The graph of function f contains the point `P(1, 2) and Q(s, r)`. The equation of the secant line through P and Q is `y=((s^2+2s-3)/(s-1))x-1-s`. The value of `f'(1)`, is

Let P S be the median of the triangle with vertices P(2,2),Q(6,-1)a n dR(7,3) Then equation of the line passing through (1,-1) and parallel to P S is

Show that the points P(1,-2), Q(5,2), R(3,-1), S(-1,-5) are the vertices of a parallelogram.

Show that points P(1,-2),Q(5,2),R(3,-1),S(-1,-5) are the vertices of a parallelogram.

There is a point (p,q) on the graph of f(x)=x^(2) and a point (r,s) on the graph of g(x)=(-8)/(x), where p>0 and r>0. If the line through (p,q) and (r,s) is also tangent to both the curves at these points,respectively, then the value of P+r is

Show that points P(1,-2), Q(5, 2), R(3, -1), S(-1, -5) are the vertices of a parallelogram.