Equations of the form `ax+by =c` and `bx+ay=d` where `a!=b`

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

The solution set of the equation tan ax=tan bx where (a!=b) constitutes

Show that the lines given by the equations ax+by+c=0 and bx-ay+d=0 (where a,b,c,d in R) are perpendicular by finding a vector inthe direction of each line and showing that these vectors are orthogonal. (Hint: Watch out for the cases in which a or b equals zero.)

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

Form the differential equation for the curve: y=Ax^2+Bx , where a and b are arbitrary constant.

If the point of intersection of the line 2ax + 4ay+c =0 and 7bx +3by-d=0 lies in the 4^(th) quadrant and is equidistant from the two axes, where a,b,c and d are non-zero numbers, then ad:bc equals to

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

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