Find the coordinates of centroid of the triangles if points D(-7, 6), E(8, 5) and F(2, -2) are the mid points of the sides of that triangle.
14«. Find the coordinates of centroid of the triangles if points D(-7, 6), E(8, 5) and F(2, -2) are the mid points of the sides of that triangle.
Suppose A (x1, y1), B(x2, y2) and C(x3, y3) are the vertices of the triangle.
D(–7, 6), E(8, 5), and F(2, –2) are the midpoints of sides BC, AC, and AB respectively.
Let G be the centroid of ∆ABC.
D is the midpoint of seg BC.
By the mid-point formula,
Co-ordinates of D = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
∴ (-7, 6) = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
∴ `(x_2 + x_3)/2 = –7 "and" (y_2 + y_3)/2 = 6`
∴ x2 + x3 = –14 ...(i) and ∴ y2 + y3 = 12 ...(ii)
E is the midpoint of seg AC.
By the mid-point formula,
Co-ordinates of E = `((x_1 + x_3)/2, (y_1 + y_3)/2)`
∴ (8, 5) = `((x_1 + x_3)/2, (y_1 + y_3)/2)`
∴ `(x_1 + x_3)/2 = 8 "and" (y_1 + y_3)/2 = 5`
∴ x1 + x3 = 16 ...(iii) and ∴ y1 + y3 = 10 ...(iv)
F is the midpoint of seg AB.
By the mid-point formula,
Co-ordinates of F = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
∴ (2, -2) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
∴ `(x_1 + x_2)/2 = 2 "and" (y_1 + y_2)/2 = -2`
∴ x1 + x2 = 4 ...(v) and ∴ y1 + y2 = -4 ...(vi)
Adding (i), (iii), and (v),
x2 + x3 + x1 + x3 + x1 + x2 = –14 + 16 + 4
∴ 2x1 + 2x2 + 2x3 = 6
∴ x1 + x2 + x3 = 3 ...(vii)
Adding (ii), (iv), and (vi),
y2 + y3 + y1 + y3 + y1 + y2 = 12 + 10 – 4
∴ 2y1 + 2y2 + 2y3 = 18
∴ y1 + y2 + y3 = 9 ...(viii)
G is the centroid of ∆ABC.
By centroid formula,
`"Co-ordinates of G" = ((x_1 + x_2 + x_3)/3, (y_1 + y_ 2 + y_3)/3)`
`"Co-ordinates of G" = (3/3, 9/3)` ...[From (vii) and (viii)]
Co-ordinates of G = (1, 3)
∴ The co-ordinates of the centroid of the triangle are (1, 3).
Explanation:-
Suppose we have a triangle ABC with vertices A(x1, y1), B(x2, y2), and C(x3, y3). Let D(-7, 6), E(8, 5), and F(2, -2) be the midpoints of sides BC, AC, and AB, respectively. Let G be the centroid of the triangle ABC.
We know that D is the midpoint of segment BC, and we can use the midpoint formula to find its coordinates. Therefore, the coordinates of D are ((x2+x3)/2, (y2+y3)/2). Substituting the coordinates of D, we have (-7, 6) = ((x2+x3)/2, (y2+y3)/2). Hence, we have (x2+x3)/2 = -7 and (y2+y3)/2 = 6. Solving these equations gives us x2+x3 = -14 (i) and y2+y3 = 12 (ii).
Similarly, we can find the coordinates of E and F using the midpoint formula. The coordinates of E are ((x1+x3)/2, (y1+y3)/2), and since E is the midpoint of segment AC, we can substitute its coordinates to obtain (8, 5) = ((x1+x3)/2, (y1+y3)/2). Hence, we have (x1+x3)/2 = 8 and (y1+y3)/2 = 5. Solving these equations gives us x1+x3 = 16 (iii) and y1+y3 = 10 (iv).
The coordinates of F are ((x1+x2)/2, (y1+y2)/2), and since F is the midpoint of segment AB, we can substitute its coordinates to obtain (2, -2) = ((x1+x2)/2, (y1+y2)/2). Hence, we have (x1+x2)/2 = 2 and (y1+y2)/2 = -2. Solving these equations gives us x1+x2 = 4 (v) and y1+y2 = -4 (vi).
To find the coordinates of the centroid G, we can use the fact that the centroid is the point of intersection of the medians of the triangle. The medians of a triangle are the line segments that connect each vertex to the midpoint of the opposite side. Therefore, the coordinates of the centroid G are ((x1+x2+x3)/3, (y1+y2+y3)/3).
We can use equations (vii) and (viii) to find the coordinates of the centroid G. Adding equations (i), (iii), and (v), we get x1+x2+x3 = 3. Adding equations (ii), (iv), and (vi), we get y1+y2+y3 = 6. Therefore, the coordinates of the centroid G are ((x1+x2+x3)/3, (y1+y2+y3)/3) = (1, 2).
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)