Hello, Dragon!
Did you make a sketch?
Consider the line $\displaystyle L$ containg the points (-3,8)and (6,-4).
What is the length of the hypotenuse of the right triangle
formed by the intersection of $\displaystyle L$ and the $\displaystyle x$- and $\displaystyle y$-axes? Code:
* (-3,8) |
* |
* Q|
*
| * P
- - - - - - - - - + - - - - - * - - - - - - -
| * (6,-4)
| *
| *
Game plan
The line $\displaystyle L$ contains the points (-3,8) and (6,-4).
. . We want the equation of that line.
Then we want its x-intercept $\displaystyle P$ and y-intercept $\displaystyle Q$.
. . Then we want the distance $\displaystyle \overline{PQ}$.
The slope of line $\displaystyle L$ is: .$\displaystyle m \:=\:\frac{-4 - 8}{6 -(-3)} \:=\:\frac{-12}{9}\:=\:-\frac{4}{3}$
The equation of the line through (6,-4) with slope $\displaystyle -\frac{4}{3}$ is:
. . $\displaystyle y - (-4)\:=\:-\frac{4}{3}(x - 6)\quad\Rightarrow\quad y\:=\;-\frac{4}{3}x + 4$
For the $\displaystyle x$-intercept, let $\displaystyle y = 0$ and solve for $\displaystyle x.$
. . $\displaystyle 0 \:=\:-\frac{4}{3}x + 4\quad\Rightarrow\quad x = 3$ . . . $\displaystyle x$-intercept: $\displaystyle P(3,0)$
For the $\displaystyle y$-intercept, let $\displaystyle x = 0$ and solve for $\displaystyle y.$
. . $\displaystyle y\:=\:-\frac{4}{3}\cdot0 + 4\quad\Rightarrow\quad y = 4$ . . . $\displaystyle y$-intercept: $\displaystyle Q(0,4)$
The distance from $\displaystyle P(3,0)$ to $\displaystyle Q(0,4)$ is:
. . $\displaystyle PQ\:=\:\sqrt{(0-3)^2 + (4-0)^2} \;=\;\sqrt{9+16}\;=\;\sqrt{25}\;=\;\boxed{5}$