--- title: "NCERT Solutions Class 9 Maths Orienting Yourself: The Use of Coordinates" url: https://www.swavid.com/maths/class/9/chapter/orienting-yourself-the-use-of-coordinates/ncert-solutions dateModified: 2026-10-07T15:39:44+00:00 --- # NCERT Solutions Class 9 Maths Orienting Yourself: The Use of Coordinates This chapter's questions cover the fundamentals of the Cartesian coordinate system, including plotting points, finding distances between points, and applying coordinate geometry to real-world scenarios like room layouts and grid maps. Free PDF (14 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-1-orienting-yourself-the-use-of-coordinates-4b656a2f59.pdf ## Exercise Set 1.1 ### Question 1 *3 marks · Short answer* Referring to Fig. 1.3, answer the following questions: (i) If $D_1R_1$ 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 $D_1$? (iii) If $R_1$ 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 $B_1(0, 1.5)$ and $B_2(0, 4)$ represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door? **Part (i)** 1. The point $D_1$ is on the x-axis with an x-coordinate of $11.5$, so the door is $11.5$ units (feet) away from the left wall (y-axis). 2. The door lies along the x-axis, so its distance from the x-axis is $0$ units. Answer (i): $11.5$ ft from the left wall and $0$ ft from the x-axis. **Part (ii)** 1. The point $D_1$ is located on the x-axis. 2. Its x-coordinate is $11.5$ and y-coordinate is $0$, so the coordinates are $(11.5, 0)$. Answer (ii): $(11.5, 0)$ **Part (iii)** 1. The door starts at $D_1(11.5, 0)$ and ends at $R_1(11.5, 0)$, but looking at the figure, the width of the door $D_1R_1$ is measured along the x-axis from $11.5$ to $12$, making it $0.5$ ft or 6 inches wide. 2. This is extremely narrow and not a comfortable width for a room door. 3. A person in a wheelchair will not be able to enter the room as standard wheelchair accessible doors require a much larger width. Answer (iii): Width is $0.5$ ft; it is not a comfortable width and a wheelchair user cannot enter easily. **Part (iv)** 1. The bathroom door has ends $B_1(0, 1.5)$ and $B_2(0, 4)$, so its width is $4 - 1.5 = 2.5$ units (feet). 2. The room door width is $0.5$ ft. 3. Comparing the two, the bathroom door ($2.5$ ft) is wider than the room door ($0.5$ ft). Answer (iv): The bathroom door is wider than the room door. **Answer:** Refer to the individual parts for the solutions. > Common mistake: Confusing the coordinates of the door ends with its actual width. ## Exercise Set 1.2 ### Question 1 *3 marks · Short answer* 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? **Part (i)** 1. Opposite sides of a rectangle are equal and parallel. 2. Given three feet at $(8, 9)$, $(11, 9)$, and $(11, 7)$, the fourth foot forms a rectangle with them. 3. The x-coordinate of the fourth foot is $8$ and the y-coordinate is $7$, giving $(8, 7)$. Answer (i): $(8, 7)$ **Part (ii)** 1. The coordinates $(8, 9)$, $(11, 9)$, $(11, 7)$, and $(8, 7)$ place the table neatly against the room boundaries in Fig. 1.5. 2. Hence, this is a good and practical spot for the study table. Answer (ii): Yes, it is a good spot as it fits neatly against the walls. **Part (iii)** 1. The distance between $(8, 9)$ and $(11, 9)$ gives the length as $3\text{ ft}$, and between $(11, 9)$ and $(11, 7)$ gives the width as $2\text{ ft}$. 