{"id":6064,"date":"2023-08-20T07:12:24","date_gmt":"2023-08-20T07:12:24","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6064"},"modified":"2023-08-20T07:12:24","modified_gmt":"2023-08-20T07:12:24","slug":"ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 7 (Coordinate Geometry)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 7 Coordinate Geometry <\/strong>Exercise 7.4 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 7 Coordinate Geometry<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 7.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 7.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 7.3<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 7.4<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Determine the ratio in which the line 2x + y \u2013 4 = 0 divides the line segment joining the points A (2, \u20132) and B (3, 7). <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Consider line 2x + y \u2013 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k : 1 ratio.<br \/>\nCoordinates of point of division can be given as follows:<br \/>\nx = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(2+3k)}{(k+1)}\" alt=\"\\frac{(2+3k)}{(k+1)}\" align=\"absmiddle\" \/><br \/>\ny = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(-2+7k)}{(k+1)}\" alt=\"\\frac{(-2+7k)}{(k+1)}\" align=\"absmiddle\" \/><br \/>\nSubstituting the values of x and y given equation, i.e. 2x + y \u2013 4 = 0, we have<br \/>\n2<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(2+3k)}{(k+1)}\" alt=\"\\frac{(2+3k)}{(k+1)}\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(-2+7k)}{(k+1)}\" alt=\"\\frac{(-2+7k)}{(k+1)}\" align=\"absmiddle\" \/> \u2013 4 = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(4+6k)}{(k+1)}&amp;space;+&amp;space;\\frac{(-2+7k)}{(k+1)}\" alt=\"\\frac{(4+6k)}{(k+1)} + \\frac{(-2+7k)}{(k+1)}\" align=\"absmiddle\" \/>= 4<br \/>\n4 + 6k \u2013 2 + 7k = 4(k+1)<br \/>\n-2 + 9k = 0<br \/>\nk = 2\/9<br \/>\nHence, the ratio is 2 : 9.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>If given points are collinear, then the area of the triangle formed by them must be zero.<br \/>\nLet (x, y), (1, 2) and (7, 0) are vertices of a triangle,<br \/>\nArea of a triangle = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 [x<sub>1<\/sub>(y<sub>2<\/sub>\u00a0\u2013 y<sub>3<\/sub>) + x<sub>2<\/sub>(y<sub>3<\/sub>\u00a0\u2013 y<sub>1<\/sub>) + x<sub>3<\/sub>(y<sub>1<\/sub>\u00a0\u2013 y<sub>2<\/sub>)] = 0<br \/>\n[x(2 \u2013 0) + 1 (0 \u2013 y) + 7( y \u2013 2)] = 0<br \/>\n2x \u2013 y + 7y \u2013 14 = 0<br \/>\n2x + 6y \u2013 14 = 0<br \/>\nx + 3y \u2013 7 = 0.<br \/>\nWhich is the required result.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3). <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Let A = (6, -6), B = (3, -7), and C = (3, 3) are the points on a circle.<br \/>\nIf O is the centre, then OA = OB = OC (radii are equal)<br \/>\nIf O = (x, y), then<br \/>\nOA = \u221a[(x \u2013 6)<sup>2\u00a0<\/sup>+ (y + 6)<sup>2<\/sup>]<br \/>\nOB =\u00a0\u221a[(x \u2013 3)<sup>2\u00a0<\/sup>+ (y + 7)<sup>2<\/sup>]<br \/>\nOC =\u00a0\u221a[(x \u2013 3)<sup>2\u00a0<\/sup>+ (y \u2013 3)<sup>2<\/sup>]<br \/>\nChoose : OA = OB, we have<br \/>\nAfter simplifying above, we get -6x = 2y \u2013 14\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\nSimilarly, OB = OC<br \/>\n(x \u2013 3)<sup>2\u00a0<\/sup>+ (y + 7)<sup>2<\/sup>\u00a0= (x \u2013 3)<sup>2<\/sup>\u00a0+ (y \u2013 3)<sup>2<br \/>\n<\/sup>(y + 7)<sup>2<\/sup>\u00a0= (y \u2013 3)<sup>2<br \/>\n<\/sup>y<sup>2\u00a0<\/sup>+ 14y + 49 = y<sup>2\u00a0<\/sup>\u2013 6y + 9<br \/>\n20y = -40<br \/>\ny = -2<br \/>\nSubstituting the value of y in equation (i), we get<br \/>\n-6x = 2y \u2013 14<br \/>\n-6x = -4 \u2013 14 = -18<br \/>\nx = 3<br \/>\nHence, the centre of the circle is located at point (3, -2).