Question

$A( - 4,4),B(x, - 1){\text{ and }}C(6,y)$ are the vertices of triangle $\Delta ABC$. If the centroid of this triangle $\Delta ABC$ is at the origin, find the values of $x{\text{ and }}y$.

Answer 1

Hint: This problem is related to the properties of a triangle. We know that the centroid of a triangle is the intersection point of three medians and medians are the lines which join a vertex of the triangle to the midpoint of the opposite side. As it is given that the centroid of the triangle is at origin then the coordinates of the centroid will be (0, 0), it will help to solve the problem. Also, you should know the general formula for point of centroid if vertices of a triangle are known.

Complete step-by-step answer:
It is given that the verticals of the triangle $\Delta ABC$ are $A( - 4,4),B(x, - 1){\text{ and }}C(6,y)$. As we know that the centroid of a triangle is the intersection point of three medians drawn from the verticals. If we say that in a triangle $\Delta ABC$ with verticals $A({x_1},{y_1}),B({x_2},{y_2}){\text{ and C}}({x_3},{y_3})$, the coordinates of centroid are $G(X,Y)$, then that point of the triangle can be expressed as
X – Coordinates of $G:$
$X = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}{\text{ }}.................{\text{ (1)}}$
Y – Coordinates of $G:$
$Y = \dfrac{{{y_1} + {y_2} + {y_3}}}{3}{\text{ }}................{\text{(2)}}$
Thus, $G(X,Y) = (\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3})$
As we know that the verticals are$A( - 4,4),B(x, - 1){\text{ and }}C(6,y)$. Therefore, the values of the variables are
 $
  {x_1} = - 4,{x_2} = x,{\text{ and }}{x_3} = 6 \\
  {y_1} = 4,{y_2} = - 1,{\text{ and }}{y_3} = y \\
$
Now, putting these values in eq. (1) and (2), we get
X – Coordinates of $G:$
$
  X = \dfrac{{ - 4 + x + 6}}{3} \\
  X = \dfrac{{x + 2}}{3}{\text{ }}...............{\text{ (3)}} \\
$
Y – Coordinates of $G:$
$
  Y = \dfrac{{4 - 1 + y}}{3} \\
  Y = \dfrac{{y + 3}}{3}{\text{ }}................{\text{(4)}} \\
$
It is known that the centroid is at origin, therefore, the values of $X{\text{ and }}Y$ is (0, 0). Now, putting this value in eq. (3) and (4) and simplifying, we will get,
$
  0 = \dfrac{{x + 2}}{3} \\
  x = - 2 \\
$
And,
$
  0 = \dfrac{{y + 3}}{3} \\
  y = - 3 \\
$

This way, values of $x = - 2,y = - 3$.

Note: Solving such types of problems, you should have a formula for centroid of the triangle handy. Calculation of the centroid of a triangle can be shown as below:Let us assume that in a triangle $\Delta ABC$, coordinates of its verticals are $A({x_1},{y_1}),B({x_2},{y_2}){\text{ and C}}({x_3},{y_3})$. Let D is the middle point of side BC, then its coordinates can be represented as $(\dfrac{{{x_2} + {x_3}}}{2},\dfrac{{{y_2} + {y_3}}}{2})$. Let $G(X,Y)$ be the centroid of the triangle. Then, from the Geometry, we know that the centroid is located on the median drawn from point A on the opposite side’s middle point and $G$ point divides this median in a 2:1 ratio. It means$AG:GD = 2:1$.
Therefore,
$X = \{ \dfrac{{2.\dfrac{{({x_2} + {x_3})}}{2} + 1.{x_1}}}{{2 + 1}}\} = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}$
 $Y = \{ \dfrac{{2.\dfrac{{({y_2} + {y_3})}}{2} + 1.{y_1}}}{{2 + 1}}\} = \dfrac{{{y_1} + {y_2} + {y_3}}}{3}$
Hence, the centroid of a triangle whose vertices are $({x_1},{y_1}),({x_2},{y_2}){\text{ and }}({x_3},{y_3})$ has the coordinates $(\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3})$.