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Consider a pair of perpendicular straight lines ax^(2)+3xy-2y^(2)-5x+5y+c=0.Distance between the orthocenter and the circumcenter of triangle ABC is |
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Answer» 4 `:.a-2=0` or `a=2` Also , `abc+2fgh-af^(2)-bg^(2)-ch^(2)=0` `:. C=-3` HENCE , the given pair of lines is `2x^(2)+3xy-2y^(2)-5x+5y-3=0` Factorizing , we get lines `x+2y-3=0and 2x-y+1=0`  The point of INTERSECTION of the lines is C `(1//5,7//5)`. The points of intersection of the lines with the x- axis are A(3,0) and B `(-1//2,0)`. The orthocenter of triangle isC `(1//5,7//5)` and the CIRCUMCENTER is the midpoint of AB which isM `(5//4,0)`.Therefore, CM`sqrt(((5)/(4)-(1)/(5))^(2)+(49)/(25))=(7)/(4)` |
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