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Let ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD.<br \/>\n<strong>To Find:<\/strong> Coordinate of points B and D.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6464\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"296\" height=\"234\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans4.png 410w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans4-300x237.png 300w\" sizes=\"auto, (max-width: 296px) 100vw, 296px\" \/><br \/>\n<strong>Step 1: Find the distance between A and C and the coordinates of point O.<br \/>\n<\/strong>We know that the diagonals of a square are equal and bisect each other.<br \/>\nAC =\u00a0\u221a[(3 + 1)<sup>2\u00a0<\/sup>+ (2 \u2013 2)<sup>2<\/sup>] = 4<br \/>\nCoordinates of O can be calculated as follows:<br \/>\nx = (3 \u2013 1)\/2 = 1<br \/>\ny = (2 + 2)\/2 = 2<br \/>\nSo, O(1, 2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step 2: Find the side of the square using the Pythagoras theorem<br \/>\n<\/strong>Let a be the side of the square and AC = 4<br \/>\nFrom the right triangle, ACD,<br \/>\na = 2\u221a2<br \/>\nHence, each side of the square = 2\u221a2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step 3: Find the coordinates of point D<br \/>\n<\/strong>Equate the length measure of AD and CD<br \/>\nSay, if the coordinates of D are (x<sub>1<\/sub>, y<sub>1<\/sub>)<br \/>\nAD =\u00a0\u221a[(x<sub>1<\/sub>\u00a0+ 1)<sup>2\u00a0<\/sup>+ (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<\/sup>]<br \/>\nSquaring both sides,<br \/>\nAD<sup>2<\/sup>\u00a0= (x<sub>1<\/sub>\u00a0+ 1)<sup>2\u00a0<\/sup>+ (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<br \/>\n<\/sup>CD<sup>2<\/sup>\u00a0=\u00a0(x<sub>1<\/sub>\u00a0\u2013 3)<sup>2\u00a0<\/sup>+ (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<br \/>\n<\/sup>Since all sides of a square are equal, which means AD = CD<br \/>\n(x<sub>1<\/sub>\u00a0+ 1)<sup>2\u00a0<\/sup>+ (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<\/sup>\u00a0=\u00a0(x<sub>1<\/sub>\u00a0\u2013 3)<sup>2\u00a0<\/sup>+ (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<br \/>\n<\/sup>x<sub>1<\/sub><sup>2<\/sup>\u00a0+ 1 + 2x<sub>1<\/sub>\u00a0= x<sub>1<\/sub><sup>2<\/sup>\u00a0+ 9 \u2013 6x<sub>1<br \/>\n<\/sub>8x<sub>1<\/sub> = 8<br \/>\nx<sub>1<\/sub>\u00a0= 1<br \/>\nThe value of y<sub>1<\/sub> can be calculated as follows by using the value of x.<br \/>\nFrom step 2: each side of the square = 2\u221a2<br \/>\nCD<sup>2<\/sup>\u00a0=\u00a0(x<sub>1<\/sub>\u00a0\u2013 3)<sup>2\u00a0<\/sup>+ (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<br \/>\n<\/sup>8 =\u00a0(1\u00a0\u2013 3)<sup>2\u00a0<\/sup>+ (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<br \/>\n<\/sup>8 = 4 + (y<sub>1<\/sub>\u00a0\u2013 2)<sup>2<br \/>\n<\/sup>y<sub>1<\/sub> \u2013 2 = 2<br \/>\ny<sub>1<\/sub> = 4<br \/>\nHence, D = (1, 4)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step 4: Find the coordinates of point B<br \/>\n<\/strong>From line segment, BOD<br \/>\nCoordinates of B can be calculated using coordinates of O, as follows:<br \/>\nEarlier, we had calculated O = (1, 2)<br \/>\nSay B = (x<sub>2<\/sub>, y<sub>2<\/sub>)<br \/>\nFor BD:<br \/>\n1 = (x<sub>2<\/sub> + 1)\/2<br \/>\nx<sub>2\u00a0<\/sub>= 1<br \/>\n2 = (y<sub>2<\/sub> + 4)\/2<br \/>\ny<sub>2<\/sub> = 0<br \/>\nTherefore, the coordinates of required points are B = (1,0) and D = (1,4)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot, as shown in fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.