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)