Originally Posted by

**ahmads121** **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