2. Since Fig. 1.1 and Fig. 1.5 show only a 2-D floor map, the height of the table cannot be determined. Answer (iii): Width = $2\text{ ft}$, Length = $3\text{ ft}$, Height cannot be made out. **Answer:** (i) $(8, 7)$, (ii) Yes, near the corner, (iii) Width = $2\text{ ft}$, Length = $3\text{ ft}$, Height cannot be determined from a 2D floor map. > Common mistake: Confusing the length and width axes or assuming height can be read from a 2-D floor plan. ### Question 2 *3 marks · Short answer* If the bathroom door has a hinge at $B_1$ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider? **Part (i)** 1. The bathroom door has a hinge at $B_1(0, 1.5)$ and opens into the bedroom towards the wardrobe. 2. The wardrobe is located further along the wall near the positive x-axis. 3. Checking the distance and position in Fig. 1.5, the door swings clear of the wardrobe. Answer (i): No, it will not hit the wardrobe. **Part (ii)** 1. If the door is made wider, the arc of the door swing will increase. 2. A wider door might collide with the wardrobe if it exceeds the available clearance space. Answer (ii): Yes, changes such as shifting the hinge or reducing the wardrobe length would be suggested. **Answer:** No, it will not hit the wardrobe. If the door is made wider, it may hit the wardrobe and the hinge position or door width should be adjusted. > Common mistake: Ignoring the direction of the door swing and the position of the wardrobe. ### Question 3 *3 marks · Short answer* 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 $3\text{ ft} \times 2\text{ ft}$ space for the washbasin and a $2\text{ ft} \times 3\text{ ft}$ space for the toilet. Write the coordinates of the corners of these spaces. **Part (i)** 1. From Fig. 1.5, point $O$ is at the origin $(0, 0)$. Answer (i): $O = (0, 0)$, $F = (0, 9)$, $R = (-5, 9)$, and $P = (-5, 0)$ **Part (ii)** 1. The showering area $SHWR$ is rectangular in shape. 2. Its four corners are $S = (-5, 5)$, $H = (-3, 5)$, $W = (-3, 9)$, and $R = (-5, 9)$. Answer (ii): Rectangle; corners are $S = (-5, 5)$, $H = (-3, 5)$, $W = (-3, 9)$, and $R = (-5, 9)$ **Part (iii)** 1. Using the scale $1\text{ cm} : 1\text{ ft}$, place the washbasin and toilet spaces against the walls. 2. The corners of the $3\text{ ft} \times 2\text{ ft}$ washbasin and $2\text{ ft} \times 3\text{ ft}$ toilet spaces are read directly from the grid. Answer (iii): Washbasin space corners are $(-5, 1)$, $(-2, 1)$, $(-2, 3)$, and $(-5, 3)$; toilet space corners are $(-5, 3)$, $(-3, 3)$, $(-3, 6)$, and $(-5, 6)$ **Answer:** The coordinates of the corners, showering area, and fixtures are determined using the given floor plan in Fig. 1.5. > Common mistake: Confusing the order of $x$ and $y$ coordinates when reading points in the second quadrant. ### Question 4 *3 marks · Short answer* Other rooms in the house: (i) Reiaan's room door leads from the dining room which has the length $18\text{ ft}$ and width $15\text{ ft}$. 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 $5\text{ ft} \times 3\text{ ft}$ dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table. **Part (i)** 1. The length of the dining room extends from point $P$ to point $A$ with a length of $18\text{ ft}$ and width of $15\text{ ft}$. 2. Sketch a rectangle starting from point $P$ and mark its four corners using the given dimensions. Answer (i): A sketched rectangle of length $18\text{ ft}$ and width $15\text{ ft}$ with labeled corner coordinates. **Part (ii)** 1. The dining table is $5\text{ ft} \times 3\text{ ft}$ and placed precisely in the centre of the dining room. 2. Calculate the centre of the dining room and find the coordinates of the four feet of the table by offsetting half the table's length and width from the centre. Answer (ii): Coordinates of the four feet of the dining table placed at the centre. **Answer:** (i) Dining room corners from $P$ to $A$ with length $18\text{ ft}$ and width $15\text{ ft}$. (ii) Centre coordinates of the dining table feet. > Common mistake: Incorrectly calculating the centre coordinates of the room when placing the table. ## End-of-Chapter Exercises ### Question 1 *2 marks · Very short answer* What are the x-coordinate and y-coordinate of the point of intersection of the two axes? **Solution** 1. The point of intersection of the x-axis and y-axis is the origin. 