Question

$A\left(4,3\right),B\left(6,5\right),C\left(5,-2\right)$ are vertices of $∆ABC$. If $D$ is the midpoint of $BC$, find the co-ordinates of $D$. Find the co-ordinates of $P$ such that $AP:PD=2:3$. Find the area of $∆PBC$ and compare it with the area of $∆ABC$.

Answer 1

Step1: Calculation of the coordinates of $D$.

The co-ordinates of the midpoint $M$ which divides the line joining $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$.is given by $M\left(x,y\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

The point $D$ is the midpoint of $BC$.

$$\begin{gathered} \Rightarrow D\left(x,y\right)=\left(\frac{6+5}{2},\frac{5-2}{2}\right) \\ \therefore D\left(x,y\right)=\left(\frac{11}{2},\frac{3}{2}\right) \end{gathered}$$

Step2: Calculation of the coordinates of $P$.

The co-ordinates of the point $P$ which divides the line joining $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ internally in the ratio $m_1:m_2$.is given by $P\left(x,y\right)=\left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)$.

The point $P$ divides $AD$ in the ratio $2:3$.

$$\begin{gathered} \Rightarrow P\left(x,y\right)=\left(\frac{2\times \frac{11}{2}+3\times 4}{2+3},\frac{2\times \frac{3}{2}+3\times 3}{2+3}\right) \\ \therefore P\left(x,y\right)=\left(\frac{23}{5},\frac{12}{5}\right) \end{gathered}$$

Step3: Calculation of the area of $∆ABC$.

The formula to calculate the area of a triangle with vertices $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$, and $\left(x_3,y_3\right)$ is $\left|\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\right|$.

$$\begin{gathered} \text{area}\left(∆ABC\right)=\left|\frac{1}{2}\left[4\left(5-\left(-2\right)\right)+6\left(-2-3\right)+5\left(3-5\right)\right]\right| \\ \Rightarrow \text{area}\left(∆ABC\right)=\left|\frac{1}{2}\left[4·7+6\left(-5\right)+5\left(-2\right)\right]\right| \\ \Rightarrow \text{area}\left(∆ABC\right)=\left|\frac{1}{2}\left[28-30-10\right]\right| \\ \Rightarrow \text{area}\left(∆ABC\right)=\left|\frac{1}{2}\left(-12\right)\right| \\ \Rightarrow \text{area}\left(∆ABC\right)=\left|-6\right| \\ \therefore \text{area}\left(∆ABC\right)=6 \end{gathered}$$

Step4: Calculation of the area of $∆PBC$.

$$\begin{gathered} \text{area}\left(∆PBC\right)=\left|\frac{1}{2}\left[\frac{23}{5}\left(5-\left(-2\right)\right)+6\left(-2-\frac{12}{5}\right)+5\left(\frac{12}{5}-5\right)\right]\right| \\ \Rightarrow \text{area}\left(∆PBC\right)=\left|\frac{1}{2}\left[\frac{23}{5}·7+6\left(-\frac{22}{5}\right)+5\left(-\frac{13}{5}\right)\right]\right| \\ \Rightarrow \text{area}\left(∆PBC\right)=\left|\frac{1}{2}\left[\frac{161}{5}-\frac{132}{5}-\frac{65}{5}\right]\right| \\ \Rightarrow \text{area}\left(∆PBC\right)=\left|\frac{1}{2}\left(-\frac{36}{5}\right)\right| \\ \Rightarrow \text{area}\left(∆PBC\right)=\left|-\frac{18}{5}\right| \\ \therefore \text{area}\left(∆PBC\right)=\frac{18}{5} \end{gathered}$$

Hence, $\frac{\text{area}\left(∆PBC\right)}{\text{area}\left(∆ABC\right)}=\frac{\frac{18}{5}}{6}=\frac{3}{5}$.

Final answer: The coordinates of $D$ and $P$ are $D\left(\frac{11}{2},\frac{3}{2}\right)$ and $P\left(\frac{23}{5},\frac{12}{5}\right)$ respectively, and $\text{area}\left(∆PBC\right)=\frac{18}{5}$ and $\frac{\text{area}\left(∆PBC\right)}{\text{area}\left(∆ABC\right)}=\frac{3}{5}$.