Analytic Geometry Help!

**Nimmy** · January 7th 2007, 08:24 PM

Show that (1/3 [2x1 + x2], 1/3[2y1+y2]) is one of the points of trisection of the line segment joining (x1,y1) and (x2,y2). Then find the second point of trisection by finding the midpoint of the segment joining

(1/3[2x1 + x2], 1/3[2y1 + y2]) and (x2, y2)

Use exercise 28 to find the points of trisection of the line segment joining the given points.

a (1 , -2) (4, -1)
b. (-2, -3) (0,0)

**topsquark** · January 8th 2007, 12:36 AM

Originally Posted by Nimmy

Show that (1/3 [2x1 + x2], 1/3[2y1+y2]) is one of the points of trisection of the line segment joining (x1,y1) and (x2,y2). Then find the second point of trisection by finding the midpoint of the segment joining

(1/3[2x1 + x2], 1/3[2y1 + y2]) and (x2, y2)

Use exercise 28 to find the points of trisection of the line segment joining the given points.

a (1 , -2) (4, -1)
b. (-2, -3) (0,0)

To do the first part I would imagine you need to show that the distance between $(x_1, y_1)$ and $\left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right )$ is 1/3 the distance between $(x_1, y_1)$ and $(x_2, y_2)$ .

So the distance between $(x_1, y_1)$ and $\left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right )$ is:
$d = \sqrt{\left ( x_1 - \frac{1}{3}(2x_1 + x_2) \right ) ^2 + \left ( y_1 - \frac{1}{3}(2y_1 + y_2) \right ) ^2 }$

$d = \sqrt{\left ( \frac{1}{3}x_1 - \frac{1}{3}x_2 \right ) ^2 + \left ( \frac{1}{3}y_1 - \frac{1}{3}y_2 \right ) ^2 }$

$d = \sqrt{\frac{1}{9}(x_1 - x_2)^2 + \frac{1}{9}(y_1 - y_2)^2 }$

$d = \sqrt{\frac{1}{9}}\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 }$

$d = \frac{1}{9} \cdot \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 }$

Note that $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 }$ is the distance between $(x_1, y_1)$ and $(x_2, y_2)$ so we have proven what we have set out to show.

For the second part we need to find the midpoint of the two points $\left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right )$ and $(x_2, y_2)$ .

$M \left (\left ( \frac{1}{3}(2x_1 + x_2), \frac{1}{3}(2y_1 + y_2) \right ) , (x_2, y_2) \right )$ $= \left ( \frac{1}{2} \left [ \frac{1}{3}(2x_1 + x_2) + x_2 \right ] , \frac{1}{2} \left [ \frac{1}{3}(2y_1 + y_2) + y_2 \right ] \right )$

$= \left ( \frac{1}{2} \left [ \frac{2}{3}x_1 + \frac{1}{3}x_2 + x_2 \right ] , \frac{1}{2} \left [ \frac{2}{3}y_1 + \frac{1}{3}y_2 + y_2 \right ] \right )$

$= \left ( \frac{1}{2} \left [ \frac{2}{3}x_1 + \frac{4}{3}x_2 \right ] , \frac{1}{2} \left [ \frac{2}{3}y_1 + \frac{4}{3}y_2 \right ] \right )$

$= \left ( \frac{1}{3}x_1 + \frac{2}{3}x_2 , \frac{1}{3}y_1 + \frac{2}{3}y_2 \right )$

$= \left ( \frac{1}{3} \left [ x_1 + 2x_2 \right ] , \frac{1}{3} \left [ y_1 + 2y_2 \right ] \right )$

This is your second point of trisection.

I'll do a) as an example. We are given the points (1 , -2) and (4, -1). The first point of trisection is:
$\left ( \frac{1}{3} \left [ 2x_1 + x_2 \right ] , \frac{1}{3} \left [ 2y_1 + y_2 \right ] \right )$

$= \left ( \frac{1}{3} \left [ 2(1) + (4) \right ] , \frac{1}{3} \left [ 2(-2) + (-1) \right ] \right )$

$= \left ( \frac{1}{3} \left [ 6 \right ] , \frac{1}{3} \left [ -5 \right ] \right )$

$= \left ( 2 , -\frac{5}{3} \right )$

The second point of trisection is:
$\left ( \frac{1}{3} \left [ x_1 + 2x_2 \right ] , \frac{1}{3} \left [ y_1 + 2y_2 \right ] \right )$

$= \left ( \frac{1}{3} \left [ (1) + 2(4) \right ] , \frac{1}{3} \left [ (-2) + 2(-1) \right ] \right )$

$= \left ( \frac{1}{3} \left [ 9 \right ] , \frac{1}{3} \left [ -4 \right ] \right )$

$= \left ( 3 , -\frac{4}{3} \right )$

-Dan

**Soroban** · January 8th 2007, 05:13 AM

Hello, Nimmy!

Show that $\left(\frac{1}{3}[2x_1 + x_2],\:\frac{1}{3}[2y_1+y_2]\right)$
is one of the points of trisection of the line segment joining $P(x_1,y_1)$ and $Q(x_2,y_2).$

Make a sketch . . .

Code:

                                                * Q(x2,y2)
                                            *   | :
                                        *       | :
                                    *           | :
                                *               | y2 - y1
                        T   *                   | :
                        *                       | :
                    *   |                       | :
                *       |                       | :
  (x1,y1) P *-----------*-----------------------* R(x2,y1)
                        U
          : - - - - - - - x2 - x1 - - - - - - - :

We have: . $\Delta PTU \sim \Delta PQR$

Since $PT = \frac{1}{3}PQ$ , then: . $\begin{array}{cc}PU \:=\:\frac{1}{3}PR\:=\:\frac{1}{3}(x_2-x_1) \\ TU \:= \:\frac{1}{3}QR\:=\:\frac{1}{3}(y_2-y_1)\end{array}$

The coordinates of $T$ are: . $\begin{array}{cc}x \:= \: x_1 + \frac{1}{3}(x_2-x_1) \: =\: \frac{1}{3}(2x_1 + x_2) \\<br /> y \:= \: y_1 + \frac{1}{3}(y_2-y_1) \: =\: \frac{1}{3}(2y_1+y_2)<br /> \end{array}$

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