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Dear sir Please help me solve these two problems- Consider a family of straight lines (x+y)+lamda(2x-y+1)=0. Find the equation of the straight line belonging to this family that is farthest from (1,-3). If the lines joining the intersection of curves ax^2+2hxy+b(y)^2+2gx=0 and a1(x^2)+2h1xy+b1(y^2)+2g1x=0 are mutually perpendicular then prove that g(a1+b1)=g1(a+b)

Dear sir 
Please help me solve these two problems-
  1. Consider a family of straight lines (x+y)+lamda(2x-y+1)=0. Find the equation of the straight line belonging to this family that is farthest from (1,-3).
  2. If the lines joining the intersection of curves ax^2+2hxy+b(y)^2+2gx=0 and a1(x^2)+2h1xy+b1(y^2)+2g1x=0 are mutually perpendicular then prove that g(a1+b1)=g1(a+b)

Grade:12

1 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
Question 1 :To find the equation of straight line farthest from (1,-3), the perpendicular distance should be the maximum..Say line is x+y + lambda(2x-y+1) = 0Write the formula for perpendicular distance.. and maximise it to find lambda..
\operatorname{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. Then put this lambda into the equation.. You will get the equation of line..
Thanks
Bharat BajajIIT Delhi
askiitians faculty

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