2. The x-coordinate and y-coordinate of the origin are both 0. **Answer:** The x-coordinate is 0 and the y-coordinate is 0. > Common mistake: Writing coordinates as a single number instead of stating both x and y coordinates separately. ### Question 2 *3 marks · Short answer* 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? **Solution** 1. Point W has x-coordinate equal to $-5$, so it lies on the vertical line $x = -5$. 2. Point H is on the line through W parallel to the y-axis, which means H also lies on the line $x = -5$. Thus, the x-coordinate of H is $-5$ and its y-coordinate can be any real number $y$. 3. Since the x-coordinate of H is negative ($x = -5$), point H can lie in Quadrant II (when $y > 0$) or Quadrant III (when $y < 0$). **Answer:** The coordinates of H are $(-5, y)$, and H can lie in Quadrant II or Quadrant III. > Common mistake: Forgetting that the y-coordinate can be any real number and giving a fixed value. ### Question 3 *3 marks · Short answer* 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. **Part (i)** 1. The points are $R(3, 0)$, $A(0, -2)$, $M(-5, -2)$, and $P(-5, 2)$. 2. Side RM is horizontal along the line $y = -2$ and side AM is vertical along the line $x = -5$, or side AP is vertical along $x = 0$ and side AM is horizontal. 3. Thus, the two sides RM and AM are perpendicular to each other. Answer (i): Sides RM and AM are perpendicular to each other. **Part (ii)** 1. The y-coordinates of points A $(0, -2)$ and M $(-5, -2)$ are both $-2$. 2. The line segment joining A and M is parallel to the x-axis. Answer (ii): Side AM is parallel to the x-axis. **Part (iii)** 1. Points R $(3, 0)$ and P $(-3, 0)$ are not a pair here, but let us check points R $(3, 0)$ and others. Point A $(0, -2)$ and point $P(-5, 2)$ have y-coordinates with opposite signs. Specifically, points on the y-axis or across axes: points R $(3, 0)$ on the x-axis remains on it. Notice points like $(x, y)$ and $(x, -y)$; here P $(-5, 2)$ and M $(-5, -2)$ have the same x-coordinate and y-coordinates with opposite signs. 2. Therefore, points P $(-5, 2)$ and M $(-5, -2)$ are mirror images of each other in the x-axis. Answer (iii): Points P $(-5, 2)$ and M $(-5, -2)$ are mirror images of each other in the x-axis. **Answer:** RAMP has perpendicular sides RM and AP, side AM parallel to the x-axis, and R and P as mirror images across the x-axis. > Common mistake: Confusing the x-axis and y-axis when identifying mirror images. ### Question 4 *3 marks · Short answer* 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.) **Solution** 1. Plot the point $Z(5, -6)$ on the Cartesian plane. 2. Choose another point I on the x-axis, say $I(5, 0)$, and point N on the horizontal line through Z, say $N(0, -6)$, to form a right-angled triangle IZN at origin or with another vertex. 3. Using the distance formula derived from the Baudhāyana-Pythagoras Theorem, calculate the lengths of the sides of the chosen right-angled triangle formed with axes or coordinate points. **Answer:** The lengths of the sides depend on the chosen triangle vertices, for instance, with origin O $(0,0)$, OZ $= \sqrt{5^2 + (-6)^2} = \sqrt{61}$ units. > Common mistake: Incorrectly calculating the squares of negative coordinates. ### Question 5 *3 marks · Short answer* 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? **Solution** 1. Without negative numbers, the coordinate system would only consist of the first quadrant (where both coordinates are non-negative). 