centroid of a triangle with equations, 13x-8y+118=0, 29x+5y-148=0, 16x+13y+31=0  

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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Let us say the three vertices's are A,B and C. So the equations of the three lines denote the legs of the triangle. 

Let us say 13x-8y+118=0 represent AB, 29x+5y-148=0 represent BC and 16x+13y+31=0 AC respectively.

Once we solve for AB and AC you will get the coordinates of vertex A. Similarly we can find B and C also.

`13x-8y+118=0 -----(AB)`

`16x+13y+31=0 ----(AC)`

`29x+5y-148=0----(BC)`

 

By solving (AB) and (AC) you will get A.

By solving (AB) and (BC) you will get B.

By solving (BC) and (AC) you will get C.

 

Once you solve them you will get;

`A = (-6,5)`

`B =(2,18)`

`C =(7,-11)`

 

Centroid of a triangle is given by the point of intersections of medians.

C = [(1/3)(x1+x2+x3),(1/3)(y1+y2+y3))]

C =[(1/3)(-6+2+7),(1/3)(5+18-11)]

C = [(1/3)(3),(1/3)(12)]

`C = (1,4)`

 

So the centroid is (1,4)

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