Triangles: Calculating Orthocentre, Circumcentre, Centroid? I need help ASAP?

(i) Centroid:

Centroid of the triangle formed by (a1,b1), (a2,b2), (a3,b3) is ( (a1+a2+a3)/3, (b1+b2+b3)/3 ).

Hence the centroid is: ( (0-15+24)/3, (24+0+0)/3 ) = (3,8).

(ii) Orthocenter

Orthocenter is the intersection of altitudes.

m(BC) = 0/39 = 0 which is parallel to the x-axis. So the altitude through A(0,24) is parallel to the y-axis, the equation is: y=24.

m(AC) = -24/24 = -1. So the altitude through B(-15,0) has a slope of 1, the equation is: y = x+15

The orthocenter is the intersection of the lines y=24 and y=x+15. Substitute y=24 to y=x+15, we get x=9. So the orthocenter is (9,24).

(iii) Circumcenter

Label the circumcenter, centroid, and orthocenter by O, G, H, respectively.

O, G, H lies on Euler line in that order, such that OG = 2GH. So O = (G+2H)/3:

( (3+2*9)/3, (8+2*24)/3 ) = (7, 56/3).

do you elect geometric or algebraic ideas? eg: if given which you have co-ordinates of three factors, A, B, C, forming vertices of a triangle, the centroid is at (A+B+C)/3; yet do you have the co-ordinates? if no longer, map them as A=(0,0), B=(x[b],0), C=(x[c],y[c]). then centroid = (A+B+C)/3 = ( (x[b]+x[c])/3, y[c]/3 ). if the circumcentre is at (u,v), then factors A,B,C are all on the circle defined by skill of (x-u)² + (y-v)² = R² that's undemanding to confirm as quickly as you have got here across it. (x[a]-u)² + (y[a]-v)² = (x[b]-u)² + (y[b]-v)² = (x[c]-u)² + (y[c]-v)² = R² yet whilst A=(0,0), as above, then (0-u)² + (0-v)² = (x[b]-u)² + (0-v)² = (x[c]-u)² + (y[c]-v)² = R² the orthocentre is slightly extra officious. yet examine the wikipedia internet site for some information approximately it, additionally. this is got here across by skill of extending perpendicular lines from the perimeters, and finding one that passes in the path of the different vertex. algebraically, those extensions have the optimum pass-product with the corresponding edges; you could hire a co-ordinate shift to discover that line. eg: for line BC, its line is defined by skill of B + ok(C-B) = ( ok(x[c]-x[b])+x[b], ky[c] ); and the set of all perpendicular lines is then defined by skill of ( hy[c] +e, h(x[c]-x[b])+x[b] +f) for some constants e,f. if (e,f) is on the line BC, then the only that intersects A is unique. x[a] = hy[c] +e y[a] = h(x[c]-x[b])+x[b] +f provided that (e,f) is online BC, x[a] = hy[c] +ok(x[c]-x[b])+x[b]; and y[a] = h(x[c]-x[b])+x[b] + ky[c] yet whilst A=(0,0), as above, then the line will fulfill those equations... 0 = kx[c] + hy[c] + x[b](a million-ok); and 0 = hx[c] + ky[c] + x[b](a million-h). the different 2 lines are got here across in addition; the orthocentre then meets at their intersection. yet i'm actually sympathetic to what you're announcing approximately calculation errors.