<br \/>\n<\/strong>(i) Taking A as the origin, find the coordinates of the vertices of the triangle.<br \/>\n(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?<br \/>\n<strong>Also, calculate the areas of the triangles in these cases. What do you observe?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6465 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans5-300x181.jpg\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"181\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans5-300x180.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans5-1024x617.jpg 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans5-768x463.jpg 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans5-850x512.jpg 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans5.jpg 1434w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> Taking A as the origin, the coordinates of the vertices P, Q and R are,<br \/>\nFrom figure: P = (4, 6), Q = (3, 2), R (6, 5)<br \/>\nHere, AD is the x-axis and AB is the y-axis.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Taking C as the origin,<br \/>\nThe coordinates of vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3), respectively.<br \/>\nHere, CB is the x-axis and CD is the y-axis.<br \/>\nFind the area of triangles:<br \/>\nArea of triangle PQR in case of origin A:<br \/>\nUsing formula: Area of a triangle =\u00a01\/2 \u00d7 [x<sub>1<\/sub>(y<sub>2<\/sub>\u00a0\u2013 y<sub>3<\/sub>) + x<sub>2<\/sub>(y<sub>3<\/sub>\u00a0\u2013 y<sub>1<\/sub>) + x<sub>3<\/sub>(y<sub>1<\/sub>\u00a0\u2013 y<sub>2<\/sub>)]<br \/>\n= \u00bd [4(2 \u2013 5) + 3 (5 \u2013 6) + 6 (6 \u2013 2)]<br \/>\n= \u00bd (- 12 \u2013 3 + 24 )<br \/>\n= 9\/2 sq unit<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> Area of triangle PQR in case of origin C:<br \/>\nArea of a triangle =\u00a01\/2 \u00d7 [x<sub>1<\/sub>(y<sub>2<\/sub>\u00a0\u2013 y<sub>3<\/sub>) + x<sub>2<\/sub>(y<sub>3<\/sub>\u00a0\u2013 y<sub>1<\/sub>) + x<sub>3<\/sub>(y<sub>1<\/sub>\u00a0\u2013 y<sub>2<\/sub>)]<br \/>\n= \u00bd [ 12(6 \u2013 3) + 13 ( 3 \u2013 2) + 10( 2 \u2013 6)]<br \/>\n= \u00bd ( 36 + 13 \u2013 40)<br \/>\n= 9\/2 sq unit<br \/>\nThis implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C.<br \/>\nThe area is the same in both cases because the triangle remains the same no matter which point is considered as the origin.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. The vertices of a \u2206 ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E, respectively, such that AD\/AB = AE\/AC = 1\/4. Calculate the area of the \u2206 ADE and compare it with the area of \u2206 ABC. (Recall Theorem 6.2 and Theorem 6.6)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given:<\/strong> The vertices of a \u2206 ABC are A (4, 6), B (1, 5) and C (7, 2)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6467\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"280\" height=\"241\" \/><br \/>\nAD\/AB = AE\/AC = 1\/4<br \/>\nAD\/(AD + BD) = AE\/(AE + EC) = 1\/4<br \/>\nPoint D and Point E divide AB and AC, respectively, in ratio 1:3.