Solution: Given vertices of the triangle are A(-5,3), B(p,-1) and C(6,q). Co-ordinates of centroid are (1,-1). Co-ordinates of the centroid of a triangle, whose vertices are (x1,y1), (x2,y2) and...
Solution: Let third vertex be C(x3,y3). Given (x1,y1) = (3,-5) (x2,y2) = (-7,4) Co-ordinates of centroid are (2,-1) Co-ordinates of the centroid of a triangle, whose vertices are (x1,y1), (x2,y2)...
Solution: Given vertices of the triangle are A(-1,3), B(1,-1) and C(5,1) Co-ordinates of the centroid of a triangle, whose vertices are (x1,y1), (x2,y2) and (x3,y3) are [(x1 + x2+ x3)/3, (y1 + y2+...
Solution: Let P(4,-3) be the midpoint of line joining the points A and B. Since A lies on X axis, its co-ordinates are (x2,0) Since B lies on Y axis, its co-ordinates are (0,y1) By midpoint formula,...
(i) Given vertices of the ABC are A(2,5), B(-1,2) and C(5,8). P and Q are points on AB and AC respectively such that AP:PB = AQ :QC = 1:2. P(x,y) divides AB in the ratio 1:2. x1= 2, y1 = 5 x2 = -1,...
Solution: Given points are A(10,5), B(6,-3) and C(2,1). Let L(x,y) be the midpoint of AB. Here x1= 10, y1 = 5 x2 = 6, y2 = -3 By midpoint formula, x = (x1+x2)/2 x = (10+6)/2 = 16/2 = 8 By midpoint...
Let A(-5,1) divides the line joining (1,p) and (4,-2) in the ratio m:n Then by section formula, x = (mx2+nx1)/(m+n) -5 = (m×4+n×1)/(m+n) -5 = (4m+n)/(m+n) -5m-5n = 4m+n -9m = 6n m/n = -9/6 = -2/3...
Solution: Let the point B(5,2) divides the line joining A(3,-2) and C(8,8) in the ratio m:n. Then by section formula, x = (mx2+nx1)/(m+n) 5 = (m×8+n×3)/(m+n) 5 = (8m+3n)/(m+n) 5m+5n = 8m+3n 2n = 3m...
Let A(x1,y1), B(x2,y2) and C(x3,y3) be the vertices of the triangle ABC. Consider AB By midpoint formula, (x1+x2)/2 = 0 x1+x2 = 0 x1 = -x2 ..(i) By midpoint formula, (y1+y2)/2 = ½ y1+y2 = 1 …(ii)...
Let A (-1,4) and B(5,2) are the vertices of the triangle and let D(0,3) is the midpoint of side AC. Let co-ordinate of C be (x,y). Consider D(0,3) as midpoint of AC By midpoint formula, (-1+x)/2 = 0...
Given points are A(-5,4), B(-1,-2) and C(5,2) are given. Since these are vertices of an isosceles triangle ABC then AB = BC. By distance formula, d(AB) = √[(x2-x1)2+(y2-y1)2] Here x1 = -5, y1 = 4...
Solution: Let A(3,2) and B(-1,0) be the two vertices of the parallelogram ABCD. Let M(2,-5) be the point where diagonals meet. Since the diagonals of the parallelogram bisect each other, M is the...
Given vertices of the parallelogram are A(-2,-1), B(1, 0), C(p,3) and D(1,q). Let M(x,y) be the midpoint of the diagonals of the parallelogram ABCD. Diagonals AC and BD bisect each other at M. When...
Solution: Let M be the midpoint of the diagonals of the parallelogram ABCD. Co-ordinate of M will be the midpoint of diagonal AC. Given points are A(1,2), B(1,0) and C(4,0). Consider line AC. x1 =...
Solution: Let M(x,y) be the median of ΔABC through A to BC. M will be the midpoint of BC. x1 = 5, y1 = 3 x2 = 3, y2 = -1 By midpoint formula, x = (x1+x2)/2 x = (5+3)/2 = 8/2 = 4 By midpoint formula,...
(iii)Let m:n be the ratio in which Y axis divide line AB. Since AB touches Y axis, its x co-ordinate will be zero. Here x1 = -4, y1 = 1 x2 = 17, y2 = 10 By section formula, x = (mx2+nx1)/(m+n) 0 =...
Solution: (i)Let P(x,y) divides the line segment joining the points A(-4,1), B(17,10) in the ratio 1:2, Here x1 = -4, y1 = 1 x2 = 17, y2 = 10 m:n = 1:2 By section formula, x = (mx2+nx1)/(m+n) x =...
(iii) By distance formula, d(AB) = √[(x2-x1)2+(y2-y1)2] d(AB) = √[(8-(-4))2+(-3-6)2] d(AB) = √[(12)2+(-9)2] d(AB) = √(144+81) d(AB) = √225 d(AB) = 15 Hence the length of AB is 15...
Solution: (i) Let m:n be the ratio in which the line segment joining A (-4,6) and B(8,-3) is divided by the Y axis. Since the line meets Y axis, its x co-ordinate is zero. Here x1 = -4, y1 = 6 x2 =...
Solution: (i) Let m:n be the ratio in which the line segment joining (3,4) and (-2,1) is divided by the Y axis. Since the line meets Y axis, its x co-ordinate is zero. Here x1 = 3, y1 = 4 x2 = -2,...
