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Consider the lines represented by equation (x^(2) + xy -x) xx (x-y) =0 forming a triangle. Then match the following lists: |
Answer»  The GIVEN lines are x (x+y-1) (x-y)=0. So, lines x =0, x+y-1=0, and x-y=0 form triangle OAB as shown in the diagram. The triangle is right-angled at pont B. Hence, the orthocenter is (1/2, 1/2). Also, the CIRCUMCENTER is the midpoint of OA which is (0, 1/2). The centroid is `((0+(1//2) +0)/(3), (0+(1//2)+1)/(3)) " or "((1)/(6), (1)/(2))` `"Also, " OA = 1, OB = OC= 1//2sqrt(2).` Hence, the incenter is `((0(1//sqrt(2))(1//2)1+0(1//sqrt(2)))/((1//sqrt(2))+1+(1//sqrt(2))),(0(1//sqrt(2))+(1//2)(1)+1(1//sqrt(2)))/((1//sqrt(2))+1+(1//sqrt(2)))) -= ((1)/(2+2sqrt(2)), (1)/(2))` |
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