<br \/>\nCoordinates of D can be calculated as follows:<br \/>\nx = (m<sub>1<\/sub>x<sub>2<\/sub>\u00a0+ m<sub>2<\/sub>x<sub>1<\/sub>)\/(m<sub>1<\/sub>\u00a0+ m<sub>2<\/sub>) and y =\u00a0(m<sub>1<\/sub>y<sub>2<\/sub>\u00a0+ m<sub>2<\/sub>y<sub>1<\/sub>)\/(m<sub>1<\/sub>\u00a0+ m<sub>2<\/sub>)<br \/>\nHere, m<sub>1<\/sub>\u00a0= 1 and m<sub>2<\/sub>\u00a0= 3<br \/>\nConsider line segment AB which is divided by point D at the ratio 1:3.<br \/>\nx = [3(4) + 1(1)]\/4 = 13\/4<br \/>\ny = [3(6) + 1(5)]\/4 = 23\/4<br \/>\nSimilarly, the coordinates of E can be calculated as follows:<br \/>\nx = [1(7) + 3(4)]\/4 = 19\/4<br \/>\ny = [1(2) + 3(6)]\/4 = 20\/4 = 5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Find the area of triangle:<br \/>\n<\/strong>Using formula: Area of a triangle =\u00a01\/2 \u00d7 [x<sub>1<\/sub>(y<sub>2<\/sub>\u00a0\u2013 y<sub>3<\/sub>) + x<sub>2<\/sub>(y<sub>3<\/sub>\u00a0\u2013 y<sub>1<\/sub>) + x<sub>3<\/sub>(y<sub>1<\/sub>\u00a0\u2013 y<sub>2<\/sub>)]<br \/>\nThe area of triangle \u2206 ABC can be calculated as follows:<br \/>\n= \u00bd [4(5 \u2013 2) + 1( 2 \u2013 6) + 7( 6 \u2013 5)]<br \/>\n= \u00bd (12 \u2013 4 + 7) = 15\/2 sq unit<br \/>\nThe area of \u2206 ADE can be calculated as follows:<br \/>\n= \u00bd [4(23\/4 \u2013 5) + 13\/4 (5 \u2013 6) + 19\/4 (6 \u2013 23\/4)]<br \/>\n= \u00bd (3 \u2013 13\/4 + 19\/16)<br \/>\n= \u00bd ( 15\/16 ) = 15\/32 sq unit<br \/>\nHence, the ratio of the area of triangle ADE to the area of triangle ABC = 1 : 16.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. Let A (4, 2), B (6, 5), and C (1, 4) be the vertices of \u2206 ABC.<br \/>\n<\/strong>(i) The median from A meets BC at D. Find the coordinates of point D.<br \/>\n(ii) Find the coordinates of the point P on AD such that AP:PD = 2:1.<br \/>\n(iii) Find the coordinates of points Q and R on medians BE and CF, respectively, such that BQ:QE = 2:1 and CR:RF = 2 : 1.<br \/>\n(iv) What do you observe?<strong><br \/>\n<\/strong><strong>[Note: The point which is common to all the three medians is called the centroid, <\/strong><strong>and this point divides each median in the ratio 2:1.]<br \/>\n<\/strong>(v) If A (x<sub>1<\/sub>, y<sub>1<\/sub>), B (x<sub>2<\/sub>, y<sub>2<\/sub>) and C (x<sub>3<\/sub>, y<sub>3<\/sub>) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle. <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6468\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans7.jpg\" alt=\"NCERT Class 10 Maths Solution\" width=\"210\" height=\"192\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) Coordinates of D can be calculated as follows:<br \/>\n<\/strong>Coordinates of D = ( (6+1)\/2, (5+4)\/2 ) = (7\/2, 9\/2)<br \/>\nSo, D is (7\/2, 9\/2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) Coordinates of P can be calculated as follows:<br \/>\n<\/strong>Coordinates of P = ( [2(7\/2) + 1(4)]\/(2 + 1), [2(9\/2) + 1(2)]\/(2 + 1))<br \/>\n= (11\/3, 11\/3)<br \/>\nSo, P is (11\/3, 11\/3)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) Coordinates of E can be calculated as follows:<br \/>\n<\/strong>Coordinates of E = ( (4+1)\/2, (2+4)\/2 ) = (5\/2, 6\/2)<br \/>\n= (5\/2, 3)<br \/>\nSo, E is (5\/2, 3)<br \/>\nPoints Q and P would be coincident because the medians of a triangle intersect each other at a common point called the centroid. Coordinate of Q can be given as follows:<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Coordinates of Q<\/strong>\u00a0=( [2(5\/2) + 1(6)]\/(2 + 1), [2(3) + 1(5)]\/(2 + 1) )<br \/>\n= (11\/3, 11\/3)<br \/>\nF is the midpoint of the side AB<br \/>\nCoordinates of F = ((4+6)\/2, (2+5)\/2 )<br \/>\n= (5, 7\/2)<br \/>\nPoint R divides the side CF in ratio 2 : 