Find the value of K in a specific case of a cartesian plane

Question

I have a linear equation of a line in a cartesian plane $r:= \{(x,y) \in \mathbb{R}^2 \mid kx-(k+1)y+k-1=0, \,\, k \in \mathbb{R}\}$ and I have to find the value of k so that the line intersects the x axis in a x positive point.

Any tips?

Answer 1

Hint: Intersection with the x axis means y=0, ie this intersection has coordinates $\left( \frac{1-k}k,0\right)$.

Answer 2

Straight line in intercepts form:

$$ \dfrac{x}{(1-k)/k} +\dfrac{y}{(k-1)/k+1} = 1 $$

If k<1 positive x- intercept

If k=1 line through origin

If k>1 negative x- intercept