A(h, -6), B(2, 3) and C(-6, k) are the co-ordinates of vertices of a triangle whose
A(h, -6), B(2, 3) and C(-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1, 5). Find h and k.
A(x1,y1) = A(h, -6),
B (x2 , y2) = B(2, 3),
C (x3 , y3) = C (-6, k)
B (x2 , y2) = B(2, 3),
C (x3 , y3) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,
x = `("x"_1 + "x"_2 + "x"_3)/3`
∴ 1 = `(h + 2 +(-6))/3`
∴ 3 = h + 2 - 6
∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7
y = `("y"_1 + "y"_2 + "y"_3)/3`
∴ 5 = `(-6 + 3 + "k")/3`
∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18
Explanation:-
The coordinates of the centroid of a triangle are given by the average of the coordinates of its vertices. That is, if the coordinates of the vertices of a triangle are (x1, y1), (x2, y2), and (x3, y3), then the coordinates of its centroid are [(x1+x2+x3)/3, (y1+y2+y3)/3].
Using this formula, we know that the coordinates of the centroid G are (1, 5), and we can set up the following two equations:
(h + 2 – 6)/3 = 1 (average of x-coordinates of A, B, C is 1) (-6 + 3 + k)/3 = 5 (average of y-coordinates of A, B, C is 5)
Simplifying these equations, we get:
h – 4 = 3 (multiply both sides by 3) k – 3 = 15 (multiply both sides by 3)
Solving for h and k, we get:
h = 7 k = 18
Therefore, the coordinates of the vertices of the triangle are A(7, -6), B(2, 3), and C(-6, 18).
Chapter 5. Co-ordinate Geometry – Practice Set 5.2 (Page 115)