`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

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

Let a,b,c be real numbers with a^2 + b^2 + c^2 =1 . Show that the equation |[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+c]|=0 represents a straight line.

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

Factorise: ax - bx+ ay- by

If the point P(x, y) is equidistant from the points A(a + b, b-a) and B(a-b, a + b). Prove that bx = ay.

Prove that: |[a, b, ax+by],[ b, c, bx+cy], [ax+by, bx+cy,0]|=(b^2-a c)(a x^2+2b x y+c y^2)

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 the lines ax+by+c=0, bx+cy+a=0 and cx+ay+b=0 (a, b,c being distinct) are concurrent, then (A) a+b+c=0 (B) a+b+c=1 (C) ab+bc+ca=1 (D) ab+bc+ca=0