Your Cartesian co-ordinaries are incorrect.
They should be $(4,0)$ and $(2 , 2\sqrt{3})$. Sorry for the bad notation.
Also you then need to use the Rsin$(θ + α)$ form to get it as the book's answer.
Your line passes through $A = (-3,7)$ and $B = (4,1).$ The gradient of the line is:
$$\frac{B_y-A_y}{B_x-A_x} = -\frac{6}{7} \, . $$
The cartesian equation of the line is thus:
$$\frac{y-B_y}{x-B_x} = -\frac{6}{7} $$
which simplifies to give $6x+7y=31.$ Next, we substitute $x=r\cos\theta$ and $y=r\sin\theta$ to give:
$$6r\cos\theta + 7r\sin\theta = 31 \iff r= \frac{31}{7\sin\theta+6\cos\theta} \, .$$
This expression is valid provided $7\sin\theta+6\cos\theta \neq 0.$ In this case, the ray joining $(0,0)$ to $(r,\theta)$ is parallel to the line. The parametrisation is thus $\theta \mapsto (r(\theta),\theta)$, where $r$ is as above and
$$-\arctan\left(\frac{6}{7}\right) < \theta < \pi - \arctan\left(\frac{6}{7}\right) . $$
Best Answer
Hint: First, find the equation of the line in Cartesian ($y=mx+b$ (shouldn't be very difficult)). You know the slope is $\frac{1}{7}=m$. They gave you another point that it goes through/satisfies. Use that to find $b$ and you'll have the equation of your line in Cartesian Coordinates. Then, use the relationship between Cartesian and Polar to get the polar equation (maybe $x=r\cos(\theta)$ and $y=r\sin(\theta)$ will ring a bell? (try to just drawing a triangle and review what Polar coordinates and Cartesian coordinates mean)) Good luck.