2. All points lying to the left of the y-axis or below the x-axis would have negative coordinates. 3. Therefore, a system without negative numbers would not allow us to locate all points on a 2-D plane. **Answer:** The system would be restricted to only the first quadrant and would not allow us to locate points in the other three quadrants of a 2-D plane. > Common mistake: Stating that all points can still be located by shifting the origin. ### Question 6 *3 marks · Short answer* 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** 1. Find the distances between the pairs of points M $(-3, -4)$, A $(0, 0)$ and G $(6, 8)$ using the distance formula. 2. Calculate $MA = \sqrt{(0 - (-3))^2 + (0 - (-4))^2} = \sqrt{9 + 16} = 5$ units, $AG = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = 10$ units, and $MG = \sqrt{(6 - (-3))^2 + (8 - (-4))^2} = \sqrt{81 + 144} = \sqrt{225} = 15$ units. 3. Since $MA + AG = 5 + 10 = 15 = MG$, the sum of the lengths of two segments equals the length of the third segment, showing that the points lie on the same straight line. **Answer:** Yes, the points are on the same straight line, verified by checking that $MA + AG = MG$. > Common mistake: Checking only the slopes without verifying the segment length relation for collinearity. ### Question 7 *3 marks · Short answer* 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. **Part (i)** 1. Find the distance RB between points R(-5, -1) and B(-2, -5) using the distance formula: RB = $\sqrt{(-2 - (-5))^2 + (-5 - (-1))^2} = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$ units. 2. Find the distance BC between points B(-2, -5) and C(4, -12): BC = $\sqrt{(4 - (-2))^2 + (-12 - (-5))^2} = \sqrt{6^2 + (-7)^2} = \sqrt{36 + 49} = \sqrt{85}$ units. 3. Find the distance RC between points R(-5, -1) and C(4, -12): RC = $\sqrt{(4 - (-5))^2 + (-12 - (-1))^2} = \sqrt{9^2 + (-11)^2} = \sqrt{81 + 121} = \sqrt{202}$ units. 4. Since RB + BC = $5 + \sqrt{85} \neq \sqrt{202}$, the sum of two segments does not equal the third, so the points are not collinear. Answer (i): Not collinear **Answer:** The points R, B, and C are not on the same straight line. > Common mistake: Adding distance values incorrectly without computing the square roots properly. ### Question 8 *3 marks · Short answer* 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. **Part (i)** 1. Let the origin O(0, 0) be the first vertex. 2. Choose the second vertex on the x-axis at A(a, 0) and the third vertex on the y-axis at B(0, a) for any non-zero real number a. 3. For example, taking a = 4, the vertices are (0, 0), (4, 0), and (0, 4), which form a right-angled isosceles triangle. Answer (i): (0, 0), (4, 0), and (0, 4) **Part (ii)** 1. Let the origin O(0, 0) be the first vertex. 2. Choose a vertex in Quadrant III, for example, P(-3, -3), and a vertex in Quadrant IV, for example, Q(3, -3). 3. The distances from the origin are OP = $\sqrt{(-3)^2 + (-3)^2} = \sqrt{18}$ and OQ = $\sqrt{3^2 + (-3)^2} = \sqrt{18}$, making it an isosceles triangle with vertices (0, 0), (-3, -3), and (3, -3). Answer (ii): (0, 0), (-3, -3), and (3, -3) **Answer:** The vertices of the requested triangles can be plotted using the origin as one vertex. > Common mistake: Failing to check that the sides originating from the origin are equal in length for the isosceles triangle. ### Question 9 *3 marks · Short answer* 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. **Part (i)** 1. For $S(-3, 0)$, $M(0, 0)$, and $T(3, 0)$, the distance $SM = 0 - (-3) = 3$ units and $MT = 3 - 0 = 3$ units. 2. Since $SM = MT$ and all three points lie on the same straight line, $M$ is the midpoint of segment $ST$. 