Plane geometry

9000149410

Level: 
B
Find all lines passing through the point \(A = [-2;-6]\) such that the distance from the point \([0.0]\) to these lines is \(2\sqrt{2}\).
\(p_{1}\colon 7x + y + 20 = 0\), \(p_{2}\colon x - y - 4 = 0\)
\(p\colon 7x - y = 0\)
\(p\colon x + y + 2\sqrt{2} = 0\)
\(p_{1}\colon x - y + 2\sqrt{2} = 0\), \(p_{2}\colon x + y - 2\sqrt{2} = 0\)

9000149409

Level: 
B
Find all lines which are parallel to \(p\colon x - 3y + 2 = 0\) and the distance from every of these lines to \(p\) is \(\sqrt{10}\).
\(p_{1}\colon x - 3y + 12 = 0\), \(p_{2}\colon x - 3y - 8 = 0\)
\(p\colon x - 3y = 0\)
\(p\colon x - 3y + \sqrt{10} = 0\)
\(p_{1}\colon x - 3y + \sqrt{10} = 0\), \(p_{2}\colon x - 3y -\sqrt{10} = 0\)

9000106806

Level: 
C
Given points \(A = [0;5]\), \(B = [6;1]\), \(C = [7;9]\), find the direction vector of the line passing through the point \(A\) and perpendicular to the segment \(BC\) (i.e. the line which contains the altitude of the triangle \(ABC\) through the point \(A\)).
\((8;-1)\)
\((1;8)\)
\((1;9)\)
\((-9;1)\)

9000106807

Level: 
C
Consider the points \(A = [0;5]\), \(B = [6;1]\), \(C = [7;9]\) and the triangle \(ABC\). Find the direction vector of the line which is the perpendicular bisector of the side \(b\) (i.e. the line through the midpoint of the side \(AC\) which is perpendicular to the segment \(AC\)).
\((4;-7)\)
\((7;4)\)
\((7;9)\)
\((7;-9)\)

9000106808

Level: 
C
Consider the points \(A = [0;5]\), \(B = [6;1]\), \(C = [7;9]\) and the triangle \(ABC\). Find the direction vector of the line which is the bisector of the angle \(ACB\) (i.e. the line which splits the internal angle at the point \(C\) into two angles with equal measures).
\((2;3)\)
\((6;-4)\)
\((7;9)\)
\((7;8)\)

9000107509

Level: 
B
In the following list identify a parametric line such that the angle between this line and the line \(q\) is \(0^{\circ }\). \[ q\colon x - 2y + 11 = 0 \]
\(\begin{aligned}[t] p\colon x& = 1 + 4t, & \\y & = 3 + 2t;\ t\in \mathbb{R} \\ \end{aligned}\)
\(\begin{aligned}[t] p\colon x& = 1 + 2t, & \\y & = 2 - t;\ t\in \mathbb{R} \\ \end{aligned}\)
\(\begin{aligned}[t] p\colon x& = 2 - t, & \\y & = 3 + 2t;\ t\in \mathbb{R} \\ \end{aligned}\)
\(\begin{aligned}[t] p\colon x& = t, & \\y & = 1 - 2t;\ t\in \mathbb{R} \\ \end{aligned}\)