Understand “distance measured parallel to a given line”

We need the length of the segment from the point $P(2,3)$ to the line $2x-3y+28=0$ where the segment is drawn parallel to the line $\sqrt{3}x-y+1=0.$

So first, find the direction of a line parallel to $\sqrt{3}x-y+1=0 \implies y=\sqrt{3}x+1.$ Hence its slope is $\sqrt{3}$.

Equation of the line through $(2,3)$ parallel to the given direction

The required line through $(2,3)$ with slope $\sqrt{3}$ is $y-3=\sqrt{3}(x-2).$

Rearranging, $y=\sqrt{3}x-2\sqrt{3}+3.$

Find its intersection with the line $2x-3y+28=0$

Substitute $y=\sqrt{3}x-2\sqrt{3}+3$ into $2x-3y+28=0.$

Then $2x-3(\sqrt{3}x-2\sqrt{3}+3)+28=0$ $2x-3\sqrt{3}x+6\sqrt{3}-9+28=0$ $(2-3\sqrt{3})x+(19+6\sqrt{3})=0.$

So x=\frac{-(19+6\sqrt{3})}{2-3\sqrt{3}}=rac{19+6\sqrt{3}}{3\sqrt{3}-2}.

Let the intersection point be $Q$.

Use line-direction distance formula instead of fully finding $Q$

Since $Q$ lies on the line through $P(2,3)$ parallel to slope $\sqrt{3}$, we can parametrize it as $Q=(2+t,\,3+\sqrt{3}t).$

Now impose that $Q$ lies on $2x-3y+28=0.$

Substituting: $2(2+t)-3(3+\sqrt{3}t)+28=0$ $4+2t-9-3\sqrt{3}t+28=0$ $23+t(2-3\sqrt{3})=0.$

Hence $t=\frac{-23}{2-3\sqrt{3}}=\frac{23}{3\sqrt{3}-2}.$

Compute the required distance

The vector from $P$ to $Q$ is $\overrightarrow{PQ}=(t,\sqrt{3}t).$

Therefore, $PQ=\sqrt{t^2+(\sqrt{3}t)^2}=\sqrt{4t^2}=2|t|.$

So $PQ=2\cdot \frac{23}{3\sqrt{3}-2}.$

Rationalize: $PQ=\frac{46}{3\sqrt{3}-2}\cdot \frac{3\sqrt{3}+2}{3\sqrt{3}+2}$ $=\frac{46(3\sqrt{3}+2)}{27-4}$ $=\frac{46(3\sqrt{3}+2)}{23}$ $=2(3\sqrt{3}+2)$ $=6\sqrt{3}+4.$

Match with options

$4+6\sqrt{3}$ which is Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.