If two vertices of a triangle are 4 3 2 6 and the centroid is the origin then find the third vertex

The Centroid of a triangle is the point of intersection of the medians of a triangle.

To find the centroid of a triangle

Let A (x$_{1}$, y$_{1}$), B (x$_{2}$, y$_{2}$) and C (x$_{3}$, y$_{3}$) are the three vertices of the ∆ABC .

Let D be the midpoint of side BC.

Since, the coordinates of B (x$_{2}$, y$_{2}$) and C (x$_{3}$, y$_{3}$), the coordinate of the point D are ($\frac{x_{2} + x_{3}}{2}$, $\frac{y_{2} + y_{3}}{2}$).

Let G(x, y) be the centroid of the triangle ABC.

Then, from the geometry, G is on the median AD and it divides AD in the ratio 2 : 1, that is AG : GD = 2 : 1.

Therefore, x = $\left \{\frac{2\cdot \frac{(x_{2} + x_{3})}{2} + 1 \cdot x_{1}}{2 + 1}\right \}$ = $\frac{x_{1} + x _{2} + x_{3}}{3}$

y = $\left \{\frac{2\cdot \frac{(y_{2} + y_{3})}{2} + 1 \cdot y_{1}}{2 + 1}\right \}$ = $\frac{y_{1} + y _{2} + y_{3}}{3}$

Therefore, the coordinate of the G are ($\frac{x_{1} + x _{2} + x_{3}}{3}$, $\frac{y_{1} + y _{2} + y_{3}}{3}$)

Hence, the centroid of a triangle whose vertices are (x$_{1}$, y$_{1}$), (x$_{2}$, y$_{2}$) and (x$_{3}$, y$_{3}$) has the coordinates ($\frac{x_{1} + x _{2} + x_{3}}{3}$, $\frac{y_{1} + y _{2} + y_{3}}{3}$).

Note: The centroid of a triangle divides each median in the ratio 2 : 1 (vertex to base).

Solved examples to find the centroid of a triangle:

1. Find the co-ordinates of the point of intersection of the medians of trangle ABC; given A = (-2, 3), B = (6, 7) and C = (4, 1).

Solution:

Here, (x$_{1}$ = -2, y$_{1}$ = 3), (x$_{2}$ = 6, y$_{2}$ = 7) and (x$_{3}$ = 4, y$_{3}$ = 1),

Let G (x, y) be the centroid of the triangle ABC. Then,

x = $\frac{x_{1} + x _{2} + x_{3}}{3}$ = $\frac{(-2) + 6 + 4}{3}$ = $\frac{8}{3}$

y = $\frac{y_{1} + y _{2} + y_{3}}{3}$ = $\frac{3 + 7 + 1}{3}$ = $\frac{11}{3}$

Therefore, the coordinates of the centroid G of the triangle ABC are ($\frac{8}{3}$, $\frac{11}{3}$)

Thus, the coordinates of the point of intersection of the medians of triangle are ($\frac{8}{3}$, $\frac{11}{3}$).

2. The three vertices of the triangle ABC are (1, -4), (-2, 2) and (4, 5) respectively. Find the centroid and the length of the median through the vertex A.

Solution:

Here, (x$_{1}$ = 1, y$_{1}$ = -4), (x$_{2}$ = -2, y$_{2}$ = 2) and (x$_{3}$ = 4, y$_{3}$ = 5),

Let G (x, y) be the centroid of the triangle ABC. Then,

x = $\frac{x_{1} + x _{2} + x_{3}}{3}$ = $\frac{1 + (-2) + 4}{3}$ = $\frac{3}{3}$ = 1

y = $\frac{y_{1} + y _{2} + y_{3}}{3}$ = $\frac{(-4) + 2 + 5}{3}$ = $\frac{3}{3}$ = 1

Therefore, the coordinates of the centroid G of the triangle ABC are (1, 1).

D is the middle point of the side BC of the triangle ABC.

Therefore, the coordinates of D are ($\frac{(-2) + 4}{2}$, $\frac{2 + 5}{2}$) = (1, $\frac{7}{2}$)

Therefore, the length of the median AD = $\sqrt{(1 - 1)^{2} + (-4 - \frac{7}{2})^{2}}$ = $\frac{15}{2}$ units.

3. Two vertices of a triangle are (1, 4) and (3, 1). If the centroid of the triangle is the origin, find the third vertex.

Solution:

Let the coordinates of the third vertex are (h, k).

Therefore, the coordinates of the centroid of the triangle ($\frac{1 + 3 + h}{3}$, $\frac{4 + 1 + k}{3}$)

According to the problem we know that the centroid of the given triangle is (0, 0)

Therefore,

$\frac{1 + 3 + h}{3}$ = 0 and $\frac{4 + 1 + k}{3}$ = 0

⟹ h = -4 and k = -5

Therefore, the third vertex of the given triangle are (-4, -5).

● Distance and Section Formulae

10th Grade Math

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Solution:

It is given the centroid is origin and two vertices are $(2,4,6)$ and $(0,-2,-5)$ Suppose the third vertex be $(x, y, z)$ The coordinates of the centroid for a triangle is given by the average of the coordinates of its vertices. $\begin{array}{l} \Rightarrow(0,0,0)=\left(\frac{2+0+x}{3}, \frac{4+(-2)+y}{3}, \frac{6+(-5)+z}{3}\right) \\ \Rightarrow \frac{2+x}{3}=0, \therefore x=-2 \\ \Rightarrow \frac{2+y}{3}=0, \therefore y=-2 \\ \Rightarrow \frac{1+x}{3}=0, \therefore x=-1 \end{array}$

So, the third vertex is $(-2,-2,-1)$.