ax+by=p and bx+ay=q

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

The base BC of a triangle ABC is bisected at the point (a, b) and equation to the sides AB and AC are respectively ax+by=1 and bx+ay=1 Equation of the median through A is:

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

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

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 length of the common chord of the circles x^2+y^2+ax+by+c=0 and x^2+y^2+bx+ay+c=0 is

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

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 .

Let a,b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay+c=0 and 5bx+2by +d=0 lies in the fourth quadrant and is equidistant from the two axes, then

if a gt b gt c and the system of equations ax + by + cz = 0, bx + cy + az 0 and cx + ay + bz = 0 has a non-trivial solution, then the quadratic equation ax^(2) + bx + c =0 has