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Question
Prove by vector method that if a line is drawn from the centre of a circle to the midpoint of a chord, then the line is perpendicular to the chord
Solution
A circle with centre at O.
AB is chord of the circle and OP bisects AB
(ie) AP = PB
To prove `bar"OP"` ⊥ `bar"AB"` O is the position vector
∴ `bar"OA" = bar"OB"` = Radius
Position vector of P
`bar"OP" = (bar"OA" + bar"OB")/2`
`bar"OP"*bar"AB" = bar"OP" * (bar"OB" - bar"OA")`
= `((bar"OB" + bar"OA")/2)* (bar"OB" - bar"OA")`
= `1/2[|bar"OB"|^2 - |bar"OA"|^2]`
= 0 ........`(∵ bar"OA" = bar"OB" = "Radius")`
∴ `bar"OP"` ⊥ `bar"AB"`
Hence proved
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