Intersection divisor of a homogeneous polynomial with a line in the projective plane

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I am trying to do an exercise from Rick's Miranda Book that goes like this

Show that if $X$ is a line in the projective plane, then the intersection divisor of any other line with $X$ has degree one. In general, show that the intersection divisor of a homogeneous polinomyal $G$ of degree $d$ with a line $X$ has degree $d$.

My attempt for the first sentence was that we know that there is only one point $p$ in the intersection of those two lines so we need to calculate $ord_p(G/H)$, but i cant seem to do this in a decent way, so any help is aprecciated. Also for the second case should i try to transform the polinomyal in lines or something ? Thanks in advance.

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Fix homogeneous coordinates $(x:y:z)$. A line $X$ is given by the zeros of a homogeneous linear form that, up to a linear change of coordinates, we may suppose to be $z$. Then we may also give a global parameterization $\mathbb{P}^1 \rightarrow \mathbb{P}^2$, $(x:y) \mapsto (x:y:0)$.

It follows that the restriction of a degree $d$ homogeneous polynomial $G$ in $x,y,z$ to $X$ is just making $G(x,y,0)$. Now this is a homogeneous polynomial in two variables and you can deduce the result using the Fundamental Theorem of Algebra.

Ex.: Let $H = 3x-y+7z$. Then $H|_X = 3x-y$ which vanishes only at $(1:3)$.

Let $G = y^2(x^2-y^2)+z^4$ then $g = G|_X = y^2(x^2-y^2)$ which has two simple zeros $(1: \pm 1)$ and a double zero at $(1:0)$. To see this we can go to the affine chart $U = \{x=1\}$ where $g|_U = y^2(1-y^2)$ and compute its order in each point.