Track: Mathematics and Computer Science | Baccalaureate exam: Part I, Exercise 5
1. Vectors in the plane (the Cartesian system)
Consider the points $A\left( x_{A},y_{A} \right)$ and $B\left( x_{B},y_{B} \right)$ and the position vectors $\overrightarrow{r_{A}} = x_{A}\overrightarrow{i} + y_{A}\overrightarrow{j}$, $\overrightarrow{r_{B}} = x_{B}\overrightarrow{i} + y_{B}\overrightarrow{j}$.
The free vector $\overrightarrow{AB}$:
$$\overrightarrow{AB} = \left( x_{B} - x_{A} \right)\overrightarrow{i} + \left( y_{B} - y_{A} \right)\overrightarrow{j}$$
The length (magnitude) of a vector $\overrightarrow{v} = a\overrightarrow{i} + b\overrightarrow{j}$:
$$|\overrightarrow{v}| = \sqrt{a^{2} + b^{2}}$$
The distance between two points $A$ and $B$:
$$AB = |\overrightarrow{AB}| = \sqrt{\left( x_{B} - x_{A} \right)^{2} + \left( y_{B} - y_{A} \right)^{2}}$$
The sum and difference of vectors:
$$\overrightarrow{u} = a_{1}\overrightarrow{i} + b_{1}\overrightarrow{j},\quad\overrightarrow{v} = a_{2}\overrightarrow{i} + b_{2}\overrightarrow{j}$$
$$\overrightarrow{u} \pm \overrightarrow{v} = \left( a_{1} \pm a_{2} \right)\overrightarrow{i} + \left( b_{1} \pm b_{2} \right)\overrightarrow{j}$$
Multiplication by a scalar:
$$\alpha \cdot \overrightarrow{u} = \left( \alpha a_{1} \right)\overrightarrow{i} + \left( \alpha b_{1} \right)\overrightarrow{j}$$
The collinearity condition for two vectors:
$$\overrightarrow{u}||\overrightarrow{v} < = > \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} < = > a_{1}b_{2} - a_{2}b_{1} = 0$$
The dot product:
$$\overrightarrow{u} \cdot \overrightarrow{v} = |\overrightarrow{u}| \cdot |\overrightarrow{v}| \cdot \cos(\theta) = a_{1}a_{2} + b_{1}b_{2}$$
The cosine of the angle between two vectors:
$$\cos(\theta) = \frac{a_{1}a_{2} + b_{1}b_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2}} \cdot \sqrt{a_{2}^{2} + b_{2}^{2}}}$$
The perpendicularity condition for two vectors:
$$\overrightarrow{u}\bot\overrightarrow{v} < = > \overrightarrow{u} \cdot \overrightarrow{v} = 0 < = > a_{1}a_{2} + b_{1}b_{2} = 0$$
2. Special points and segments
The midpoint $M$ of the segment $\lbrack AB\rbrack$:
$$x_{M} = \frac{x_{A} + x_{B}}{2},\quad y_{M} = \frac{y_{A} + y_{B}}{2} = > \overrightarrow{r_{M}} = \frac{\overrightarrow{r_{A}} + \overrightarrow{r_{B}}}{2}$$
The point $M$ dividing $\lbrack AB\rbrack$ in the ratio $k = \frac{MA}{MB}$:
$$x_{M} = \frac{x_{A} + kx_{B}}{1 + k},\quad y_{M} = \frac{y_{A} + ky_{B}}{1 + k} = > \overrightarrow{r_{M}} = \frac{\overrightarrow{r_{A}} + k\overrightarrow{r_{B}}}{1 + k}$$
The centroid $G$ of $\bigtriangleup ABC$:
$$x_{G} = \frac{x_{A} + x_{B} + x_{C}}{3},\quad y_{G} = \frac{y_{A} + y_{B} + y_{C}}{3} = > \overrightarrow{r_{G}} = \frac{\overrightarrow{r_{A}} + \overrightarrow{r_{B}} + \overrightarrow{r_{C}}}{3}$$
3. Equations of a line in the plane
