{"id":31492,"date":"2026-05-11T16:37:10","date_gmt":"2026-05-11T11:07:10","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31492"},"modified":"2026-05-11T16:38:08","modified_gmt":"2026-05-11T11:08:08","slug":"the-use-of-coordinates-ncert-solutions-class-9-ganita-manjari","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/the-use-of-coordinates-ncert-solutions-class-9-ganita-manjari\/","title":{"rendered":"The Use of Coordinates &#8211; NCERT Solutions Class 9 Ganita Manjari"},"content":{"rendered":"\n<p>The Use of Coordinates &#8211; NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 <a href=\"https:\/\/ncert.nic.in\/textbook.php?iemh1=0-8\">Ganita Manjari<\/a> textbook.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">NCERT Solutions Class 9<\/h2>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-english-kaveri\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >English Kaveri<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-hindi-ganga\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Hindi Ganga<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-sanskrit-sharada\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Sanskrit Sharada<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-maths-ganita-manjari\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Maths Ganita Manjari<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Science Exploration<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-social-understanding-society\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Social Understanding Society<\/a>\n\n\n\n<h2 class=\"wp-block-heading\">The Use of Coordinates \u2013 NCERT Solutions<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.1:<\/p>\n\n\n\n<p>Figure shows Reiaan\u2019s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777009349-n8unp5.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>If {tex}D_1 R_1{\/tex} represents the door to Reiaan&#8217;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? <strong>(1)<\/strong><\/li>\n\n\n\n<li>What are the coordinates of {tex}D_1{\/tex}? <strong>(1)<\/strong><\/li>\n\n\n\n<li>If {tex}R_1{\/tex} is the point {tex}(11.5,0){\/tex}, 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?\u00a0<strong> (2)<\/strong><br><strong>OR<\/strong><br>If {tex}{B}_1(0,1.5){\/tex} and {tex}{B}_2(0,4){\/tex} represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?\u00a0<strong> (2)<\/strong><\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The room door lies on the {tex}x{\/tex}-axis, so its distance from the {tex}x{\/tex}-axis {tex}=0{\/tex} units.<br>From the figure, {tex} {D}_1=(8,0) \\text { and } {R}_1=(11.5,0) \\text {. }{\/tex} So the door begins 8 units from the {tex}y{\/tex}-axis.<\/li>\n\n\n\n<li>The coordinates of {tex}{D}_1{\/tex} are {tex}(8,0){\/tex}.<\/li>\n\n\n\n<li>Coordinates of {tex}{D}_1=(8,0){\/tex} and {tex}{R}_1=(11.5,0){\/tex}. So, the width of the door<br>{tex}={\/tex} distance between {tex}{D}_1{\/tex} and {tex}{R}_1{\/tex} {tex} \\text { = } 11.5-8{\/tex} {tex}=3.5{\/tex} units If the unit represents feet, then {tex}3.5 {ft} \\approx 42{\/tex} inches, which is quite comfortable for a room door. Standard residential doors are usually 30-36 inches wide, so this is actually slightly wider than average, which is good. <strong>For wheelchair accessibility:<\/strong> A wheelchair typically needs at least 32 inches {tex}(\\sim 2.7 {ft}){\/tex} clear width. Since {tex}3.5 {ft}>2.7 {ft}{\/tex}, the door is wide enough. So, we can say yes, this door width is comfortable and a person using a wheelchair should be able to enter easily without difficulty.<br><strong>OR<\/strong> Coordinates of {tex}{B}_1=(0,1.5){\/tex} and {tex}{B}_2=(0,4){\/tex}. So, the width of the bathroom door<br>{tex}={\/tex} distance between {tex}{B}_1{\/tex} and {tex}{B}_2{\/tex}<br>{tex}=4-1.5=2.5{\/tex} units<br>Since {tex}2.5&lt;3.5{\/tex}, the bathroom door is narrower than the room door.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.2:<\/p>\n\n\n\n<p>On a graph sheet, mark the x -axis and y -axis and the origin O. Mark points from {tex}(-7,0){\/tex} to {tex}(13,0){\/tex} on the {tex}x{\/tex}-axis and from {tex}(0,-15){\/tex} to {tex}(0,12){\/tex} on the y-axis. (Use the scale {tex}1 {~cm}=1{\/tex} unit.) Using Figure, answer the given question-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777282982-xxeywm.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>Place Reiaan&#8217;s rectangular study table with three of its feet at the points {tex}(8,9),(11,9){\/tex} and {tex}(11,7){\/tex}.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Where will the fourth foot of the table be?<\/li>\n\n\n\n<li>Is this a good spot for the table?<\/li>\n\n\n\n<li>What is the width of the table? The length? Can you make out the height of the table?<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777012448-dzu25m.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The given three points form three corners of a rectangle: {tex}A=(8,9){\/tex}<br>{tex}B=(11,9){\/tex}<br>{tex}C=(11,7){\/tex} To complete the rectangle, the fourth point must have:\n<ul class=\"wp-block-list\">\n<li>same {tex}x{\/tex}-coordinate as {tex}A \\rightarrow 8{\/tex}<\/li>\n\n\n\n<li>same y -coordinate as {tex}{C} \\rightarrow 7{\/tex} Therefore, the fourth foot is at: {tex}(8,7){\/tex}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Yes, this is a good spot because:\n<ul class=\"wp-block-list\">\n<li>The table is placed neatly inside the room.<\/li>\n\n\n\n<li>It does not block doors or pathways.<\/li>\n\n\n\n<li>It is positioned near the wall, which is practical for study.