3. Yes, $M$ is the midpoint. Answer (i): Yes; $SM = MT = 3$ units. **Part (ii)** 1. For $S(2, 3)$, $M(3, 4)$, and $T(4, 5)$, calculate the distances using the distance formula: $SM = \sqrt{(3-2)^2 + (4-3)^2} = \sqrt{1 + 1} = \sqrt{2}$. 2. Calculate $MT = \sqrt{(4-3)^2 + (5-4)^2} = \sqrt{1 + 1} = \sqrt{2}$ and $ST = \sqrt{(4-2)^2 + (5-3)^2} = \sqrt{4 + 4} = 2\sqrt{2}$. 3. Since $SM + MT = \sqrt{2} + \sqrt{2} = 2\sqrt{2} = ST$ and $SM = MT$, $M$ is the midpoint. 4. Yes, $M$ is the midpoint. Answer (ii): Yes; $SM = MT = \sqrt{2}$ units and $ST = 2\sqrt{2}$ units. **Part (iii)** 1. For $S(0, 0)$, $M(0, 5)$, and $T(0, -10)$, the distance $SM = 5 - 0 = 5$ units while $MT = 5 - (-10) = 15$ units. 2. Since $SM \neq MT$, $M$ is not the midpoint of $ST$. 3. No, $M$ is not the midpoint. Answer (iii): No; $SM = 5$ units and $MT = 15$ units. **Answer:** Completed table for midpoint verification > Common mistake: Checking only distances without verifying that the points are collinear. ### Question 10 *3 marks · Short answer* Use the connection you found to find the coordinates of B given that $M(-7, 1)$ is the midpoint of $A(3, -4)$ and $B(x, y)$. **Solution** 1. From the midpoint connection, the x-coordinate of the midpoint is the average of the x-coordinates of the endpoints: $\frac{3 + x}{2} = -7$. 2. Solving for $x$: $3 + x = -14$, which gives $x = -17$. 3. The y-coordinate of the midpoint is the average of the y-coordinates of the endpoints: $\frac{-4 + y}{2} = 1$. 4. Solving for $y$: $-4 + y = 2$, which gives $y = 6$. 5. Therefore, the coordinates of point $B$ are $(-17, 6)$. **Answer:** $B(-17, 6)$ > Common mistake: Forgetting to multiply the midpoint coordinates by 2 before finding the endpoint. ### Question 11 *3 marks · Short answer* 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 $B(16, -2)$. **Solution** 1. Understand that points of trisection P and Q divide the segment AB into three equal parts, meaning P is the midpoint of AQ or Q is the midpoint of PB, or using the section concept. 2. Using the midpoint method described in the chapter: P divides AB in the ratio 1:2 and Q divides AB in the ratio 2:1. 3. For A(4, 7) and B(16, -2), find the x-coordinate of P by adding one-third of the difference in x-coordinates to the x-coordinate of A: $x_p = 4 + \frac{1}{3}(16 - 4) = 4 + 4 = 8$. 4. Find the y-coordinate of P: $y_p = 7 + \frac{1}{3}(-2 - 7) = 7 - 3 = 4$. So P(8, 4). 5. Find the coordinates of Q by adding two-thirds of the differences or using Q as midpoint of PB: $x_q = 4 + \frac{2}{3}(12) = 12$ and $y_q = 7 + \frac{2}{3}(-9) = 1$. So Q(12, 1). **Answer:** P(8, 4) and Q(12, 1) > Common mistake: Mixing up the ratios 1:2 and 2:1 for P and Q. ### Question 12 *3 marks · Short answer* (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. **Part (i)** 1. Find the distance of point $A(1, -8)$ from the origin $O(0, 0)$: $OA = \sqrt{(1-0)^2 + (-8-0)^2} = \sqrt{1 + 64} = \sqrt{65}$. 2. Find the distance $OB$ for $B(-4, 7)$: $OB = \sqrt{(-4-0)^2 + (7-0)^2} = \sqrt{16 + 49} = \sqrt{65}$. 3. Find the distance $OC$ for $C(-7, -4)$: $OC = \sqrt{(-7-0)^2 + (-4-0)^2} = \sqrt{49 + 16} = \sqrt{65}$. 4. Since the distances of $A$, $B$, and $C$ from the origin are equal, they lie on a circle with center $O$ and radius $\sqrt{65}$ units. Answer (i): All points are at a distance of $\sqrt{65}$ units from the origin, so the radius of circle $K$ is $\sqrt{65}$ units. **Part (ii)** 1. Calculate the distance of $D(-5, 6)$ from the origin: $OD = \sqrt{(-5-0)^2 + (6-0)^2} = \sqrt{25 + 36} = \sqrt{61}$. 2. Since $\sqrt{61} < \sqrt{65}$, point $D$ lies within the circle. 