Find the circumcenter of a triangle with vertices - id3554562, javonteoshamccaliste

Ответ:

westes0376

08.01.2022

Mathematics

Circumcenter is (-5,13) .

Step-by-step explanation:

Given:

The three vertices of triangle OVW are O(0,0),V(0,26),W(-10,0) .

Circumcenter is a point inside the triangle which is equidistant from each of the vertices of the triangle.

Let C(x,y) be the circumcenter.

Distance between two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given as:

$D=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

So, as per the definition of circumcenter, distance of point O(0,0) from point C(x,y) is equal to distance of point V(0,26) from point C(x,y) .

So, OC = VC

or $(OC)^{2}=(VC)^{2}$

$(x-0)^{2}+(y-0)^{2}=(x-0)^{2}+(y-26)^{2}\\x^{2}+y^{2}=x^{2}+y^2+26^{2}-(2\times 26y)\\x^{2}+y^{2}-x^{2}-y^2=676-52y\\0=676-52y\\52y=676\\y=\frac{676}{52}=13$ .

Similarly, distance of point O(0,0) from point C(x,y) is equal to distance of point W(-10,0) from point C(x,y) .

OC=WC or

$(OC)^{2}=(WC)^{2}$

$(x-0)^{2}+(y-0)^{2}=(x-(-10))^{2}+(y-0)^{2}\\x^{2}+y^{2}=(x+10)^{2}+y^2\\ x^{2}+y^{2}=x^{2}+100+20x+y^2\\x^{2}+y^{2}-x^{2}-y^2=100+20x\\0=100+20x\\ 20x=-100\\x=-\frac{100}{20}=-5$ .

Therefore, the circumcenter of the triangle with the given vertices is (-5,13) .

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Find the circumcenter of a triangle with vertices o(0, 0), v(0, 26), and w(−10,0).

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