Krdytkyu

How to find the third cordinate of a triangle , where as other two points are known and we know sides?

Let's say, the two points are (3, 30) and (5, 35) and we need to find the third point considering its side is perpendicular to line. And we know the distance from (3, 30) and this point.

For example

We know (x,y), (x',y'), the distance between than and the side "d". I need to discovery (a,b).

This problem arose when I needed to calculate a polygon from a polyline. Like in this map.

Thanks,

Not very clear which points you know, and where is the perpendicular. I assume $A=(3,30)$, $B=(5,35)$, and $CBperp AB$.
The first step is to find the slope of $AB$
$$m_{AB}=frac{y_B-y_A}{x_B-xA}$$
Then the slope of $BC$ is $m_{BC}=-frac1{m_{AB}}$. Note that if $x_B=x_A$ then $y_C=y_B$, and if $y_A=y_B$ then $x_C=x_B$. Let's solve these easy cases first.
$x_B=x_A$ means that the line is vertical, so the perpendicular is horizontal. That's why $y_C=y_B$. Now for $x_C$ we have two options $x_C=x_Bpm d$. Similar reasoning for $y_A=y_B$ means $x_C=x_B$ and $y_C=y_Bpm d$.

Now going back to the general case, we have calculated $m_{BC}$. That means that this line makes an angle $theta=arctan m_{BC}$ with respect to the horizontal axis. We can move $d$ from $B$ in either direction on such line. If we go in the positive $x$ direction, we can write $x_C=x_B+dcostheta$ and $y_C=y_B+dsintheta$. If we move in the negative direction, then $x_C=x_B-dcostheta$ and $y_C=y_B-dsintheta$.