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(D) Let the vertices of the triangle be $A(4,6), B(5,1),$ and $C(4,0)$. Let the circumcentre be $P(x,y)$.Since the circumcentre is equidistant from al

Vedclass · Accepted Answer

$(4,3)$

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(D) Let the vertices of the triangle be $A(4,6), B(5,1),$ and $C(4,0)$. Let the circumcentre be $P(x,y)$.Since the circumcentre is equidistant from all vertices,$PA^2 = PB^2 = PC^2$.First,equate $PA^2 = PC^2$:$(x-4)^2 + (y-6)^2 = (x-4)^2 + (y-0)^2$$(y-6)^2 = y^2$$y^2 - 12y + 36 = y^2$$12y = 36 \implies y = 3$.Now,equate $PA^2 = PB^2$ with $y=3$:$(x-4)^2 + (3-6)^2 = (x-5)^2 + (3-1)^2$$(x-4)^2 + (-3)^2 = (x-5)^2 + (2)^2$$x^2 - 8x + 16 + 9 = x^2 - 10x + 25 + 4$$-8x + 25 = -10x + 29$$2x = 4 \implies x = 2$.Thus,the circumcentre is $(2,3)$.

Find the circumcentre of the triangle with vertices $(4,6), (5,1),$ and $(4,0)$.

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