Question

The position of an open-water swimmer is shown in the graph. The shortest route to the shoreline Ay 10 8. 6. water 4 shore (2, 1) • swimmer 19 2 1 12 3 4 5x -2 An equation that represents the shortest path is y=0.

Answer 1

First, we have to find the slope of the given line.

$m=\frac{y_2-y_1}{x_2-x_1}$

Now, we select two points on the line: (0,3) and (1,7).

Then, we replace these points in the slope formula above.

$m=\frac{7-3}{1-0}=\frac{4}{1}=4$

If the shorter path is perpendicular, then its slope can be found using the rule for perpendicularity:

$\begin{gathered} m\cdot m_1=-1 \\ m\cdot4=-1 \\ m=-\frac{1}{4} \end{gathered}$

Assuming that the shorter path passes through the point (2,1), we use the point-slope formula to find the equation:

$y-y_1=m(x-x_1)_{}_{}$

But, we have to replace the slope m = -1/4, and the point (2,1) in the formula above:

$y-1=-\frac{1}{4}(x-2)$

Now, we use the distributive property to get rid of the parenthesis.

$y-1=-\frac{1}{4}x-2(-\frac{1}{4})$

We solve the product.

$y-1=-\frac{1}{4}x+\frac{2}{4}$

Then, we simplify the fraction 2/4.

$y-1=-\frac{1}{4}x+\frac{1}{2}$

Now, we add 1 on each side.

$\begin{gathered} y-1+1=-\frac{1}{4}x+\frac{1}{2}+1 \\ y+0=-\frac{1}{4}x+\frac{1}{2}+1 \end{gathered}$

We use the least common factor 2 to solve the sum of the independent terms:

$\begin{gathered} y=-\frac{1}{4}x+\frac{1\cdot1+1\cdot2}{2} \\ y=-\frac{1}{4}x+\frac{1+2}{2} \\ y=-\frac{1}{4}x+\frac{3}{2} \end{gathered}$

Finally, as you can observe, the equation that represents the shortest route, perpendicular to the given line, is

$y=-\frac{1}{4}x+\frac{3}{2}$

This line has a slope of -1/4, and its y-intercept is at 3/2, or (0, 1.5).