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Consider the triangle whose vertices are...

Consider the triangle whose vertices are (0,6) , (8,12) and (8,0). Find the centroid of the triangle.

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The correct Answer is:
a to s;b to p;c to q;d to s
(a). Centroid is `((0+5+16)/(3),(0+12+12)/(3))-=(7,8)`
(b). Let the coordinates of circumcenter be `O(x,y)` therefore,
`OA=OB=OC`
` therefore x^2+y^2=(x-5)^2+(y-12)^2=(x-16)^2+(y-12)^2`
` therefore x^2+y^2=(x-5)^2+(y-12)^2`
or `10x+24y=169`
and `(x-5)^2+(y-12)^2=(x-16)^2+(y-12)^2`
or `2x=21`
Solving, we get
`x=(21)/(2),y=(8)/(3)`
(c). `I-=((0xx11+5xx20+16xx13)/(13+20+11),(0xx11+12xx20+13xx12)/(13+20+11))-=(7,9)`
(d). `I_2-=((-5xx20+13xx16+11xx0)/(-20+13+11),(-12xx20+0xx11+13xx12)/(-20+13+11))-=(27,-21)`.
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