# Thread: Geometry: True or False

1. ## Geometry: True or False

Need help with this, if it is true or false and why.

"If a circle with its centre at the origin has a chord with slope m, the equation of the right bisector of the chord is "y=mx".
Is this statement true or false, and can you give me detailed reasons for your answer.

I have been re reading my notes and going thru my books and I can not figure this out at all.
Joanne

2. Hello, Joanne!

Did you make a sketch?

If a circle with its centre at the origin has a chord with slope $m$,
the equation of the right bisector of the chord is: $y \,=\,mx$

Is this statement true or false? . . . . false
Code:
             A  |
o * *
*     *     *
*       | *     *
*        |   *    *
|   * *
*         | *     * *
- - * - - - - * - - - - o - -
*         |         * B
|
*        |        *
*       |       *
*     |     *
* * *
|
Chord $AB$ has slope $m.$

The perpendicular bisector of the chord does pass through the origin,
. . but its slope is $-\frac{1}{m}$

Its equation is: . $y \:=\:-\frac{1}{m}\,y$

3. A chord that is a perpendicular bisector of another chord would have the negative reciprocal slope of the other chord.

If a chord has slope = m, its perpendicular bisector will have the $slope = -\frac{1}{m}$

4. Why does the circle matter? I guess it helps to talk about the "chord".

Slope of chord is m.
Supposed slope of right bisector is m, from y = mx
Supposed slope of right bisector, perpendicular to chord, is -1/m.

Is m = -1/m?

Exploration: Look up the "Normal Form" of a line. Interesting stuff.

No, wait! The circle keeps the chord away from the Origin. Why does that matter, or does it? More exploration.

Need help with this, if it is true or false and why.

"If a circle with its centre at the origin has a chord with slope m, the equation of the right bisector of the chord is "y=mx".
Is this statement true or false, and can you give me detailed reasons for your answer.

I have been re reading my notes and going thru my books and I can not figure this out at all.