# Lines

• Oct 26th 2006, 07:39 PM
Dragon
Lines
Consider the line L containg the points (-3,8)and (6,-4) what is the length of the hypotenuse of the right triangle formed b the intersection of L and the x and Y axes
• Oct 26th 2006, 08:36 PM
Soroban
Hello, Dragon!

Did you make a sketch?

Quote:

Consider the line $L$ containg the points (-3,8)and (6,-4).
What is the length of the hypotenuse of the right triangle
formed by the intersection of $L$ and the $x$- and $y$-axes?

Code:

  *    (-3,8)      |         *          |               *    Q|                     *                     |    *      P   - - - - - - - - - + - - - - - * - - - - - - -                     |                *    (6,-4)                     |                      *                     |                            *

Game plan
The line $L$ contains the points (-3,8) and (6,-4).
. . We want the equation of that line.
Then we want its x-intercept $P$ and y-intercept $Q$.
. . Then we want the distance $\overline{PQ}$.

The slope of line $L$ is: . $m \:=\:\frac{-4 - 8}{6 -(-3)} \:=\:\frac{-12}{9}\:=\:-\frac{4}{3}$

The equation of the line through (6,-4) with slope $-\frac{4}{3}$ is:
. . $y - (-4)\:=\:-\frac{4}{3}(x - 6)\quad\Rightarrow\quad y\:=\;-\frac{4}{3}x + 4$

For the $x$-intercept, let $y = 0$ and solve for $x.$
. . $0 \:=\:-\frac{4}{3}x + 4\quad\Rightarrow\quad x = 3$ . . . $x$-intercept: $P(3,0)$

For the $y$-intercept, let $x = 0$ and solve for $y.$
. . $y\:=\:-\frac{4}{3}\cdot0 + 4\quad\Rightarrow\quad y = 4$ . . . $y$-intercept: $Q(0,4)$

The distance from $P(3,0)$ to $Q(0,4)$ is:

. . $PQ\:=\:\sqrt{(0-3)^2 + (4-0)^2} \;=\;\sqrt{9+16}\;=\;\sqrt{25}\;=\;\boxed{5}$