Most Important Selected Qs for JEE AdvancedMathematicsStraight Lines

On the straight line y=x+2 , a point (a, b) is such that the sum of the square of distances from the straight lines 3 x-4 y+8=0 and 3 x-y-1=0 is least, then find value of 11(a+b) .

Correct answer

Step-by-step solution

Point be ( x , y ) but it lies on y = x +2 So, ( x , x +2)F(x)= [ 3 x-4(x+2)+8 3^2+4^2 ]^2+ [ 3 x-(x+2)-1 3^2+1^2 ]^2= 2 x^2+5 [4 x^2-12 x+9 ] 50 = 22 [ (x- 30 22 )^2- 900 484 ]+45 50 F(x) is minimum at x= 15 11 . So point is ( 15 11 , 37 11 )=(a, b)11(a+b)=52

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