[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

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}

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

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, 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

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

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

If a ne b , then the system of equations ax+by+bz=0, bx + ay + bz = 0, bx + by + ax = 0 will have a non– trivial solution if