3. Calculate the distance of $E(0, 9)$ from the origin: $OE = \sqrt{(0-0)^2 + (9-0)^2} = \sqrt{81} = 9$. 4. Since $9 > \sqrt{65}$ (as $81 > 65$), point $E$ lies outside the circle. Answer (ii): $D$ lies within the circle and $E$ lies outside the circle. **Answer:** Radius is $\sqrt{65}$ units; $D$ lies on the circle and $E$ lies outside. > Common mistake: Comparing square roots incorrectly by not squaring both values. ### Question 13 *3 marks · Short answer* 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. **Part (i)** 1. Let the vertices of triangle ABC be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. 2. Using the midpoint relations with given midpoints $D(5, 1)$, $E(6, 5)$, and $F(0, 3)$, we get $x_1 + x_2 = 10$, $x_2 + x_3 = 12$, and $x_3 + x_1 = 0$. 3. Adding all three equations gives $2(x_1 + x_2 + x_3) = 22$, so $x_1 + x_2 + x_3 = 11$. 4. Subtracting each side equation from the sum gives $x_1 = -1$, $x_2 = 11$, and $x_3 = 1$. 5. Similarly for y-coordinates, $y_1 + y_2 = 2$, $y_2 + y_3 = 10$, and $y_3 + y_1 = 6$, giving $y_1 + y_2 + y_3 = 9$. 6. Solving gives $y_1 = -1$, $y_2 = 3$, and $y_3 = 9$. 7. Therefore, the coordinates are $A(-1, -1)$, $B(11, 3)$, and $C(1, 9)$. Answer (i): A(-1, -1), B(11, 3), C(1, 9) **Answer:** The coordinates of vertices A, B, and C are $(-1, -1)$, $(11, 3)$, and $(1, 9)$ respectively. > Common mistake: Mixing up the indices of vertices while setting up the midpoint equations. ### Question 14 *3 marks · Short answer* 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 $200\text{ m}$ apart. There are 10 streets in each direction. (i) Using $1\text{ cm} = 200\text{ m}$, 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 $(3, 4)$. **Part (i)** 1. Draw two perpendicular main roads representing the x-axis and y-axis intersecting at the origin. 2. Draw 10 parallel lines for streets in the North-South direction and 10 parallel lines for the East-West direction, spaced $1\text{ cm}$ apart. Answer (i): Model drawn with parallel lines $1\text{ cm}$ apart. **Part (ii)** 1. Each street intersection is uniquely determined by the crossing of one N-S street and one E-W street. 2. Therefore, (a) only 1 intersection can be referred to as $(4, 3)$, and (b) only 1 intersection can be referred to as $(3, 4)$. Answer (ii): (a) 1 intersection, (b) 1 intersection **Answer:** (i) A grid with 10 streets in each direction drawn using scale $1\text{ cm} = 200\text{ m}$. (ii) (a) Exactly 1 intersection. (b) Exactly 1 intersection. > Common mistake: Interpreting grid coordinates as representing multiple physical crossings. ### Question 15 *3 marks · Short answer* 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. **Part (i)** 1. For circle A with centre $(100, 150)$ and radius $80$, the leftmost point is $100 - 80 = 20$ and lowest is $150 - 80 = 70$, both greater than 0. 2. The rightmost point is $100 + 80 = 180 < 800$ and highest is $150 + 80 = 230 < 600$. Thus circle A is fully inside. 3. For circle B with centre $(250, 230)$ and radius $100$, bounds are $250 \pm 100$ ($150$ to $350$) and $230 \pm 100$ ($130$ to $330$), which lie well within the $800 \times 600$ screen. Answer (i): Neither circle lies outside the screen. **Part (ii)** 1. Find the distance between centres A$(100, 150)$ and B$(250, 230)$ using the distance formula. 2. $AB = \sqrt{(250 - 100)^2 + (230 - 150)^2} = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900} = 170$ pixels. 3. The sum of radii is $80 + 100 = 180$ pixels, and the difference is $100 - 80 = 20$ pixels. 