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Width:<br>Distance between {tex}(8,9){\/tex} and {tex}(11,9){\/tex} {tex} =11-8=3 \\text { units }{\/tex} Length:<br>Distance between (11, 9) and (11, 7) {tex} =9-7=2 \\text { units }{\/tex} Height:<br>The height of the table cannot be determined from the given diagram because the figure represents only a top view (2D), not vertical dimensions.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.3:<\/p>\n\n\n\n<p>On a graph sheet, mark the x -axis and y -axis and the origin O. Mark points from {tex}(-7,0){\/tex} to {tex}(13,0){\/tex} on the {tex}x{\/tex}-axis and from {tex}(0,-15){\/tex} to {tex}(0,12){\/tex} on the y -axis. (Use the scale {tex}1 {~cm}=1{\/tex} unit.) Using Figure, answer the given question-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777011668-qd7jrd.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>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?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>From Figure:<\/p>\n\n\n\n<p>{tex}-B_1=(0,1.5){\/tex}<br>{tex}-B_2=(0,4){\/tex}<\/p>\n\n\n\n<p>So, the bathroom door has width:<\/p>\n\n\n\n<p>{tex}=4-1.5{\/tex}<br>{tex}=2.5 \\text { units }{\/tex}<\/p>\n\n\n\n<p>If the door is hinged at {tex}{B}_1{\/tex} and opens into the bedroom, it will sweep an arc of radius 2.5 units from {tex}B_1{\/tex}.<\/p>\n\n\n\n<p>Now the wardrobe begins at:<\/p>\n\n\n\n<p>{tex}W_1=(3,0){\/tex}<br>{tex}W_4=(3,2){\/tex}<\/p>\n\n\n\n<p>The nearest point of the wardrobe from {tex}{B}_1(0,1.5){\/tex} is around {tex}{x}=3{\/tex}, which is farther than the door width 2.5 units.<\/p>\n\n\n\n<p>Therefore, the bathroom door will not hit the wardrobe.<br>Suggestion if the door is made wider:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If the door becomes much wider, it may come close to or hit the wardrobe.<\/li>\n\n\n\n<li>In that case, the door could be made to open inward into the bathroom, or<\/li>\n\n\n\n<li>the wardrobe could be shifted slightly to the right or<\/li>\n\n\n\n<li>the door width could be kept limited for comfortable movement.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.4:<\/p>\n\n\n\n<p>On a graph sheet, mark the x -axis and y -axis and the origin O. Mark points from {tex}(-7,0){\/tex} to {tex}(13,0){\/tex} on the {tex}x{\/tex}-axis and from {tex}(0,-15){\/tex} to {tex}(0,12){\/tex} on the y -axis. (Use the scale {tex}1 {~cm}=1{\/tex} unit.) Using Figure, answer the given question-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777011668-qd7jrd.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>Look at Reiaan&#8217;s bathroom.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>What are the coordinates of the four corners {tex}{O}, {F}, {R}{\/tex}, and P of the bathroom?<\/li>\n\n\n\n<li>What is the shape of the showering area SHWR in Reiaan&#8217;s bathroom? Write the coordinates of the four corners.<\/li>\n\n\n\n<li>Mark off a {tex}3 {ft} \\times 2 {ft}{\/tex} space for the washbasin and a {tex}2 {ft} \\times 3 {ft}{\/tex} space for the toilet. Write the coordinates of the corners of these spaces.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>From the figure:<br>The coordinates of the four corners {tex}{O}, {F}, {R}{\/tex}, and P of the bathroom are: {tex}O=(0,0){\/tex}<br>{tex}F=(0,9){\/tex}<br>{tex}R=(-6,9){\/tex}<br>{tex}P=(-6,0){\/tex}<\/li>\n\n\n\n<li>From the figure: {tex}{S}=(-6,5){\/tex}<br>{tex}{H}=(-3,5){\/tex}<br>{tex}{W}=(-2,9){\/tex}<br>{tex}{R}=(-6,9){\/tex} Since one pair of opposite sides is parallel, SHWR is a trapezium.<br>Shape of SHWR = Trapezium<br>Coordinates of the corners: {tex}{S}=(-6,5){\/tex}<br>{tex}{H}=(-3,5){\/tex}<br>{tex}{W}=(-2,9){\/tex}<br>{tex}{R}=(-6,9){\/tex}<\/li>\n\n\n\n<li>Washbasin space ({tex}3 {ft} \\times 2 {ft}{\/tex}):<br>Take the rectangle at the bottom-left corner of the bathroom.<br>Coordinates of Corners: {tex} (-6,0),(-3,0),(-3,2) \\text { and }(-6,2){\/tex} Toilet space ( {tex}2 {ft} \\times 3 {ft}{\/tex} ):<br>Take a rectangle above the washbasin.<br>Coordinates of Corners: {tex} (-6,2),(-4,2),(-4,5) \\text { and }(-6,5){\/tex} Coordinates of Washbasin corners: {tex} (-6,0),(-3,0),(-3,2) \\text { and }(-6,2){\/tex} Coordinates of Toilet corners: {tex} (-6,2),(-4,2),(-4,5) \\text { and }(-6,5){\/tex}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.5:<\/p>\n\n\n\n<p>On a graph sheet, mark the x -axis and y -axis and the origin O. Mark points from {tex}(-7,0){\/tex} to {tex}(13,0){\/tex} on the {tex}x{\/tex}-axis and from {tex}(0,-15){\/tex} to {tex}(0,12){\/tex} on the y -axis. (Use the scale {tex}1 {~cm}=1{\/tex} unit.) Using Figure, answer the given question-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777011668-qd7jrd.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>Other rooms in the house:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Reiaan&#8217;s room door leads from the dining room which has the 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.<\/li>\n\n\n\n<li>Place a rectangular {tex}5 {ft} \\times 3 {ft}{\/tex} dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>From the Figure:<br>Coordinates of {tex}{P}=(-6,0){\/tex}<br>Coordinates of {tex}{A}=(12,0){\/tex}<br>So, the length of {tex}{PA}=12-(-6)=18 {ft}{\/tex}, which matches the given length.