Question

If the median of a triangle meet at $\left( 0,-3 \right)$ and the two vertices are at $\left( -1,4 \right)$ and $\left( 5,2 \right)$ then the third vertex is at _____. \[\]

Answer 1

Hint: We first find the coordinates of the midpoint M of side joining the given vertices $B\left( -1,4 \right),C\left( 5,2 \right)$ 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 where the medians meet O$\left( 0,-3 \right)$ divides AM with ratio 2:1. We use the section formula again and find the unknowns $x,y$. \[\]

Complete step by step answer:
A median is the line joining the vertex of a 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 $B\left( -1,4 \right),C\left( 5,2 \right)$. We denote the midpoint of BC as M. We join AM and 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
\[M=\left( \dfrac{1\left( -1 \right)+1\cdot 5}{2},\dfrac{1\cdot 4+1\times 2}{2} \right)=\left( \dfrac{4}{2},\dfrac{6}{2} \right)=\left( 2,3 \right)\]
We denote the centroid as O whose coordinates are given in the question as$\left( 0,-3 \right)$. We also know that the centroid divides the median in a ratio 2:1 where the distance from the vertex to the centroid is twice the distance from 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 $\left( 2,3 \right)=\left( {{x}_{2}},{{y}_{2}} \right)$ with ratio $AO:OM=m:n=2:1$. We use section formula and have,
\[\begin{align}
  & 0=\dfrac{2\times 2+1\times x}{3},-3=\dfrac{2\times 3+1\times y}{3} \\
 & \Rightarrow 0=4+x,-9=6+y \\
 & \Rightarrow x=-4,y=-15 \\
\end{align}\]

So the coordinates of the third vertex A is $A\left( -4,-15 \right)$\[\]

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.