हल कीजिए :
`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=a+b bx+ay=b^2

Solve: ax + by = a - b bx - ay = a + b

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

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 lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 be concurrent, then:

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