<br>If the dining room is 15 ft wide and lies below PA, then its upper side is PA and it extends downward 15 units. Hence the coordinates of four corners are:\n<ul class=\"wp-block-list\">\n<li>{tex}{P}=(-6,0){\/tex}<\/li>\n\n\n\n<li>{tex}{A}=(12,0){\/tex}<\/li>\n\n\n\n<li>{tex}{Q}=(12,-15){\/tex}<\/li>\n\n\n\n<li>{tex}{S}=(-6,-15){\/tex} The coordinates of the dining room corners are: {tex} (-6,0),(12,0),(12,-15),(-6,-15){\/tex}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The dining room extends:<br>From {tex}{x}=-6{\/tex} to {tex}{x}=12{\/tex}<br>From {tex}y=0{\/tex} to {tex}y=-15{\/tex}<br>Centre of the dining room:<br>x -coordinate of centre {tex}=(-6+12) \/ 2=3{\/tex}<br>{tex}y{\/tex}-coordinate of centre {tex}=(0+(-15)) \/ 2=-7.5{\/tex}<br>Now place a {tex}5 {ft} \\times 3 {ft}{\/tex} table at the centre.<br>Taking length {tex}=5{\/tex} units along the {tex}x{\/tex}-axis and width {tex}=3{\/tex} units along the {tex}y{\/tex}-axis:<br>Half-length = 2.5<br>Half-width = 1.5 So the coordinates of corners (feet) of the table are:\n<ul class=\"wp-block-list\">\n<li>{tex}(3-2.5,-7.5-1.5)=(0.5,-9){\/tex}<\/li>\n\n\n\n<li>{tex}(3+2.5,-7.5-1.5)=(5.5,-9){\/tex}<\/li>\n\n\n\n<li>{tex}(3+2.5,-7.5+1.5)=(5.5,-6){\/tex}<\/li>\n\n\n\n<li>{tex}(3-2.5,-7.5+1.5)=(0.5,-6){\/tex} The coordinates of the four feet of the dining table are:<br>{tex}(0.5,-9),(5.5,-9),(5.5,-6),(0.5,-6){\/tex}<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.6:<\/p>\n\n\n\n<p>What are the x-coordinate and y-coordinate of the point of intersection of the two axes?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The {tex}x{\/tex}-axis and {tex}y{\/tex}-axis intersect at the origin.<br>Therefore:<br>The x -coordinate {tex}=0{\/tex}<br>The y -coordinate {tex}=0{\/tex}<br>So, the point of intersection is {tex}(0,0){\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.7:<\/p>\n\n\n\n<p>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?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>If point {tex}W{\/tex} has {tex}x{\/tex}-coordinate -5, then any point on the line through {tex}W{\/tex} parallel to the {tex}y{\/tex}-axis will also have {tex}x{\/tex}-coordinate -5.<\/p>\n\n\n\n<p>Therefore, the coordinates of H will be of the form:<br>{tex}H=(-5, y){\/tex}, where {tex}y{\/tex} can be any real number.<\/p>\n\n\n\n<p>Now, depending on the value of {tex}y{\/tex}:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If {tex}y>0{\/tex}, then H lies in Quadrant II.<\/li>\n\n\n\n<li>If {tex}y&lt;0{\/tex}, then {tex}H{\/tex} lies in Quadrant III.<\/li>\n\n\n\n<li>If {tex}y=0{\/tex}, then {tex}H{\/tex} lies on the {tex}x{\/tex}-axis.<\/li>\n<\/ul>\n\n\n\n<p>So, H can lie in:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Quadrant II<\/li>\n\n\n\n<li>Quadrant III<\/li>\n\n\n\n<li>or on the x-axis<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.8:<\/p>\n\n\n\n<p>Consider the points {tex}{R}(3,0), {A}(0,-2), {M}(-5,-2){\/tex} and {tex}{P}(-5,2){\/tex}. If they are joined in the same order, predict:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Two sides of RAMP that are perpendicular to each other.<\/li>\n\n\n\n<li>One side of RAMP that is parallel to one of the axes.<\/li>\n\n\n\n<li>Two points that are mirror images of each other in one axis. Which axis will this be?<\/li>\n<\/ol>\n\n\n\n<p>Now plot the points and verify your predictions.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Let us observe:<br>AM joins {tex}{A}(0,-2){\/tex} to {tex}{M}(-5,-2){\/tex}, so it is horizontal.<br>MP joins {tex}{M}(-5,-2){\/tex} to {tex}{P}(-5,2){\/tex}, so it is vertical.<br>A horizontal line and a vertical line are perpendicular.<br>Therefore:<br>{tex}{AM} \\perp {MP}{\/tex}<br>The two perpendicular sides are AM and MP.<\/li>\n\n\n\n<li>{tex}A M{\/tex} is parallel to the {tex}x{\/tex}-axis because both points {tex}A{\/tex} and {tex}M{\/tex} have the same {tex}y{\/tex}-coordinate {tex}(-2){\/tex}. MP is parallel to the {tex}y{\/tex}-axis because both points {tex}M{\/tex} and {tex}P{\/tex} have the same {tex}x{\/tex}-coordinate {tex}(-5){\/tex}. One side parallel to an axis is:<br>AM, which is parallel to the x -axis or MP, which is parallel to the y -axis<\/li>\n\n\n\n<li>Compare {tex}{M}(-5,-2){\/tex} and {tex}{P}(-5,2){\/tex}:<br>They have the same x -coordinate.<br>Their {tex}y{\/tex}-coordinates are equal in magnitude but opposite in sign.<br>So, they are mirror images of each other in the x-axis.<br>The points M and P are mirror images of each other.<br>The axis is the x-axis.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.9:<\/p>\n\n\n\n<p>Plot point {tex}Z(5,-6){\/tex} on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.<br>(Comment: Answers may differ from person to person.)<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Point Z is {tex}(5,-6){\/tex}.<\/p>\n\n\n\n<p>To form a right-angled triangle easily, take:<br>{tex}I=(5,0){\/tex} on the {tex}x{\/tex}-axis<br>{tex}{N}=(0,-6){\/tex} on the y-axis<\/p>\n\n\n\n<p>Then triangle {tex}I Z N{\/tex} is right-angled at {tex}Z{\/tex}? Let&#8217;s check:<br>IZ is vertical<br>ZN is horizontal<br>So, triangle IZN is right-angled at Z.<\/p>\n\n\n\n<p>Coordinates of the points:<\/p>\n\n\n\n<p>{tex}I=(5,0){\/tex}<br>{tex}Z=(5,-6){\/tex}<br>{tex}N=(0,-6){\/tex}Now find the lengths of the sides:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>IZ: Distance between {tex}(5,0){\/tex} and {tex}(5,-6){\/tex}<br>{tex}=0-(-6)=6{\/tex} units<\/li>\n\n\n\n<li>ZN: Distance between {tex}(5,-6){\/tex} and {tex}(0,-6){\/tex} {tex} =5-0=5 \\text { units }{\/tex}<\/li>\n\n\n\n<li>IN: Using distance formula: {tex}\\text { IN }=\\sqrt{ \\left[(5-0)^2+(0-(-6))^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left(5^2+6^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (25+36)}{\/tex}<br>{tex}=\\sqrt{61 } \\text { units }{\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Therefore, one possible right-angled triangle is formed by:<\/p>\n\n\n\n<p>{tex}I=(5,0){\/tex}<br>{tex}Z=(5,-6){\/tex}<br>{tex}N=(0,-6){\/tex}<\/p>\n\n\n\n<p>Lengths of the sides:<br>IZ {tex}=6{\/tex} units<br>{tex}{ZN}=5{\/tex} units<br>IN {tex}=\\sqrt{ 61} {\/tex} units<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777348525-4sqrym.