[" triangle formed by these graph lines with "Y" -axis."],[" Solve: "ax+by=c" and "bx+ay=1+c]

Solve: ax+by=c,quad bx+ay=1+c

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

Solve: ax - by = 1 bx + ay = 0

Solve: ax+by=a-b,quad bx-ay=a+b

Solve: ax + by = a - b bx - ay = a + b

The algebraic sum of distances of the line ax + by + 2 = 0 from (1,2), (2,1) and (3,5) is zero and the lines bx - ay + 4 = 0 and 3x + 4y + 5=0 cut the coordinate axes at concyclic points. Then (a) a+b=-2/7 (b) area of triangle formed by the line ax+by+2=0 with coordinate axes is 14/5 (c) line ax+by+3=0 always passes through the point (-1,1) (d) max {a,b}=5/7

The algebraic sum of distances of the line ax + by + 2 = 0 from (1,2), (2,1) and (3,5) is zero and the lines bx - ay + 4 = 0 and 3x + 4y + 5=0 cut the coordinate axes at concyclic points. Then (a) a+b=-2/7 (b) area of triangle formed by the line ax+by+2=0 with coordinate axes is 14/5 (c) line ax+by+3=0 always passes through the point (-1,1) (d) max {a,b}=5/7

Solve by any method: ax+by=c, bx+ay=1+c