical lines for the N-S streets and 10 horizontal lines for the E-W streets. Space them 1 cm apart. This creates a square grid that represents the city at the given scale.

Exam Tip: Use a ruler to ensure lines are straight and evenly spaced. Accuracy in the grid spacing shows you understand scale and coordinate systems.

Question 14(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: In this naming system, a street intersection is identified by (N-S street number, E-W street number). So (4, 3) refers to the point where the 4th N-S street meets the 3rd E-W street. Similarly, (3, 4) refers to the point where the 3rd N-S street meets the 4th E-W street. Since each N-S street and each E-W street can meet at only one point, there is exactly one location for each intersection. Therefore: (a) Only one street intersection can be called (4, 3). (b) Only one street intersection can be called (3, 4). Note that (4, 3) and (3, 4) are different intersections - the order of the numbers matters.
In simple words: Each intersection is where exactly one N-S street crosses exactly one E-W street. The pair of numbers (a, b) always describes just one point - so (4, 3) is different from (3, 4).

Exam Tip: Emphasize that street intersections follow the ordered pair convention - (4, 3) is not the same as (3, 4). This parallels the coordinate system in geometry.

Question 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.
Answer: (i) For circle A with center at (100, 150) and radius 80: the leftmost point is at 100 - 80 = 20 pixels (on-screen), the rightmost point is at 100 + 80 = 180 pixels (on-screen), the bottom is at 150 - 80 = 70 pixels (on-screen), and the top is at 150 + 80 = 230 pixels (on-screen). All points fall within the 800 × 600 screen, so circle A stays fully inside. For circle B with center at (250, 230) and radius 100: the leftmost point is at 250 - 100 = 150 pixels (on-screen), the rightmost point is at 250 + 100 = 350 pixels (on-screen), the bottom is at 230 - 100 = 130 pixels (on-screen), and the top is at 230 + 100 = 330 pixels (on-screen). All points remain within the screen boundaries, so circle B also stays fully inside. Therefore: no part of either circle lies outside the screen. (ii) To check intersection, calculate the distance between the two centers: distance AB = √((250 - 100)² + (230 - 150)²) = √(150² + 80²) = √(22,500 + 6,400) = √28,900 = 170 pixels. The sum of the two radii is 80 + 100 = 180 pixels. Since the distance between centers (170) is less than the sum of the radii (180), the circles overlap. Therefore: the two circles intersect each other.
In simple words: Check if each circle reaches beyond the screen edges - if the farthest point of each circle is still on-screen, it stays inside. To check if two circles intersect, find the distance between their centers and compare it to the sum of their radii - if the distance is less, they overlap.

Exam Tip: Show the boundary calculations for each circle explicitly. For intersection, always state the distance between centers and compare it to both the sum and difference of radii - this demonstrates complete understanding of the relationship between circles.

Question 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?
Answer: Yes, ABCD is a square. To confirm this, check that all four sides are equal and the diagonals are equal. Calculating all 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 equal √10. Calculating the diagonals: AC = √((-2 - 2)² + (-1 - 1)²) = √(16 + 4) = √20. BD = √((1 - (-1))² + (-2 - 2)²) = √(4 + 16) = √20. Both diagonals equal √20. Since all four sides are equal and both diagonals are equal, ABCD satisfies the definition of a square. The area equals (side)² = (√10)² = 10 square units.
In simple words: Measure all four sides using the distance formula - if they are all the same length, the figure could be a square or a rhombus. Then check the diagonals - if they are also equal, it is definitely a square. The area is just the side length squared.

Exam Tip: Always show all six distance calculations (four sides and two diagonals) - this proves you have checked all the necessary conditions for a square, not just assumed it. State clearly that equal sides and equal diagonals confirm a square.