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Coordinates of R<\/strong>\u00a0= ( [2(5) + 1(1)]\/(2 + 1), [2(7\/2) + 1(4)]\/(2 + 1) )<br \/>\n= (11\/3, 11\/3)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv)<\/strong> Coordinates of P, Q and R are the same, which shows that medians intersect each other at a common point, i.e. centroid of the triangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(v)<\/strong> If A (x<sub>1<\/sub>, y<sub>1<\/sub>), B (x<sub>2<\/sub>, y<sub>2<\/sub>) and C (x<sub>3<\/sub>, y<sub>3<\/sub>) are the vertices of triangle ABC, the coordinates of the centroid can be given as follows:<br \/>\nx = (x<sub>1<\/sub>\u00a0+ x<sub>2<\/sub>\u00a0+ x<sub>3<\/sub>)\/3 and y = (y<sub>1<\/sub>\u00a0+ y<sub>2<\/sub>\u00a0+ y<sub>3<\/sub>)\/3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. ABCD is a rectangle formed by the points A (-1, \u2013 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA, respectively. Is the quadrilateral PQRS a square, a rectangle or a rhombus? Justify your answer.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6469 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans8-300x155.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"155\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans8-300x155.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.4-Ans8.png 515w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n<\/strong>P id the midpoint of side AB,<br \/>\nCoordinate of P = ( (-1 \u2013 1)\/2, (-1 + 4)\/2 ) = (-1, 3\/2)<br \/>\nSimilarly, Q, R and S are (As Q is the midpoint of BC, R is the midpoint of CD and S is the midpoint of AD)<br \/>\nCoordinate of Q = (2, 4)<br \/>\nCoordinate of R = (5, 3\/2)<br \/>\nCoordinate of S = (2, -1)<br \/>\nNow,<br \/>\nLength of PQ = \u221a[(-1 \u2013 2)<sup>2<\/sup>\u00a0+ (3\/2 \u2013 4)<sup>2<\/sup>]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2<br \/>\nLength of SP =\u00a0\u221a[(2 + 1)<sup>2<\/sup>\u00a0+ (-1 \u2013 3\/2)<sup>2<\/sup>]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2<br \/>\nLength of QR =\u00a0\u221a[(2 \u2013 5)<sup>2<\/sup>\u00a0+ (4 \u2013 3\/2)<sup>2<\/sup>]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2<br \/>\nLength of RS =\u00a0\u221a[(5 \u2013 2)<sup>2<\/sup>\u00a0+ (3\/2 + 1)<sup>2<\/sup>]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2<br \/>\nLength of PR (diagonal) =\u00a0\u221a[(-1 \u2013 5)<sup>2<\/sup>\u00a0+ (3\/2 \u2013 3\/2)<sup>2<\/sup>]\u00a0= 6<br \/>\nLength of QS (diagonal) = \u221a[(2 \u2013 2)<sup>2<\/sup>\u00a0+ (4 + 1)<sup>2<\/sup>] = 5<br \/>\nThe above values show that PQ = SP = QR = RS = \u221a61\/2, i.e. all sides are equal.<br \/>\nBut PR \u2260 QS, i.e. diagonals are not of equal measure.<br \/>\nHence, the given figure is a rhombus.<\/span><\/p>\n<p><span style=\"color: #0000ff; font-family: Georgia, Palatino;\"><b><i><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\">Go Back To Chapters<\/a><\/i><\/b><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 7 (Coordinate Geometry)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 7 Coordinate Geometry Exercise 7.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 7 Coordinate Geometry NCERT Class 10 Maths Solution Ex &#8211; 7.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1086,1090,1044,1049,1048],"class_list":["post-6064","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-7-coordinate-geometry-solutions","tag-ncert-class-10-mathematics-exercise-7-4-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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