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

ax + by = c bx+ ay =1 +c

Factorise : (v) (ax + by )^(2) + (bx - ay)^(2)

Find the value of x and y: ax +by =a-b bx-ay =a+b

A, B C and D are the points of intersection with the coordinate axes of the lines ax+by=ab and bx+ay=ab, then

If a !=b , then the system of equation ax + by + bz = 0 bx + ay + bz = 0 and bx + by + az = 0 will have a non-trivial solution, if

The equation of the circle and its chord are respectively x^2 + y^2 = a^2 and x + y = a . The equation of circle with this chord as diameter is : (A) x^2 + y^2 + ax + ay + a^2 = 0 (B) x^2 + y^2 + 2ax + 2ay = 0 (C) x^2 + y^2 - ax - ay = 0 (D) ax^2 + ay^2 + x + y = 0

If a gt b gt c and the system of equtions ax +by +cz =0 , bx +cy+az=0 , cx+ay+bz=0 has a non-trivial solution then both the roots of the quadratic equation at^(2)+bt+c are

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

Factorise : xy - ay - ax + a^2 + bx - ab

If the equation ax^2 +2hxy + by^2 = 0 and bx^2 - 2hxy + ay^2 =0 represent the same curve, then show that a+b=0 .