InterviewSolution
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InterviewSolution
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If two members of family (2+lambda)x+(1+2lambda)y-3(1+lambda) = 0 and line x+y=0 make an equilateral triangle, the the incentre of triangle so formed is |
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Answer» `((1)/(3), (1)/(3))` `(2x+y-3)+lambda(x+2y-3) =0, lambda in R.` The family of lines is concurrent at point of intersection of 2x+y-3 = 0 and x+2y-3=0 which is A(1,1). Two members `L_(1)" and "L_(2)` of this family and line x+y=0 form an equilateral TRIANGLE. Let foot of perpendicular from A(1, 1) on the line x+y=0 be M(h,K). `(h-1)/(1) = (k-1)/(1) = (-(1+1))/(2)` `therefore M-=(h,k)-=(0,0)` SINCE triangle is equilateral, AM is median, altitude and angle bisector. `AM= sqrt((1-0)^(2)+(1-0)^(2)) = sqrt(2)` Incentre coincides with centroid G such that AG `=(2)/(3)sqrt(2)`. `"Slope of " AM is 1 = "TAN" 45^(@)`. `therefore G-=(1-(2sqrt(2))/(3)"cos" 45^(@), 1-(2sqrt(2))/(3)"sin" 45^(@))` `-=((1)/(3),(1)/(3))` |
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