`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

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

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 the area bounded by the curve |x| + |y| = c (c > 0) is 2c^(2) , then the area bounded by |ax + by| + |bx - ay| = c where b^(2) - a^(2) = 1 is

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

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

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:

For a gt b gt c gt 0 , if the distance between (1,1) and the point of intersection of the line ax+by-c=0 and bx+ay+c=0 is less than 2sqrt2 then, (A) a+b-cgt0 (B) a-b+clt0 (C) a-b+cgt0 (D) a+b-clt0

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

If the system of linear equations ax +by+ cz =0 cx + ay + bz =0 bx + cy +az =0 where a,b,c, in R are non-zero and distinct, has a non-zero solution, then: