Question

If the two vertices of the triangle are $\left( -4,4 \right),\left( -1,5 \right)$ and the centroid is $\left( 6,-5 \right)$, then find the third vertex.

Answer 1

Hint: We first find the coordinates of the midpoint M of side joining the given vertices $B\left( -4,4 \right),C\left( -1,5 \right)$ using the section formula for internal division with ratio 1:1. We then assume the coordinate of the third vertex as $A\left( x,y \right)$ . We know that centroid O$\left( 6,-5 \right)$ divides AM with ratio 2:1. We use the section formula again and find the unknowns $x,y$.\[\]

Complete step-by-step solution:
A median is a line joining the vertex of the triangle to the midpoint of the opposite side. The centroid of a triangle is the point of intersection of the medians.
Section Formula: Any point $P(x,y)$ which divides a line segment internally $\overline{AB}$ in a ratio$AP:PB=m:n$ with endpoints $A({{x}_{1}},{{y}_{2}})\text{ and B(}{{\text{x}}_{2}}\text{,}{{\text{y}}_{2}}\text{)}$ then the coordinates of P are
\[x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]

We denote the given vertices as $B\left( -4,4 \right),C\left( -1,5 \right)$. We denote the midpoint of BC as M. We join AM, now AM is median. We know that midpoint divides any line segment in a ratio 1:1. So M divides BC with ratio 1:1. We have the coordinates of M using section formula as
\[M=\left( \dfrac{-4+\left( -1 \right)}{2},\dfrac{4+5}{2} \right)=\left( \dfrac{-5}{2},\dfrac{9}{2} \right)\]
We denote the centroid as O whose coordinates are given in the question as $\left( 6,-5 \right)$. We also know that the centroid divides the median in a ratio 2:1 where distance from vertex to the centroid is twice the distance from the to the midpoint on the opposite side to the said vertex. We assume the coordinate of A as $\left( x,y \right)$. We have O dividing A$\left( x,y \right)=\left( {{x}_{1}},{{y}_{1}} \right)$ and M $\left( \dfrac{-5}{2},\dfrac{9}{2} \right)=\left( {{x}_{2}},{{y}_{2}} \right)$ with ratio $m:n=2:1$. We use section formula and have,
\[\begin{align}
  & 6=\dfrac{2\times \left( \dfrac{5}{2} \right)+1\times x}{3},-5=\dfrac{2\times \dfrac{9}{2}+1\times y}{3} \\
 & \Rightarrow 18=5+x,-15=9+y \\
 & \Rightarrow x=13,y=-24 \\
\end{align}\]
So the coordinates of the third vertex A is $A\left( 13,-24 \right)$

Note: We need to be careful of the confusion of internal division for external division which is given by $x=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n},y=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}$. We can alternatively solve directly using the coordinates of centroid $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$ where $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ are coordinates of the vertices of the triangle.