[(bx)/(a)-(ay)/(b)+a+b=0],[bx-ay+2ab=0]

Solve the following system of equations b(x)/(a)-a(y)/(b)+a+b=0,bx-ay+2ab=0

(bx)/(a)-(ay)/(b)+a+b=0 and bx-ay+2ab

The equation of a line perpendicular to line ax+by+c=0 and passing through (a,b) is equal to ................ A) bx-ay=0 B) bx+ay-2ab=0 C) bx+ay=0 D) bx-ay+ 2ab=0

(bx ) /(a) + ( ay ) /( b) = a ^(2) + b ^(2) , x + y = 2ab

ax+by=a-b;bx-ay=a+b

ax+by=1 bx+ay=2

A circle with center (a,b) passes through the origin.The equation of the tangent to the circle at the origin is ax-by=0 (b) ax+by=0bx-ay=0 (d) bx+ay=0

(ax+by)^(2)-(ay+bx)^(2)

2(ax-by)+(a+4b)=0,2(bx+ay)+(b-4a)=0

The straight line x-y-2=0 cuts the axis of x at A. It is rotated about A in such a manner that it is perpendicular to ax+by+c=0. Its equation is: (a) bx-ay-2b=0 (b) ax-by-2a=0 (c) bx+ay-2b=0 (d) ax+by+2a=0