[" d "2(ax-ly)+a+4b=0],[2(bx+ay)-(b-4a)=0]

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

Solve the following pair of linear equations 2(ax-by)+(a+4b)=02(bx+ay)+(b-40=0

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

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

If a!=b and differences between the roots of the equations x^2+ax+b=0 and x^2+bx+a=0 is the same then (A) a+b+4=0 (B) a+b-4=0 (C) a-b+4=0 (D) a-b-4=0

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

ax+by=a+b bx+ay=b^2

If a!=b, then the system of equations ax+by+bz=0,bx+ay+bz=0,bx+by+az=0 will have a non-trivial solution if (1)a+b=0 (2) a+2b=0 (3) 2a+b=0(4)a+4b=0

If ax^2+bx+c=0 has equal roots,then,c= a) b/(2a) b) -b/(2a) c) b^2/(4a) d) -b^2/(4a)

If the quadratic equations ax^(2)+cx-b=0 and ax^(2)-2bx+(c)/(2)=0,(b+(c)/(2)!=0) have a common root,then the value of a-4b+2c is