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.10:<\/p>\n\n\n\n<p>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?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>If we did not have negative numbers, then coordinates could only be zero or positive.<br>So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>On the {tex}x{\/tex}-axis, we could mark only the points to the right of the origin.<\/li>\n\n\n\n<li>On the {tex}y{\/tex}-axis, we could mark only the points above the origin.<\/li>\n<\/ul>\n\n\n\n<p>This means we could locate points only in:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Quadrant I<\/li>\n\n\n\n<li>the positive part of the {tex}x{\/tex}-axis<\/li>\n\n\n\n<li>the positive part of the {tex}y{\/tex}-axis<\/li>\n\n\n\n<li>and the origin<\/li>\n<\/ul>\n\n\n\n<p>We would not be able to locate:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>points in Quadrant II<\/li>\n\n\n\n<li>points in Quadrant III<\/li>\n\n\n\n<li>points in Quadrant IV<\/li>\n\n\n\n<li>points on the negative parts of the axes<\/li>\n<\/ul>\n\n\n\n<p>Therefore, such a system would not allow us to locate all the points on a 2-D plane.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.11:<\/p>\n\n\n\n<p>Are the points {tex}{M}(-3,-4), {A}(0,0){\/tex} and {tex}{G}(6,8){\/tex} on the same straight line? Suggest a method to check this without plotting and joining the points.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To check whether the points {tex}{M}(-3,-4), {A}(0,0){\/tex}, and {tex}{G}(6,8){\/tex} lie on the same straight line using the Distance formula:<\/p>\n\n\n\n<p>{tex}d=\\sqrt{ \\left[\\left(x_2-x_1\\right)^2+\\left(y_2-y_1\\right)^2\\right]}{\/tex}<br>{tex}M A=\\sqrt{ \\left[(0+3)^2+(0+4)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left(3^2+4^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (9+16)=\\sqrt{ } 25=5}{\/tex}<\/p>\n\n\n\n<p>{tex}A G=\\sqrt{ \\left[(6-0)^2+(8-0)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ (36+64)}{\/tex}<br>{tex}=\\sqrt{ } 100=10{\/tex}<\/p>\n\n\n\n<p>{tex}M G=\\sqrt{ \\left[(6+3)^2+(8+4)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left(9^2+12^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (81+144)}{\/tex}<br>{tex}=\\sqrt{ 225=15} {\/tex}<\/p>\n\n\n\n<p>Now checking: {tex}{MA}+{AG}=5+10=15={MG}{\/tex}<br>Since the sum of two distances equals the third, the points {tex}{M}, {A}{\/tex} and G lie on the same straight Line.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.12:<\/p>\n\n\n\n<p>Use your method (from Problem 6) to check if the points {tex}{R}(-5,-1), {B}(-2,-5){\/tex} and {tex}{C}(4,-12){\/tex} are on the same straight line. Now plot both sets of points and check your answers.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To check whether the points {tex}{R}(-5,-1), {B}(-2,-5){\/tex} and {tex}{C}(4,-12){\/tex} lie on the same straight line using the distance formula: {tex}d=\\sqrt{ \\left[\\left(x_2-x_1\\right)^2+\\left(y_2-y_1\\right)^2\\right]}{\/tex}<\/p>\n\n\n\n<p>{tex}R B=\\sqrt{ \\left[(-2+5)^2+(-5+1)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left(3^2+(-4)^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (9+16)=\\sqrt{ } 25=5}{\/tex}<\/p>\n\n\n\n<p>{tex}B C=\\sqrt{ \\left[(4+2)^2+(-12+5)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left(6^2+(-7)^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (36+49)=\\sqrt{ } 85}{\/tex}<\/p>\n\n\n\n<p>{tex}{RC}=\\sqrt{ \\left[(4+5)^2+(-12+1)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left(9^2+(-11)^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (81+121)=\\sqrt{ } 202}{\/tex}<\/p>\n\n\n\n<p>Now: {tex}{RB}+{BC}=5+\\sqrt{85} \\neq \\sqrt{202}{\/tex}<br>Since the sum of two distances is not equal to the third, the points {tex}{R}, {B}{\/tex} and C do not lie on the same straight line.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777015078-fk3w64.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.13:<\/p>\n\n\n\n<p>Using the origin as one vertex, plot the vertices of:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>A right-angled isosceles triangle.<\/li>\n\n\n\n<li>An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>One possible set of vertices of triangle {tex}O A B{\/tex} is: {tex}O=(0,0){\/tex}<br>{tex}A=(4,0){\/tex}<br>{tex}B=(0,4){\/tex} Because:<br>{tex}{OA}=4{\/tex} units<br>{tex}O B=4{\/tex} units<br>OA is perpendicular to OB<br>So triangle OAB is a right-angled isosceles triangle. <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777015540-kcfxqk.jpg\"><\/li>\n\n\n\n<li>One possible set of vertices is: {tex} O=(0,0), P=(-3,-4), Q=(3,-4){\/tex} Explanation:<br>P lies in Quadrant III<br>Q Lies in Quadrant IV<br>{tex}O P=O Q=5{\/tex} units<br>So triangle OPQ is an isosceles triangle. <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777015582-hnr9e5.jpg\"><\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.14:<\/p>\n\n\n\n<p>The following table shows the coordinates of points {tex}{S}, {M}{\/tex} and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>S<\/strong><\/td><td><strong>M<\/strong><\/td><td><strong>T<\/strong><\/td><td><strong>Is M the midpoint of ST? Yes or No<\/strong><\/td><td><strong>Reason for your answer<\/strong><\/td><\/tr><tr><td>{tex}(-3,0){\/tex}<\/td><td>{tex}(0,0){\/tex}<\/td><td>{tex}(3,0){\/tex}<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><\/tr><tr><td>{tex}(2,3){\/tex}<\/td><td>{tex}(3,4){\/tex}<\/td><td>{tex}(4,5){\/tex}<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><\/tr><tr><td>{tex}(0,0){\/tex}<\/td><td>{tex}(0,5){\/tex}<\/td><td>{tex}(0,-10){\/tex}<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><\/tr><tr><td>{tex}(-8,7){\/tex}<\/td><td>{tex}(0,-2){\/tex}<\/td><td>{tex}(6,-3){\/tex}<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>When M is the mid-point of ST, can you find any connection between the coordinates of {tex}{M}, {S}{\/tex} and T?