Lines

**Dragon** · October 26th 2006, 06:39 PM

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

**Soroban** · October 26th 2006, 07:36 PM

Hello, Dragon!

Did you make a sketch?

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:

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

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