Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.0 AMC 10/12 Find the answer

Given the vertices of triangle $ABC$ are $A(4,1)$ , $B(1,5)$ , and $C(-3,2)$ ;

$(1)$ Find the general form of the equation of line $AB$ ;

$(2)$ Prove that $\triangle ABC$ is a right-angled triangle;

$(3)$ Find the equation of the circumcircle of $\triangle ABC$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

$(1)$ The equation of line $AB$ is $\frac{y-1}{5-1}=\frac{x-4}{1-4}$ , simplifying gives $4x+3y-19=0$ ;

$(2)$ The slope of $AB$ , $k_{AB}=\frac{5-1}{1-4}=-\frac{4}{3}$ , and the slope of $BC$ , $k_{BC}=\frac{5-2}{1-(-3)}=\frac{3}{4}$ , thus $k_{AB}k_{BC}=-1$ , which means $AB \perpendicular BC$ ; therefore, $\triangle ABC$ is a right-angled triangle.

$(3)$ Since $\triangle ABC$ is a right-angled triangle, the center of the circumcircle of $\triangle ABC$ is the midpoint of $AC$ , $M(\frac{1}{2},\frac{3}{2})$ , and the radius is $r=\frac{|AC|}{2}=\frac{\sqrt{(4+3)^2+(1-2)^2}}{2}=\frac{5\sqrt{2}}{2}$ ; therefore, the equation of the circumcircle of $\triangle ABC$ is $\boxed{\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{25}{2}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.