Question

7 two boats head out on straight paths. a sailboat starts at point s, a motorboat starts at point m and they both pass through the point p. approximately how far apart were they when they started their trips (the distance between points s and m)? round the answer you calculated to the nearest km. y (km) y = 1/2 x + 8 p (12,14) s m x (km) my calculations

Explanation:

Step1: Find the equation of the line perpendicular to $y = \frac{1}{2}x+8$ passing through $P(12,14)$.

The slope of the perpendicular line is $- 2$ (since the product of slopes of perpendicular lines is -1). Using the point - slope form $y - y_1=m(x - x_1)$ with $m=-2$ and $(x_1,y_1)=(12,14)$, we get $y - 14=-2(x - 12)$, which simplifies to $y=-2x + 38$.

Step2: Find the coordinates of $S$ and $M$.

For point $S$, it is on the $y$-axis, so $x = 0$. Substituting $x = 0$ into $y=-2x + 38$, we get $y=38$, so $S=(0,38)$.
For point $M$, it is on the $x$-axis, so $y = 0$. Substituting $y = 0$ into $y=-2x + 38$, we get $0=-2x + 38$, then $x = 19$, so $M=(19,0)$.

Step3: Calculate the distance between $S(0,38)$ and $M(19,0)$ using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Here, $x_1 = 0,y_1 = 38,x_2 = 19,y_2 = 0$. Then $d=\sqrt{(19 - 0)^2+(0 - 38)^2}=\sqrt{19^2+(-38)^2}=\sqrt{361 + 1444}=\sqrt{1805}\approx42.5$.

Answer:

43 km