Question

If two vertices of a triangle are $\left( {5,4} \right)$ and $\left( { - 2,4} \right)$ and centroid is $\left( {5,6} \right)$, then third vertex is
A. $\left( {12,10} \right)$
B. $\left( {10,12} \right)$
C. $\left( { - 10,12} \right)$
D. $\left( {12, - 10} \right)$

Answer 1

Hint: A triangle has three vertices among which two is given in the question. Centroid of a triangle is a point where three medians of a triangle meet. In the question, the coordinate of the centroid is also given. You have to find the third vertex using the centroid formula.

Formula Used:
Centroid of a triangle having vertices $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)$ is given by $\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$

Complete step by step solution:
Let the third vertex be $\left( {h,k} \right)$
In the given question, ${x_1} = 5,{y_1} = 4,{x_2} = - 2,{y_2} = 4$
And according to our assumption, ${x_3} = h,{y_3} = k$
Now, find the centroid using the centroid formula.
The centroid is $\left( {\dfrac{{5 - 2 + h}}{3},\dfrac{{4 + 4 + k}}{3}} \right) = \left( {\dfrac{{3 + h}}{3},\dfrac{{8 + k}}{3}} \right)$
And the given centroid is $\left( {5,6} \right)$
Compare the centroid which we obtain using the formula and which is given.
$\dfrac{{3 + h}}{3} = 5$ and $\dfrac{{8 + k}}{3} = 6$
Now, solve these two equations to find the values of $h$ and $k$.
The first equation is $\dfrac{{3 + h}}{3} = 5$
$\begin{array}{l} \Rightarrow 3 + h = 15\\ \Rightarrow h = 15 - 3\\ \Rightarrow h = 12\end{array}$
The second equation is $\dfrac{{8 + k}}{3} = 6$
$\begin{array}{l} \Rightarrow 8 + k = 18\\ \Rightarrow k = 18 - 8\\ \Rightarrow k = 10\end{array}$
Finally, we get $h = 12$ and $k = 10$
$\therefore $The centroid is $\left( {12,10} \right)$

Option ‘A’ is correct

Note: You can also use an alternative method to solve the given question. For this, it is essential to know that a median of a triangle is a line segment joining a vertex and the midpoint of its opposite side and the centroid of a triangle divides each of the three medians in the ratio $2:1$.
Coordinates of the midpoint of the line segment joining $\left( {5,4} \right)$ and $\left( { - 2,4} \right)$ is $\left( {\dfrac{{5 - 2}}{2},\dfrac{{4 + 4}}{2}} \right) = \left( {\dfrac{3}{2},4} \right)$.
Let the third vertex be $\left( {h,k} \right)$.
Since the centroid of a triangle divides a median in the ratio $2:1$.
So, using the section formula the third vertex can be obtained.
If a point divides a line segment joining the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ in the ratio $m:n$ then coordinates of the point are $\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$
Here ${x_1} = h,{y_1} = k,{x_2} = \dfrac{3}{2},{y_2} = 4$
So, the coordinates of the centroid is $\left( {\dfrac{{2 \times \dfrac{3}{2} + 1 \times h}}{{2 + 1}},\dfrac{{2 \times 4 + 1 \times k}}{{2 + 1}}} \right) = \left( {\dfrac{{3 + h}}{3},\dfrac{{8 + k}}{3}} \right)$
Given coordinates of centroid is $\left( {5,6} \right)$
So, $\dfrac{{3 + h}}{3} = 5$ and $\dfrac{{8 + k}}{3} = 6$
$ \Rightarrow 3 + h = 15$ and $8 + k = 18$
$ \Rightarrow h = 12$ and $k = 10$
$\therefore $The centroid is $\left( {12,10} \right)$