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If (b2-b1)(b3-b1)+(a2-a1)(a3-a1)=0, then...

If `(b_2-b_1)(b_3-b_1)+(a_2-a_1)(a_3-a_1)=0`, then prove that the circumcenter of the triangle having vertices `(a_1,b_1),(a_2,b_2)` and `(a_3,b_3)` is `((a_2+a_3)/(2),(b_2+b_3)/(2))`.

Text Solution

Verified by Experts

Given `(b_2-b_1)(b_3-b_1)+(a_2-a_1)(a_3-a_1)=0`

or `(b_2-b_1)(b_3-b_1)=-(a_2-a_1)(a_3-a_1)`
or `((b_2-b_1)/(a_2-a_1))((b_3-b_1)/(a_3-a_1))=-1`
or `("slope of AB")xx("slope of AC") =-1`
Therefore, the triangle is right -angled at point `(a_1,b_2)`. Hence, the circumcenter is the midpoint of `(a_2b_2)` and `(a_3,b_3)` which is `((a_2+a_3)/(2),(b_2+b_3)/(2))`
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