Question

Let $A(4,2),B(6,5)$and $C(1,4)$ be the vertices of $∆ABC$.

(i) The median from $A$ meets $BC$ at $D$. Find the coordinates of point $D$.

(ii) Find the coordinates of the point $P$ on $AD$ such that $AP:PD=2:1$.

(iii) Find the coordinates of point $Q$ and $R$ on medians $BE$ and $CF$ respectively such that $BQ:QE=2:1$ and $CR:RF=2:1$.

(iv) What do you observe? [Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio $2:1$].

(v) If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices of triangle $ABC$, find the coordinates of the centroid of the triangle.

Answer 1

Step 1 : Find co-ordinates of given point $D$:

Given, $A(4,2),B(6,5)$and $C(1,4)$

Also, the median from $A$ meets $BC$ at $D$

So, by using the midpoint formula

$\mathbf{D}\mathbf{=}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$D=\left(\frac{6+1}{2},\frac{5+4}{2}\right)=\left(\frac{7}{2},\frac{9}{2}\right)$

Step 2: Find co-ordinates of point $P$:

We need to find the coordinates of the point $P$ on $AD$ such that $AP:PD=2:1$

So, by using section formula

$\mathbf{P}\mathbf{=}\left(\frac{m_2{x}_1+m_1{x}_2}{m_1+m_2},\frac{m_2{y}_1+m_1{y}_2}{m_1+m_2}\right)$

Here, $m_1=2,m_2=1and(x_1,y_1)=\left(4,2\right),(x_2,y_2)=\left(\frac{7}{2},\frac{9}{2}\right)$

$P=\left(\frac{2\times x_2+1\times x_1}{2+1},\frac{2\times y_2+1\times y_1}{2+1}\right)$

$P=\left(\frac{11}{3},\frac{11}{3}\right)$

Step 3: Find co-ordinates of point $Q$ and $R$

Given that, $Q$ and $R$ are on $BE$ and $CF$ respectively such that $BQ:QE=2:1$ and $CR:RF=2:1$.

So, by using midpoint formula, we will first find the coordinates of $E$ and $F$

Here,$(x_A,y_A)=\left(4,2\right)and(x_C,y_C)=\left(1,4\right)$

$\mathbf{E}\mathbf{=}\left(\frac{x_A+x_C}{2},\frac{y_A+y_C}{2}\right)\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{F}\mathbf{=}\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right)$

$E=\left(\frac{5}{2},\frac{6}{2}\right),F=\left(\frac{10}{2},\frac{7}{2}\right)$

$\therefore$ $E=\left(\frac{5}{2},3\right),F=\left(5,\frac{7}{2}\right)$

Now,

$(x_E,y_E)=\left(\frac{5}{2},3\right)$ and $(x_F,y_F)=\left(5,\frac{7}{2}\right),\left(x_B,y_B\right)=\left(6,5\right)$

By using section formula

$Q=\left(\frac{2\times x_E+1\times x_B}{2+1},\frac{2\times y_E+1\times y_B}{2+1}\right)$

$Q=\left(\frac{11}{3},\frac{11}{3}\right)$

and,

$R=\left(\frac{2\times x_F+1\times x_C}{2+1},\frac{2\times y_F+1\times y_C}{2+1}\right)$

$R=\left(\frac{11}{3},\frac{11}{3}\right)$

Step 4: Draw an observation from the results obtained

We can observe that the co-ordinates of the points $P,Q,R$ are same, i.e. $\left(\frac{11}{3},\frac{11}{3}\right)$

So, we can conclude that the point is the intersection point of all medians and divides them in the ratio $2:1$ and is called centroid of the triangle.

Step 5: Find co-ordinates of the centroid

$A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the co-ordinates of the triangle then the centroid is given as,

Also, the median from $A$ meets $BC$ at $D$

So, by using the midpoint formula

$\mathbf{D}\mathbf{=}\mathbf{(}\frac{{\mathbf{x}}_{\mathbf{3}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}}{\mathbf{2}}\mathbf{,}\frac{{\mathbf{y}}_{\mathbf{3}}\mathbf{+}{\mathbf{y}}_{\mathbf{2}}}{\mathbf{2}}\mathbf{)}$

As we know from the above results centroid divide medain in ratio of $2:1$

So,by section formula

$G=\left(\frac{x_1+2x_D}{3},\frac{y_1+2y_D}{3}\right)$

$G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

The centroid of the given triangle is

$G=\left(\frac{11}{3},\frac{11}{3}\right)$

Hence, the results obtained are,

(i) Co-ordinates of point $\mathit{D}\mathbf{=}\left(\frac{7}{2},\frac{9}{2}\right)$

(ii) Co-ordinates of point $\mathit{P}\mathbf{=}\left(\frac{11}{3},\frac{11}{3}\right)$

(iii) Co-ordinates of point $\mathit{Q}$ and $\mathit{R}$ are, $\mathit{Q}\mathbf{=}\left(\frac{11}{3},\frac{11}{3}\right)$ and $\mathit{R}\mathbf{=}\left(\frac{11}{3},\frac{11}{3}\right)$

(iv) We can conclude that the point is the intersection point of all medians and divides them in the ratio $\mathbf{2}\mathbf{:}\mathbf{1}$ and is called centroid of the triangle.

(v) Co-ordinates of the centroid $\mathit{G}\mathbf{=}\left(\frac{11}{3},\frac{11}{3}\right)$