Solution: (i) Given points are A(-4,3) and B(8,-6). Here x1 = -4, y1 = 3 x2 = 8, y2 = -6 By distance formula, d(AB) = √[(x2-x1)2+(y2-y1)2] d(AB) = √[(8-(-4))2+(-6-3)2] d(AB) = √[(12)2+(-9)2] d(AB) =...
Solution: Since the point K is on X axis, its y co-ordinate is zero. Let the point K be (x,0). Let the point K divides the line segment joining A(2,3) and B(6,-5) in the ratio m:n. Here x1 = 2 ,...
Solution: (i) Let the ratio that the point (5,4) divide the line segment joining the points (2,1) and (7,6) be m:n, Here x1 = 2 , y1 = 1 , x2 = 7, y2 = 6, x = 5, y = 4 By section formula, x =...
Solution: Let the co-ordinates of B be (x2,y2). Given co-ordinates of A = (2,-2) Co-ordinates of P = (-4,1) Ratio m:n = 3:5 x1 = 2, y1 = -2, x = -4, y = 1 P divides AB in the ratio 3:5 By section...
Solution: (i) Let P(x,y) divides the line segment joining the points A(-1,5/3), B(a,5) in the ratio 1:3, Here m:n = 1:3 x1 = -1 , y1 = 5/3 , x2 = a, y2 = 5 By Section formula x = (mx2+nx1)/(m+n) x =...
Solution: Let the co-ordinates of the image of the point P(5,-3) be P1(x, y) in the point (-1, 3) then the point (-1, 3) will be the midpoint of PP1. By midpoint formula, x = (x1+x2)/2 -1 = (5+x2)/2...
(i) Length of radius AC = d(A,C) Co-ordinates of A = (3,-7) Co-ordinates of C = (-2,5) Here x1 = 3, y1 = -7, x2 = -2, y2 = 5 By distance formula, d(A,C) = √[(x2-x1)2+(y2-y1)2] = √[(-2-3)2+(5-(-7))2]...
Solution: Let the co-ordinates of Q be (x2, y2). Given co-ordinates of P = (-3,2) Co-ordinates of midpoint = (1,-2) Here x1 = -3, y1 = 2 , x = 1 , y = -2 By Midpoint formula, x = (x1+x2)/2 1 =...
Solution: Let the midpoint of line joining the points A(3m,6) and B(-4,3n) be C(1,2m-1). Here x1 = 3m, y1 = 6 , x2 = -4, y2 = 3n x = 1 , y = 2m-1 By Midpoint formula, x = (x1+x2)/2 1 = (3m+-4)/2...
(v) In quadrilateral ABC1B1, ABB1C1 Hence the quadrilateral ABC1B1 is a trapezium.
(iii) The quadrilateral ABA1B1 will be a rhombus. (iv) Let C be midpoint of AB. Co-ordinate of C = ((3+0)/2 , (0+4)/2) = (3/2, 2) [Midpoint formula] In a point reflection in the origin, the image of...
(i) Co-ordinates of point A are (3,0). When you reflect a point across the Y-axis, the y-coordinate remains the same, but the x-coordinate is transformed into its opposite (its sign is changed)...
(iii) Co-ordinates of P’ = (3,5) Co-ordinates of P’’ = (-3,-5) Here x1 = 3, y1 = 5 , x2 = -3, y2 = -5 Let Q(x,y) be the midpoint of P’P’’ By midpoint formula, x = (x1+x2)/2 y = (y1+y2)/2 x =...
Solution: (i) The image of P(3,-5) when reflected in X-axis will be (3,5). When you reflect a point across the X-axis, the x-coordinate remains the same, but the y-coordinate is transformed into its...
Let P be the point which is three-fourth of the way from A(3,1) to B(-2,5). AP/AB = 3/ 4 AB = AP+PB AP/AB = AP/(AP+PB) = 3/4 4AP = 3AP+3PB 4AP-3AP = 3PB AP = 3PB AP/PB = 3/1 The ratio m:n = 3:1 x1 =...
Solution: (i) Let the co-ordinates of P(x, y) divides AB in the ratio m:n. A(3,2) and B(5,1) are the given points. Given m:n = 1:2 x1 = 3 , y1 = 2 , x2 = 5 , y2 = 1 , m = 1 and n = 2 By Section...
Let P and Q be the points of trisection of AB i.e., AP = PQ = QB Given A(3,-3) and B(6,9) x1 = 3, y1 = -3, x2 = 6, y2 = 9 P(x, y) divides AB internally in the ratio 1 : 2. m:n = 1:2 By applying the...
Solution: Let the co-ordinates of P(x, y) divides AB in the ratio m:n. A(-2,1) and B(1,4) are the given points. Given m:n = 2:1 x1 = -2 , y1 = 1 , x2 = 1 , y2 = 4 , m = 2 and n = 1 By Section...
Solution: Let the co-ordinates of P(x, y) divides AB in the ratio m:n. A(-3,3) and B(12,-7) are the given points. Given m:n = 2:3 x1 = -3 , y1 = 3 , x2 = 12 , y2 = -7 , m = 2 and n = 3 By Section...
(iii) Co-ordinates of midpoint of line joining the points (x1,y1) and (x2,y2) = {(x1+x2)/2 ,(y1+y2)/2} Co-ordinates of midpoint of line joining the points (a+3, 5b) and (2a-1,3b+4) = {(a+3+2a-1)/2,...
Solution: Co-ordinates of midpoint of line joining the points (x1,y1) and (x2,y2) = {(x1+x2)/2 ,(y1+y2)/2} (i) Co-ordinates of midpoint of line joining the points (2, -3) and (-6,7) = {(2+-6)/2,...