# Thread: Right Bisector of a Chord?

1. ## Right Bisector of a Chord?

Right Bisector of a Chord?

The equation of a circle with centre O (0,0) is x2 +y2 = 169
the points A(0,13) and B(-12,-5) are endpoints of the chord AB. DE right bisects chord AB at F. Verify that the centre of the circle lies on the right bisector of chord AB

1) Find midpoint of AB -- igot (-6,0) and (0,4) for midpoint
2) Find the slopes of AB-- igot 3/2 for slope of AB
**3) Find Equation of the Line DE --- idont noe how to do this
a) Use the negative reciprocal slope of AB for Slope of DE
b) Use point slope form to find equation of DE

igot help for numebr 3

and igot y= (-2/3)x +b

idont get how they got (-2/3)
can yu explain tht process

Right Bisector of a Chord?

The equation of a circle with centre O (0,0) is x2 +y2 = 169
the points A(0,13) and B(-12,-5) are endpoints of the chord AB. DE right bisects chord AB at F. Verify that the centre of the circle lies on the right bisector of chord AB

1) Find midpoint of AB -- igot (-6,0) and (0,4) for midpoint How can a line segment have two midpoints?! Remember the formula:
$\displaystyle Midpoint(m)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2} {2}\right)$
2) Find the slopes of AB-- igot 3/2 for slope of AB Very good.
**3) Find Equation of the Line DE --- idont noe how to do this Use the midpoint that you obtain in 1. to find the equation using the point slope form of a line.
a) Use the negative reciprocal slope of AB for Slope of DE. Because the slope of a perpindicular line will be the negative reciprocal of the line it is perpindicular to.
b) Use point slope form to find equation of DE. Done.

igot help for numebr 3

and igot y= (-2/3)x +b

idont get how they got (-2/3) Draw 2 perpindicular lines at the origin and analyze their slopes. You will be enlightened.
can yu explain tht process
..

Right Bisector of a Chord?

The equation of a circle with centre O (0,0) is x2 +y2 = 169
the points A(0,13) and B(-12,-5) are endpoints of the chord AB. DE right bisects chord AB at F. Verify that the centre of the circle lies on the right bisector of chord AB

1) Find midpoint of AB -- igot (-6,0) and (0,4) for midpoint
You mean that the midpoint is (-6, 4).

2) Find the slopes of AB-- igot 3/2 for slope of AB
**3) Find Equation of the Line DE --- idont noe how to do this
The equation of a line through point (a, b) with slope m is y= m(x- a)+ b.
Here, this is y= (3/2)(x- (-6))+ 4 or y= (3/2)x+ 9+ 4= (3/2)x+ 13.
You don't actually need the equation of this line to answer the question.

a) Use the negative reciprocal slope of AB for Slope of DE
b) Use point slope form to find equation of DE

igot help for numebr 3

and igot y= (-2/3)x +b

idont get how they got (-2/3)
can yu explain tht process
If two lihes, given by $\displaystyle y= m_1 x+ b_1$ and $\displaystyle y= m_2x+ b_2$ are perpendicular then $\displaystyle m_1\times m_2= -1$. Here $\displaystyle m_1= \frac{3}{2}$ so you have $\displaystyle \frac{3}{2} m_2= -1$. Multiplying both sides by 2, $\displaystyle 3m_2= -2$. Dividing both sides by 3, $\displaystyle m_2= -\frac{2}{3}$.

Now, you want that line to go through the midpoint of AB which was (-6, 4) so you must have 4= (-2/3)(-6)+ b= 4+ b. Adding b to both sides, b= 0.
The equation of the perpendicular to the chord through its midpoint is y= (-2/3).
You could also have used the formula I gave before: the equation of the line through (-6, 4) with slope -2/3 is y= -2/3(x-(-6))+ 4= -2/3(x+ 6)+ 4= -(2/3)x- 4+ 4= (-2/3) x again. Now do you see that the origin, (0, 0), satisfies that equation?

,
,

,

,

,

,

,

,

,

,

,

,

,

,

# centre on right bisectora of chord

Click on a term to search for related topics.