4. Since the distance between centres ($170$) is less than the sum of radii ($180$) and greater than their difference ($20$), the two circles intersect each other. Answer (ii): The two circles intersect each other. **Answer:** (i) No part of either circle lies outside the screen. (ii) The two circles do not intersect each other. > Common mistake: Comparing the distance between centres only with the individual radii instead of their sum and difference. ### Question 16 *3 marks · Short answer* 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? **Part (i)** 1. Plot the points $A(2, 1)$, $B(-1, 2)$, $C(-2, -1)$, and $D(1, -2)$ on the coordinate plane. 2. Calculate the lengths of all four sides using the distance formula: $AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{9+1} = \sqrt{10}$. 3. $BC = \sqrt{(-2-(-1))^2 + (-1-2)^2} = \sqrt{1+9} = \sqrt{10}$, $CD = \sqrt{(1-(-2))^2 + (-2-(-1))^2} = \sqrt{9+1} = \sqrt{10}$, and $DA = \sqrt{(2-1)^2 + (1-(-2))^2} = \sqrt{1+9} = \sqrt{10}$. 4. Since all four sides are equal ($AB = BC = CD = DA = \sqrt{10}$) and diagonals calculated as $AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{16+4} = \sqrt{20}$ and $BD = \sqrt{(1-(-1))^2 + (-2-2)^2} = \sqrt{4+16} = \sqrt{20}$ are equal, ABCD is a square. 5. Area of square = $(\text{side})^2 = (\sqrt{10})^2 = 10$ square units. Answer (i): Yes, it is a square with an area of $10$ square units. **Answer:** Yes, ABCD is a square. Area of the square is $10$ square units. > Common mistake: Proving only that all four sides are equal without checking that diagonals are also equal (rhombus vs square). ## Frequently asked questions ### How many exercises and questions are there in Class 9 Maths Chapter 1 Orienting Yourself The Use of Coordinates? This chapter contains Exercise Set 1.1 with 1 question, Exercise Set 1.2 with 4 questions, and an End-of-Chapter Exercise section with 16 questions. You can find step-by-step solutions for all these questions in the free PDF available on this SwaVid page. ### What topics are covered in the exercises of this chapter? Exercise Set 1.1 covers coordinate geometry and distance along axes, while Exercise Set 1.2 focuses on coordinates of points, geometric shapes in a Cartesian plane, and sketching rooms. The End-of-Chapter exercises explore advanced concepts like the distance formula, midpoint coordinates, collinearity, and circle intersections. ### What are the hardest question types in this chapter and how should I approach them? The most challenging questions involve checking geometric properties like squares using the distance formula or finding triangle vertices from side midpoints. To solve these, you should carefully plot the points, clearly state formulas such as $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, and proceed step-by-step. ### How should I write my answers to score full marks in Class 9 coordinate geometry? You should always begin by writing the given coordinates and the relevant mathematical formula clearly. Drawing a rough Cartesian sketch and showing every intermediate calculation will help you secure full marks. ### Is a free PDF of these NCERT solutions available for the 2026-27 session? Yes, complete solutions based on the new NCERT book for the 2026-27 session are provided in the free PDF on this SwaVid page. These solutions are designed to help students understand each concept thoroughly. ## Related pages - [Exercise 1.1 solutions](https://www.swavid.com/maths/class/9/chapter/orienting-yourself-the-use-of-coordinates/ncert-solutions/exercise-1-1) - [Exercise 1.2 solutions](https://www.swavid.com/maths/class/9/chapter/orienting-yourself-the-use-of-coordinates/ncert-solutions/exercise-1-2) - [Class 9 Maths chapters](https://www.swavid.com/maths/class/9) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. 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