Question

The gradient of a curve y+px+qy=0 at (1,1) is 1/2. then values of p and q is

Answer 1

Step-by-step explanation:

y+px+qy=0

y+qy=-px

(1+q)y=-px

y=-[p/(1+q)]x

This is a linear function, so has constant gradient at all points on the curve. Hence

-p/(1+q)=1/2

2p=-(1+q)

But there is an issue: you have stated that the curve's gradient is 1/2 at the point (1,1) but the curve does not cross through this point! Regardless of our choices for p and q satisfying the expressions above this paragraph, the equation of the curve will always simplify to y=0.5x, which crosses through the origin (0,0), as well as (1,0.5) and (2,1) - but not (1,1).

For your curve to pass through (1,1), we would need to add a constant term, like so:

y+px+qy=1/2