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777016335-huf3hk.jpg\"><br>Using Distance Formula: {tex}d=\\sqrt{ \\left[\\left(x_2-x_1\\right)^2+\\left(y_2-y_1\\right)^2\\right]}{\/tex}<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Row 1: {tex}{S}(-3,0), {M}(0,0), {T}(3,0){\/tex} {tex}S M=\\sqrt{ \\left[(0-(-3))^2+(0-0)^2\\right]}={\/tex}{tex}\\sqrt{ [9+0]=3}{\/tex}<br>{tex}M T=\\sqrt{ \\left[(3-0)^2+(0-0)^2\\right]}={\/tex}{tex}\\sqrt{ [9+0]}=3{\/tex}<br>{tex}S M=M T=3{\/tex} Yes, {tex}M{\/tex} is the midpoint<\/li>\n\n\n\n<li>Row 2: {tex}{S}(2,3), {M}(3,4), {T}(4,5){\/tex} {tex}S M=\\sqrt{ \\left[(3-2)^2+(4-3)^2\\right]}={\/tex}{tex}\\sqrt{ [1+1]}=\\sqrt{ 2}{\/tex}<br>{tex}M T=\\sqrt{ \\left[(4-3)^2+(5-4)^2\\right]}={\/tex}{tex}\\sqrt{ [1+1]}=\\sqrt{ 2}{\/tex}<br>{tex}S M=M T=\\sqrt{ } 2{\/tex} Yes, {tex}M{\/tex} is the midpoint<\/li>\n\n\n\n<li>Row 3: {tex}{S}(0,0), {M}(0,5), {T}(0,-10){\/tex} {tex}S M=\\sqrt{ \\left[(0-0)^2+(5-0)^2\\right]}={\/tex}{tex}\\sqrt{ [0+25]}=5{\/tex}<br>{tex}M T=\\sqrt{ \\left[(0-0)^2+(-10-5)^2\\right]}={\/tex}{tex}\\sqrt{ [0+225]}=15{\/tex}<br>{tex}S M \\neq M T(5 \\neq 15){\/tex} No, M is NOT the midpoint<\/li>\n\n\n\n<li>{tex}{S}(-8,7), {M}(0,-2), {T}(6,-3){\/tex} {tex}S M=\\sqrt{ \\left[(0-(-8))^2+(-2-7)^2\\right]}={\/tex}{tex}\\sqrt{ [64+81]}=\\sqrt{ }145 {\/tex}<br>{tex}M T=\\sqrt{ \\left[(6-0)^2+(-3-(-2))^2\\right]}={\/tex}{tex}\\sqrt{ [36+1]}=\\sqrt{ 37} {\/tex}<br>{tex}S M \\neq M T(\\sqrt{ 14} 5 \\neq \\sqrt{37 } ){\/tex} No, M is NOT the midpoint.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.15:<\/p>\n\n\n\n<p>Use the connection you found to find the coordinates of B given that {tex}{M}(-7,1){\/tex} is the midpoint of {tex}{A}(3,-4){\/tex} and {tex}{B}(x, y){\/tex}.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: {tex}M(-7,1){\/tex} is the midpoint of {tex}A(3,-4){\/tex} and {tex}B(x, y){\/tex}.<br>So, using the concept of distance formula, point {tex}M{\/tex} will be equidistant from {tex}A{\/tex} and {tex}B{\/tex}.<br>Using distance formula: {tex}{AM}=\\sqrt{ }\\left[(-7-3)^2+(1-(-4))^2\\right]{\/tex}<\/p>\n\n\n\n<p>{tex}=\\sqrt{ \\left[(-10)^2+(5)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ (100+25)}{\/tex}<br>{tex}=\\sqrt{ 125}{\/tex}<\/p>\n\n\n\n<p>Now, {tex}M B=\\sqrt{ \\left[(x+7)^2+(y-1)^2\\right]}{\/tex}<br>Since Distance {tex}{AM}={\/tex} Distance MB<br>So, {tex}\\sqrt{ \\left[(x+7)^2+(y-1)^2\\right]=\\sqrt{ } 125}{\/tex}<br>Squaring both sides:<\/p>\n\n\n\n<p>{tex} \\begin{equation*} (x+7)^2+(y-1)^2=125 {1} \\end{equation*}{\/tex}<\/p>\n\n\n\n<p>Also, since {tex}M{\/tex} is the midpoint, it lies between {tex}A{\/tex} and {tex}B{\/tex}, so coordinates of {tex}B{\/tex} will be such that {tex}M{\/tex} divides AB into two equal parts. From symmetry (or balancing coordinates):<\/p>\n\n\n\n<p>From {tex}x{\/tex}-coordinates:<br>Distance from A to M is {tex}-7-3=-10{\/tex}<br>So from {tex}M{\/tex} to {tex}B{\/tex} must also be -10<\/p>\n\n\n\n<p>{tex} x=-7-10=-17{\/tex}<\/p>\n\n\n\n<p>From y-coordinates:<br>Distance from {tex}A{\/tex} to {tex}M{\/tex} is {tex}1-(-4)=5{\/tex}<br>So from {tex}M{\/tex} to {tex}B{\/tex} must also be 5<\/p>\n\n\n\n<p>{tex} y=1+5=6{\/tex}<\/p>\n\n\n\n<p>Therefore, the coordinates of {tex}B=(-17,6){\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.16:<\/p>\n\n\n\n<p>Let {tex}{P}, {Q}{\/tex} 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 {tex}P{\/tex} and {tex}Q{\/tex}? Do this for the case when the points are {tex}A(4,7){\/tex} and B {tex}(16,-2){\/tex}.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To trisect {tex}{AB}, {P}{\/tex} and Q divides the segment into 3 equal parts.<br>Given: {tex}{A}(4,7), {B}(16,-2){\/tex}<br>Let {tex}{P}\\left({x}_1, {y}_1\\right){\/tex} and {tex}{Q}\\left({x}_2, {y}_2\\right){\/tex}<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777017036-jtxz62.jpg\" alt=\"\" style=\"width:352px;height:auto\"\/><\/figure>\n\n\n\n<p>Here, {tex}P{\/tex} is midpoint of {tex}A{\/tex} and {tex}Q{\/tex}, so<\/p>\n\n\n\n<p>{tex}x_1=\\left(4+x_2\\right) \/ 2\\ &#8230;(1){\/tex}<br>{tex}y_1=\\left(7+y_2\\right) \/ 2\\ &#8230;(2){\/tex}<\/p>\n\n\n\n<p>So, Q is midpoint of P and B, so<\/p>\n\n\n\n<p>{tex}x_2=\\left(x_1+16\\right) \/ 2 \\ldots(3){\/tex}<br>{tex}y_2=\\left(y_1-2\\right) \/ 2 \\ldots(4){\/tex}<\/p>\n\n\n\n<p>Solving for x-coordinates, from (1):<\/p>\n\n\n\n<p>{tex} x_1=\\left(4+x_2\\right) \/ 2 \\text { and }{\/tex}<\/p>\n\n\n\n<p>From (3): {tex}{x}_2=\\left({x}_1+16\\right) \/ 2{\/tex}<\/p>\n\n\n\n<p>Substituting (1) into (3), we get<\/p>\n\n\n\n<p>{tex}x_2=\\left(\\left(4+x_2\\right) \/ 2+16\\right) \/ 2{\/tex}<br>{tex}=\\left(4+x_2+32\\right) \/ 4{\/tex}<br>{tex}=\\left(x_2+36\\right) \/ 4{\/tex}<\/p>\n\n\n\n<p>So, {tex}4 {x}_2={x}_2+36{\/tex}<\/p>\n\n\n\n<p>{tex}\\Rightarrow 3 x_2=36{\/tex}<br>{tex}\\Rightarrow x_2=12{\/tex}<\/p>\n\n\n\n<p>Then from (1): {tex}{x}_1=(4+12) \/ 2=8{\/tex}<\/p>\n\n\n\n<p>Solving for y -coordinates, from (2):<br>{tex}y_1=\\left(7+y_2\\right) \/ 2{\/tex} and<br>From (4): {tex}{y}_2=\\left({y}_1-2\\right) \/ 2{\/tex}<br>Substituting (2) into (4):<\/p>\n\n\n\n<p>{tex}y_2=\\left(\\left(7+y_2\\right) \/ 2-2\\right) \/ 2{\/tex}<br>{tex}=\\left(7+y_2-4\\right) \/ 4{\/tex}<br>{tex}=\\left(y_2+3\\right) \/ 4{\/tex}<br>So, {tex}4 y_2=y_2+3{\/tex}<\/p>\n\n\n\n<p>{tex} \\Rightarrow 3 y_2=3{\/tex}<\/p>\n\n\n\n<p>{tex} \\Rightarrow y_2=1{\/tex}<\/p>\n\n\n\n<p>Then from (2): {tex}{y}_1=(7+1) \/ 2=4{\/tex}<br>Therefore, {tex}P=(8,4){\/tex} and {tex}Q=(12,1){\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.17:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Given the points {tex}{A}(1,-8), {B}(-4,7){\/tex} and {tex}{C}(-7,-4){\/tex}, show that they lie on a circle K whose center is the origin {tex}{O}(0,0){\/tex}. What is the radius of circle K ?