Chapter 13.1, Problem 10PSA

To determine

To find: The probability both contain the point $\left(2,6\right)$ if two of the lines are selected at random.

Answer to Problem 10PSA

The required probability is $\frac{1}{3}$ .

Explanation of Solution

Given:

The equations are $y=x+4$ , $y=3x$ and $y=2x-2$ , and point is $\left(2,6\right)$ .

Calculation:

Substitute $2$ for $x$ , and $6$ for $y$ in the equation $y=x+4$ .

  $\begin{array}{c}6=2+4\\ 6=6\end{array}$

Substitute $2$ for $x$ , and $6$ for $y$ in the equation $y=x+4$ .

  $\begin{array}{c}6=3\left(2\right)\\ 6=6\end{array}$

Substitute $2$ for $x$ , and $6$ for $y$ in the equation $y=2x-2$ .

  $\begin{array}{c}6=2\left(2\right)-2\\ 6\ne 2\end{array}$

If $y=x+4$ , $y=3x$ and $y=2x-2$ are named $l_1,l_2$ , and $l_3$ , respectively, only $l_1$ and $l_2$ contain the point $\left(2,6\right)$ . Possible combinations are $\left\{l_1,l_2,l_1{l}_3,l_3{l}_2\right\}$ . Only $l_1{l}_2$ both contain the point, so probability is $\frac{1}{3}$ .

Therefore, the required probability is $\frac{1}{3}$ .