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Consider the triangle whose vetices are (0,0) , (5,12) and (16,12). |
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Answer» (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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