<\/li>\n\n\n\n<li>Given the points {tex}{D}(-5,6){\/tex} and {tex}{E}(0,9){\/tex}, check whether D and E lie within the circle, on the circle, or outside the circle K.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Distance OA: {tex}=\\sqrt{ \\left(1^2+(-8)^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (1+64)}{\/tex}<br>{tex}=\\sqrt{ 65}{\/tex} Distance OB: {tex}=\\sqrt{ \\left((-4)^2+7^2\\right) }{\/tex}<br>{tex}=\\sqrt{ (16+49)}{\/tex}<br>{tex}=\\sqrt{65 }{\/tex} Distance {tex}O C{\/tex}: {tex}=\\sqrt{ \\left((-7)^2+(-4)^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (49+16)}{\/tex}<br>{tex}=\\sqrt{ 65}{\/tex} Since all three points are at the same distance from origin, they lie on a circle centred at {tex}(0,0){\/tex}.<br>Radius {tex}=\\sqrt{ 65}{\/tex}<br>Points {tex}A, B{\/tex} and {tex}C{\/tex} lie on circle {tex}K{\/tex} with center {tex}(0,0){\/tex} and radius {tex}\\sqrt{ 65}{\/tex}. <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777019796-6yrf82.jpg\"><\/li>\n\n\n\n<li>Distance OD: {tex}=\\sqrt{ \\left((-5)^2+6^2\\right)}{\/tex}<br>{tex}=\\sqrt{ (25+36)}{\/tex}<br>{tex}=\\sqrt{ 61}{\/tex} Distance OE: {tex}=\\sqrt{ \\left(0^2+9^2\\right)}{\/tex}<br>{tex}=9{\/tex} Compare with radius {tex}\\sqrt{ } 65 \\approx 8.06{\/tex}<br>{tex}-\\sqrt{ 61} \\approx 7.81&lt;\\sqrt{ 65} \u00a0\\rightarrow{\/tex} D lies inside the circle {tex}-9>\\sqrt{ 65} \\rightarrow{\/tex} E Lies outside the circle<br>D lies inside the circle.<br>E lies outside the circle.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.18:<\/p>\n\n\n\n<p>The midpoints of the sides of triangle ABC are the points {tex}{D}, {E}{\/tex}, and {tex}F{\/tex}. Given that the coordinates of {tex}D, E{\/tex}, and {tex}F{\/tex} are {tex}(5,1),(6,5){\/tex}, and {tex}(0,3){\/tex}, respectively, find the coordinates of A, B and C.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the vertices of triangle {tex}A B C{\/tex} be {tex}A\\left(x_1, y_1\\right), B\\left(x_2, y_2\\right){\/tex}, and {tex}C\\left(x_3, y_3\\right){\/tex}.<br>Given midpoints: {tex}{D}(5,1), {E}(6,5), {F}(0,3){\/tex}<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777020120-rv7vb6.jpg\" alt=\"\" style=\"width:273px;height:auto\"\/><\/figure>\n\n\n\n<p>D is midpoint of BC, so using midpoint formula:<\/p>\n\n\n\n<p>{tex}x_2+x_3=10\\&nbsp; \\ldots(1){\/tex}<br>{tex}y_2+y_3=2 \\ \\ldots(2){\/tex}<\/p>\n\n\n\n<p>E is midpoint of CA, therefore:<\/p>\n\n\n\n<p>{tex}x_2+x_1=12 \\ \\ldots(3){\/tex}<br>{tex}y_2+y_1=10\\ \\ldots(4){\/tex}<\/p>\n\n\n\n<p>F is midpoint of AB, therefore:<\/p>\n\n\n\n<p>{tex}x_1+x_2=0 \\ \\ldots(5){\/tex}<br>{tex}y_1+y_2=6 \\ \\ldots(6){\/tex}<\/p>\n\n\n\n<p>Solving for x -coordinates, from (5):<\/p>\n\n\n\n<p>{tex} x_2=-x_1{\/tex}<\/p>\n\n\n\n<p>Substitute into (1): {tex}-{x}_1+{x}_2=10{\/tex}<\/p>\n\n\n\n<p>{tex} \\Rightarrow x_3=10+x_1 \\ &#8230;{7}{\/tex}<\/p>\n\n\n\n<p>Substituting into (3): {tex}\\left(10+x_1\\right)+x_1=12{\/tex}<\/p>\n\n\n\n<p>{tex}\\Rightarrow 2 {x}_1+10=12{\/tex}<br>{tex}\\Rightarrow 2 {x}_1=2{\/tex}<br>{tex}\\Rightarrow {x}_1=1{\/tex}<\/p>\n\n\n\n<p>Then: {tex}{x}_2=-1{\/tex}<\/p>\n\n\n\n<p>{tex} \\Rightarrow x_3=10+1=11{\/tex}<\/p>\n\n\n\n<p>Solving {tex}y{\/tex}-coordinates, from (6):<\/p>\n\n\n\n<p>{tex} y_2=6-y_1{\/tex}<\/p>\n\n\n\n<p>Substituting into (2): (6-y<sub>1<\/sub>) {tex}+y_3=2{\/tex}<\/p>\n\n\n\n<p>{tex}\\Rightarrow y_2=y_1-4 \\ &#8230;{8}{\/tex}<\/p>\n\n\n\n<p>Substituting into (4): {tex}\\left(y_1-4\\right)+y_1=10{\/tex}<\/p>\n\n\n\n<p>{tex}\\Rightarrow 2 y_1-4=10{\/tex}<br>{tex}\\Rightarrow 2 y_1=14{\/tex}<br>{tex}\\Rightarrow y_1=7{\/tex}<\/p>\n\n\n\n<p>Then: {tex}y=6-7=-1{\/tex}<\/p>\n\n\n\n<p>{tex} \\Rightarrow y_3=7-4=3{\/tex}<\/p>\n\n\n\n<p>Therefore, the coordinates of the points are {tex}{A}(1,7), {B}(-1,-1){\/tex} and {tex}{C}(11,3){\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.19:<\/p>\n\n\n\n<p>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.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Using {tex}1 {~cm}=200 {~m}{\/tex}, draw a model of the city in your notebook. Represent the roads\/streets by single lines.<\/li>\n\n\n\n<li>There are street intersections in the model. Each street intersection is formed by two streets-one running in the {tex}{N}-{S}{\/tex} direction and another in the E-W direction. Each street\u00a0intersection 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:\n<ol start=\"1\" class=\"wp-block-list\">\n<li>how many street intersections can be referred to as {tex}(4,3){\/tex}.<\/li>\n\n\n\n<li>how many street intersections can be referred to as {tex}(3,4){\/tex}.