Question

Suppose $ABC$ be a triangle , whose centroid is $G$. If the coordinates of $B,CandG$ are $(-2,0),(3,1),(1,5)$, then the equations of the sides $AB$ and $AC$ are respectively.

• A. $7x-2y+14=0,13x+y-40=0$
• B. $7x+2y+14=0,x+13y-40=0$
• C. $7x-2y-14=0,13x+13y-40=0$
• D. none of these

Answer 1

The correct option is A

$7x-2y+14=0,13x+y-40=0$

Explanation of correct option:

Finding the coordinates of $A:$

Given that ,$ABC$ is a triangle and $G$ is centroid. Coordinates of B, C and G are $(-2,0),(3,1)and(1,5)$

we need to the Equation of sides $AB$ and $AC$.

First we will find the coordinates of point $A$using centroid formula,

Coordinates of Centroid $G$ is given By $\left(\frac{{\mathrm{x}}_1\hspace{0.17em}+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2\hspace{0.17em}+{\mathrm{y}}_3}{3}\right)$

The coordinates of centroid are given as $(1,5)$

$$\begin{gathered} \text{centroid, G =}\left(\frac{{\mathrm{x}}_1\hspace{0.17em}+{\mathrm{x}}_2+{\mathrm{x}}_3}{3}\text{,}\frac{{\mathrm{y}}_1+\hspace{0.17em}{\mathrm{y}}_2\hspace{0.17em}+{\mathrm{y}}_3}{3}\right) \\ \Rightarrow (1,5)=\left(\frac{{\mathrm{x}}_1-2+\hspace{0.17em}3}{3},\frac{{\mathrm{y}}_1+0+\hspace{0.17em}1}{3}\right) \\ \Rightarrow (1,5)=\left(\frac{{\mathrm{x}}_1+1}{3},\frac{{\mathrm{y}}_1\hspace{0.17em}+1}{3}\right) \end{gathered}$$

On comparing coordinates on both sides, we get

$\frac{{\mathrm{x}}_1+1}{3}=1\mathrm{and}\frac{{\mathrm{y}}_1+1}{3}=5$

${\mathrm{x}}_1=2\mathrm{and}{\mathrm{y}}_1=14$

Therefore, the coordinates of $A(2,14)$

Finding the equation of line $AB$ and $AC$:

The equation of Line is given by $\left(\mathrm{y}-{\mathrm{y}}_1\right)=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right)$

Equation of line $AB$ which is passing through$(2,14)and(-2,0)$

$$\begin{gathered} \underset{}{\overset{}{}} \\ \Rightarrow \stackrel{}{} \\ \Rightarrow \stackrel{}{} \\ \Rightarrow \stackrel{}{} \\ \Rightarrow \stackrel{}{} \\ \Rightarrow \end{gathered}$$$\begin{array}{rcl}\left(\mathrm{y}-14\right)& =& \frac{0-14}{-2-2}\left(\mathrm{x}-2\right)\\ \left(\mathrm{y}-14\right)& =& \frac{-14}{-4}\left(\mathrm{x}-2\right)\\ \left(\mathrm{y}-14\right)& =& \frac{7}{2}(\mathrm{x}-2)\\ 2(\mathrm{y}-14)& =& 7\hspace{0.17em}(\mathrm{x}-2)\\ 2\mathrm{y}-28& =& 7\mathrm{x}-14\\ 7\mathrm{x}-2\mathrm{y}+\hspace{0.17em}14& =& 0\end{array}$

Equation of $AC$ which is passing through $(2,14)and(3,1)$

$$\begin{gathered} \text{⇒}\left(\mathrm{y}-14\right)=\frac{1-14}{3-2}\left(\mathrm{x}-2\right) \\ \Rightarrow \left(\mathrm{y}-14\right)=\frac{-13}{1}\left(\mathrm{x}-2\right) \\ \Rightarrow \left(\mathrm{y}-14\right)=\frac{-13}{1}(\mathrm{x}-2) \\ \Rightarrow (\mathrm{y}-14)=-13(\mathrm{x}-2) \\ \Rightarrow \mathrm{y}-14=-13\mathrm{x}+26 \\ \Rightarrow 13\mathrm{x}+\hspace{0.17em}\mathrm{y}-40=0 \end{gathered}$$

Therefore, the equation of sides $AB$ and $AC$ are $7x-\; 2y+\; 14\; =\; 0,\; 13x+y-\; 40\; =\; 0$ respectively.

Therefore, option (A) is correct.