[2(ax-by)+(x+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

{:(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):}

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

Factorise the following expressions : (i) ax-ay+bx-by " " (ii) x^(2)-x-ax+a " " (iii) x^(4)+x^(3)+x^(2)+x (iv) 16(a+b)^(2)-4a-4b " " (v) x^(2)+(1)/(x^(2))+2-3x-(3)/(x) " " (vi) x^(2)-((a)/(b)+(b)/(a))x+1 (vii) x^(2)+(a-(1)/(a))x-1 " " (viii)ab(x^(2)+y^(2)+xy(a^(2)+b^(2)) " "(ix) (ax+by)^(2)+(bx-ay)^(2)

The line parallel to the x-axis and passing through the point of intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) ne (0, 0) is

If x^(2) + ax + b = 0, x^(2) + bx + a = 0 ( a != 0 ) have a common root, then a + b =

If x^(2) + ax + b = 0, x^(2) + bx + a = 0 ( a != 0 ) have a common root, then a + b =