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Drawing the model of the city<br>Since: Scale is {tex}1 \\ {cm}=200 \\ {m}{\/tex}<br>Each pair of adjacent streets is 200 m apart<br>Given that:<br>10 streets in the {tex}{N}-{S}{\/tex} direction<br>10 streets in the E-W direction Hence, the model will consist of:<br>10 vertical parallel lines (for N-S streets)<br>10 horizontal parallel lines (for E-W streets)<br>Each consecutive line 1 cm apart, which forms a square grid. <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777020652-p26mfr.jpg\"><\/li>\n\n\n\n<li>A street intersection is named by: (first number, second number)<br>{tex}=({N}-{S}{\/tex} street number, {tex}{E}-{W}{\/tex} street number{tex}){\/tex}<br>So:<br>{tex}(4,3){\/tex} means the intersection of the 4th {tex}{N}-{S}{\/tex} street and the 3rd {tex}{E}-{W}{\/tex} street {tex}(3,4){\/tex} means the intersection of the 3rd N-S street and the 4th E-W street Now, one {tex}{N}-{S}{\/tex} street and one {tex}{E}-{W}{\/tex} street can meet at only one point.<br>Therefore:<br>(a) Only one intersection can be called {tex}(4,3){\/tex}.<br>(b) Only one intersection can be called {tex}(3,4){\/tex}. <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777020710-95umz3.jpg\"><\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.20:<\/p>\n\n\n\n<p>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 {tex}{A}(100,150){\/tex}. Another circular icon of radius 100 pixels is drawn with its centre at the point {tex}{B}(250,230){\/tex}. Determine:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>whether any part of either circle lies outside the screen.<\/li>\n\n\n\n<li>whether the two circles intersect each other.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>No. Each of the two circles lies inside the screen.<\/li>\n\n\n\n<li>Yes. The two circles intersect each other as shown in the picture.<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777021426-4zc2re.jpg\" alt=\"\" style=\"width:217px;height:auto\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.21:<\/p>\n\n\n\n<p>Plot the points {tex}{A}(2,1), {B}(-1,2), {C}(-2,-1){\/tex}, and {tex}{D}(1,-2){\/tex} in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, {tex}A B C D{\/tex} is a square.<\/p>\n\n\n\n<figure class=\"wp-block-image is-resized\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777021563-d47k9a.jpg\" alt=\"\" style=\"width:230px;height:auto\"\/><\/figure>\n\n\n\n<p>We can check it finding the lengths of sides and diagonals.<br>Finding the lengths of all sides:<\/p>\n\n\n\n<p>{tex}A B=\\sqrt{ \\left[(-1-2)^2+(2-1)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left[(-3)^2+1^2\\right]}{\/tex}<br>{tex}=\\sqrt{ (9+1)}{\/tex}<br>{tex}=\\sqrt{ 10}{\/tex}<\/p>\n\n\n\n<p>{tex}B C=\\sqrt{ \\left[(-2-(-1))^2+(-1-2)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left[(-1)^2+(-3)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ (1+9)}{\/tex}<br>{tex}=\\sqrt{10 }{\/tex}<\/p>\n\n\n\n<p>{tex}C D={\/tex}{tex}\\sqrt{ \\left[(1-(-2))^2+(-2-(-1))^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left[3^2+(-1)^2\\right] }{\/tex}<br>{tex}=\\sqrt{ (9+1)}{\/tex}<br>{tex}=\\sqrt{10}{\/tex}<\/p>\n\n\n\n<p>{tex}D A=\\sqrt{ \\left[(2-1)^2+(1-(-2))^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left[1^2+3^2\\right]}{\/tex}<br>{tex}=\\sqrt{ (1+9)}{\/tex}<br>{tex}=\\sqrt{ 10}{\/tex}<\/p>\n\n\n\n<p>All four sides are equal.<\/p>\n\n\n\n<p>Now finding diagonals:<\/p>\n\n\n\n<p>{tex}A C=\\sqrt{ \\left[(-2-2)^2+(-1-1)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left[(-4)^2+(-2)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ (16+4)}{\/tex}<br>{tex}=\\sqrt{ 20}{\/tex}<\/p>\n\n\n\n<p>{tex}B D=\\sqrt{ \\left[(1-(-1))^2+(-2-2)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ \\left[2^2+(-4)^2\\right]}{\/tex}<br>{tex}=\\sqrt{ (4+16)}{\/tex}<br>{tex}=\\sqrt{ 20}{\/tex}<\/p>\n\n\n\n<p>Diagonals are equal.<br>All sides are equal and diagonals are equal, so ABCD is a square.<\/p>\n\n\n\n<p>Area of {tex}{ABCD}=(\\text {side})^2{\/tex}<\/p>\n\n\n\n<p>{tex}=(\\sqrt{ 10^2}){\/tex}<br>{tex}=10 \\text { square units. }{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.22:<\/p>\n\n\n\n<p>What is the x-coordinate of a point on the y-axis?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The {tex}{x}{\/tex}-coordinate of any point located on the {tex}{y}{\/tex}-axis is always {tex}{0}{\/tex}. This is a fundamental property of the Cartesian coordinate system because the {tex}y{\/tex}-axis itself is defined by the vertical line where no horizontal displacement has occurred from the origin.<\/p>\n\n\n\n<p>In a coordinate pair {tex}(x, y){\/tex}, the {tex}x{\/tex} value represents the horizontal distance from the center. Since points on the {tex}y{\/tex}-axis do not move left or right, the value remains at the starting point of the scale. Therefore, every point on this axis follows the general form {tex}(0, y){\/tex}, where {tex}y{\/tex} can be any real number.<\/p>\n\n\n\n<p>This concept is essential when identifying intercepts in geometry. For instance, when you are looking for the y-intercept of a line or a circle&#8217;s equation, you mathematically set {tex}x=0{\/tex} to find where the shape crosses the vertical axis. Understanding this &#8220;zero&#8221; placement helps in accurately plotting graphs and solving algebraic equations involving coordinate geometry.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.23:<\/p>\n\n\n\n<p>Is there a similar generalization for a point on the x-axis?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, there is a mirror generalization for the horizontal axis. The {tex}{y}{\/tex}-coordinate of any point located on the x -axis is always {tex}{0}{\/tex}. Just as the y -axis represents the line where {tex}x{\/tex} is zero, the x axis represents the horizontal boundary where there is no vertical movement.