Question

Determine, algebraically, the vertices of the triangle formed by the lines

$3\mathrm{x}-\mathrm{y}=3,2\mathrm{x}-3\mathrm{y}=2,\mathrm{x}+2\mathrm{y}=8$

Answer 1

As we are given with the equation of lines

$$\begin{gathered} 3x–y=3\dots \left(1\right) \\ 2x-3y=2\dots \left(2\right) \\ x+2y=8\dots \left(3\right) \end{gathered}$$

Let the equations of the line $\left(1\right),\left(2\right)and\left(3\right)$ represent the side of an$∆ABC.$

On solving the equations$\left(1\right)and\left(2\right)$ by elimination method, we obtain

At First, will multiply and then subtract we obtain

$\Rightarrow (9x-3y)-(2x-3y)=9-2$

$\Rightarrow 7x=7$

$\Rightarrow x=1$

On substituting$x=1inEq.\left(1\right)$, we obtain

$\Rightarrow 3\times \left(1\right)-y=3$

$\Rightarrow y=0$

So, the coordinate of point B is$(1,0)$

On solving lines $\left(2\right)and\left(3\right),$

We obtain,

Multiply$Eq.\left(3\right)by2$ and then subtract

$\Rightarrow (2x+4y)-(2x-3y)=16-2$

$\Rightarrow 7y=14$

$\Rightarrow y=2$

Substituting$y=2inEq.\left(3\right)$, we obtain

$$\begin{gathered} \Rightarrow x+2\times 2=8 \\ \Rightarrow x+4=8 \\ \Rightarrow x=4 \end{gathered}$$

Hence, the coordinate of point $Cis(4,2).$

On solving lines $\left(3\right)and\left(1\right),$

We obtain,

Multiply in$Eq.\left(1\right)by2$ and then add

$$\begin{gathered} \Rightarrow (6x-2y)+(x+2y)=6+8 \\ \Rightarrow 7x=14 \\ \Rightarrow x=2 \end{gathered}$$

Substituting$x=2inEq.\left(1\right)$, we obtain

$$\begin{gathered} \Rightarrow 3\times 2–y=3 \\ \Rightarrow y=3 \end{gathered}$$

So, the coordinate of point $Ais(2,3).$

Hence, the vertices of the $∆ABC$ formed by the given lines are as follows,

$A(2,3),B(1,0)andC(4,2).$