Chapter 6.1, Problem 45PPS

To determine

Find a system of linear equations for the given graph and if it has one solution then name it.

Answer to Problem 45PPS

The system of linear equation:

  $y=2-x$ and $y=2x-1$

The name of one solution is $(1,1)$ .

Explanation of Solution

Given information:

It is given in the question that,

The graph is:

  

Formula used:

Here,

Use the formula of two point form of a line that is,

  $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

(where $\left(x_1,y_1\right)\&(x_2,y_2)$ are the co-ordinate of the two point on line)

Calculation: Co-ordinate on the $1^{st}$ line: $(0,2)\&(2,0)$ .

Co-ordinate on the $2^{nd}$ line: $(1,1)\&(0,-1)$ .

So, for finding the equation of $1^{st}$ line,

By putting $(0,2)\&(2,0)$ in place of $\left(x_1,y_1\right)\&(x_2,y_2)$ in the formula that is

  $\Rightarrow \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}\Rightarrow \frac{y-2}{x-0}=\frac{0-2}{2-0}\\ \Rightarrow \frac{y-2}{x}=\frac{-2}{2}\\ \Rightarrow \frac{y-2}{x}=\frac{-1}{1}\\ \Rightarrow (y-2)=-x\\ \Rightarrow y=2-x\end{array}$

So, the equation is $y=2-x$

And for finding the equation of $2^{nd}$ line,

By putting $(1,1)\&(0,-1)$ in place of $\left(x_1,y_1\right)\&(x_2,y_2)$ in the formula that is,

  $\Rightarrow \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}\Rightarrow \frac{y-1}{x-1}=\frac{-1-1}{0-1}\\ \Rightarrow \frac{y-1}{x-1}=\frac{-2}{-1}\\ \Rightarrow \frac{y-1}{x-1}=\frac{2}{1}\\ \Rightarrow )(y-1)=2(x-1)\\ \Rightarrow (y-1)=2x-2\\ \Rightarrow y=2x-1\end{array}$

So, the equation is $y=2x-1$.

And it is clear from the above graph is the point of intersection is $(1,1)$ .