<\/p>\n\n\n\n<p>In the Cartesian system, a point is written as {tex}(x, y){\/tex}. For a point to lie exactly on the x -axis, it cannot have any upward or downward displacement from the origin. This means its vertical position must be exactly at the starting level of the grid, which corresponds to {tex}y=0{\/tex}. Consequently, all points on this axis follow the general form ({tex}x, 0{\/tex}), where {tex}x{\/tex} can be any positive or negative value.<\/p>\n\n\n\n<p>This rule is particularly useful when solving for the x-intercepts of a circle or a line. To find where a curve crosses the horizontal axis, you simply set {tex}y=0{\/tex} in the equation and solve for {tex}x{\/tex}. This symmetry between the two axes forms the basis for navigating and plotting data in a two dimensional plane.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.24:<\/p>\n\n\n\n<p>Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Point {tex}Q(y, x){\/tex} coincides with point {tex}P(x, y){\/tex} only when the coordinates are equal, meaning {tex}x=y{\/tex}. This occurs for every point located on the identity line.<\/p>\n\n\n\n<p>Mathematically, for two points to be identical, their corresponding coordinates must match exactly. This requires the {tex}x{\/tex}-value of {tex}P{\/tex} to equal the {tex}x{\/tex}-value of {tex}Q(x=y){\/tex} and the {tex}y{\/tex}-value of {tex}P{\/tex} to equal the {tex}y{\/tex}-value of {tex}Q(y=x){\/tex}.<\/p>\n\n\n\n<p>Both conditions lead to the same requirement: the horizontal and vertical distances from the origin must be identical. If {tex}x \\neq y{\/tex}, the points are reflections of each other across the line {tex}y=x{\/tex} but occupy different positions in the plane. For example, {tex}(2,3){\/tex} and {tex}(3,2){\/tex} are distinct points, whereas {tex}(4,4){\/tex} remains the same regardless of which coordinate you read first. Thus, the points coincide if and only if they lie on the diagonal line passing through the origin at a {tex}45^{\\circ}{\/tex} angle.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.25:<\/p>\n\n\n\n<p>If {tex}x \\neq y{\/tex}, then {tex}(x, y) \\neq(y, x){\/tex}; and {tex}(x, y)=(y, x){\/tex} if and only if {tex}x=y{\/tex}. Is this claim true?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The claim is entirely true because ordered pairs are defined by the specific position of each element within the parentheses. For two points ({tex}x, y{\/tex}) and ({tex}y, x{\/tex}) to be identical, the first components must match ({tex}x=y{\/tex}) and the second components must match ({tex}y=x{\/tex}).<\/p>\n\n\n\n<p>When {tex}x \\neq y{\/tex}, the horizontal displacement of the first point differs from that of the second, placing them at different locations. These points act as reflections of one another across the diagonal line {tex}y=x{\/tex} but remain distinct. For example, the point {tex}(1,5){\/tex} is located in a completely different position than {tex}(5,1){\/tex} on the coordinate grid.<\/p>\n\n\n\n<p>The &#8220;if and only if&#8221; condition establishes that {tex}x=y{\/tex} is both necessary and sufficient for the points to coincide. This means the only way to swap coordinates without changing the point&#8217;s location is if the values are identical. This fundamental principle ensures that every unique location in a plane corresponds to a specific, unique ordered pair.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.26:<\/p>\n\n\n\n<p>In moving from {tex}A(3,4){\/tex} to {tex}D(7,1){\/tex}, what distance has been covered along the {tex}x{\/tex}-axis? What about the distance along the {tex}y{\/tex}-axis?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The distance covered along the {tex}x{\/tex}-axis is 4 units and along the {tex}y{\/tex}-axis is 3 units, found by calculating the differences between coordinates.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.27:<\/p>\n\n\n\n<p>Would these observations be the same if {tex}\\triangle {ADM}{\/tex} is reflected in the x-axis (instead of the {tex}y{\/tex}-axis)?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Reflecting {tex}\\triangle {ADM}{\/tex} in the {tex}x{\/tex}-axis would yield similar observations regarding magnitude, but the vertical direction and specific coordinates would change. In an {tex}x{\/tex}-axis reflection, the horizontal distance ({tex}x{\/tex}-distance) between points remains exactly the same as the original because x coordinates are unaffected.<\/p>\n\n\n\n<p>However, the {tex}y{\/tex}-coordinates swap signs, meaning the distance along the {tex}y{\/tex}-axis remains 3 units in magnitude but moves in the opposite vertical direction. If the original movement from {tex}A{\/tex} to {tex}D{\/tex} was &#8220;downward,&#8221; the reflected movement would appear &#8220;upward&#8221; relative to the new points.<\/p>\n\n\n\n<p>Essentially, the &#8220;gap&#8221; between the values stays constant, but the orientation flips. Just as a circle maintains its radius regardless of where it is reflected, the absolute distances covered along each axis remain identical to your previous calculations. Only the signs of the {tex}y{\/tex}-values and the final positions in the quadrants would differ from a {tex}y{\/tex}-axis reflection.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Use of Coordinates &#8211; NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari textbook. NCERT Solutions Class 9 The Use of Coordinates \u2013 NCERT Solutions Q.1: Figure shows Reiaan\u2019s room with points OABC marking its corners. 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