The slope of a line ($m$):
From the angle $\alpha$ formed with the $Ox$ axis: $m = {tg}(\alpha)$
Through two points $A$ and $B$: $m = \frac{y_{B} - y_{A}}{x_{B} - x_{A}}$ (where $x_{A} \neq x_{B}$)
Forms of the equation of a line:
Explicit: $y = mx + n$ ($m$ = the slope, $n$ = the $Oy$ intercept)
General: $aX + bY + c = 0$ (slope $m = - \frac{a}{b}$)
Through the point $A\left( x_{0},y_{0} \right)$ with slope $m$:
$$y - y_{0} = m\left( x - x_{0} \right)$$
Through two points $A\left( x_{A},y_{A} \right)$ and $B\left( x_{B},y_{B} \right)$:
$$\frac{x - x_{A}}{x_{B} - x_{A}} = \frac{y - y_{A}}{y_{B} - y_{A}}\quad$$
Intercept form (intersections with the axes): $\frac{x}{a} + \frac{y}{b} = 1$
4. Relative positions of two lines
Let $d_{1}:y = m_{1}x + n_{1}$ and $d_{2}:y = m_{2}x + n_{2}$ (or, in general form, $a_{1}x + b_{1}y + c_{1} = 0$, $a_{2}x + b_{2}y + c_{2} = 0$):
Parallel lines:
$$d_{1}||d_{2} < = > m_{1} = m_{2} < = > \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
Perpendicular lines:
$$d_{1}\bot d_{2} < = > m_{1} \cdot m_{2} = - 1 < = > a_{1}a_{2} + b_{1}b_{2} = 0$$
Coincident lines:
$$d_{1} \equiv d_{2} < = > m_{1} = m_{2}\text{~and~}n_{1} = n_{2} < = > \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$
Intersecting lines (intersection at one point):
$$m_{1} \neq m_{2} < = > \text{The~system~of~the~two~line~equations~has~a~unique~solution.}$$
5. Distances and areas
The distance from a point $P\left( x_{0},y_{0} \right)$ to a line $d:aX + bY + c = 0$:
$${dist}(P,d) = \frac{|ax_{0} + by_{0} + c|}{\sqrt{a^{2} + b^{2}}}$$
6. Frequent ideas and strategies for the BAC (Part I, Exercise 5)
| Problem type | Solution strategy |
|---|---|
| Line parallel to $d$ through $A$ | Find the slope $m_{d}$. The slope of the required line is $m = m_{d}$. Apply $y - y_{A} = m\left( x - x_{A} \right)$. |
| Line perpendicular to $d$ through $A$ | Compute $m_{d}$. The required slope is $m = - \frac{1}{m_{d}}$. Apply $y - y_{A} = m\left( x - x_{A} \right)$. |
| Reflection of the point $A$ with respect to $B$ | The point $B$ is the midpoint of the segment $\lbrack AA'\rbrack$. Apply $x_{B} = \frac{x_{A} + x_{A'}}{2}$, $y_{B} = \frac{y_{A} + y_{A'}}{2}$. |
| Reflection of the point $A$ across a line $d$ | 1. Construct $d'\bot d$ through $A$. 2. Find $M = d \cap d'$ (the projection of $A$). 3. $M$ is the midpoint of $\lbrack AA'\rbrack$. |
| Computing the altitude from $A$ in $\bigtriangleup ABC$ | 1. Compute the slope $m_{BC}$. 2. The slope of the altitude is $m_{h_{A}} = - \frac{1}{m_{BC}}$. 3. Write the equation $y - y_{A} = m_{h_{A}}\left( x - x_{A} \right)$. |
| Vector equilibrium (Parallelogram) | $ABCD$ is a parallelogram $\overrightarrow{AB} = \overrightarrow{DC} < = > \overrightarrow{r_{A}} + \overrightarrow{r_{C}} = \overrightarrow{r_{B}} + \overrightarrow{r_{D}}$. |