bisect the chord theorem ppt

Bisect the chord theorem ppt, Perpendicular dropped from the center of circle on a chord bisects the chord.

Proof:

Suppose that x² + y² = a² be a circle, in which AB is a chord with end point A(x ₁, y ₁) B(x ₂, y ₂) on the circle   and OM is perpendicular from the center to the chord. We need to show that OM bisects the chord AB.

SLOPE OF AB = y ₂ – y ₁ / x ₂ – x ₁

Slope of perpendicular to AB = −(x ₂ – x ₁) /y ₂ – y ₁

= x ₁ − x ₂ / y ₂ − y ₁    =  m 

So the equation of OM with the slope m and point on origin O(0, 0)on it, is given by

y  – 0 = x ₁ − x ₂ / y ₂ − y ₁  (x-0)  

y   =( x ₁ − x ₂ / y ₂ − y ₁) x……… (a)

(a) is the equation of the perpendicular OM from center to the chord. We will show that it bisect the chord That is the intersection of OM and AB is the mid-point of AB 

y –  y ₁  = y ₁ – y ₂ / x ₁ – x ₂ (x – x ₁)  ….(b)

The foot of the perpendicular OM is the mid-point of intersection  (a) and (b) putting the value of y from (a) into (b) we have 

  –    

or

After algebraic simplification, we get the result

x ₁ + x ₂ (a² – x ₁  x ₂ – y ₁ y ₂)

The point (x ₁, y ₁), (x ₂, y ₂)lies on the circle.

Or

x = x ₁ + x ₂  /  2

putting x = x ₁ + x ₂  /  2 into equation (a) we get

  y   =( x ₁ − x ₂ / y ₂ − y ₁)  ×   x ₁ + x ₂  /  2

 y =   x² ₁ − x² ₂ /  2(y ₂ − y ₁)

⇒ 

  is the point of intersection OM and AB, which is the

mid-point of AB.(which is bisected the chord theorem ppt)