# Formulas

## Analytic Geometry # Two-Dimensional Coordinate System

Points on a plane: $$A,$$ $$B,$$ $$C,$$ $$D$$
Point coordinates: $$\left( {{x_0},{y_0}} \right),$$ $$\left( {{x_1},{y_1}} \right),$$ $$\left( {{x_2},{y_2}} \right),$$ $$\left( {{x_3},{y_3}} \right)$$
Centroid: $$M\left( {{x_0},{y_0}} \right)$$
Incenter: $$I\left( {{x_0},{y_0}} \right)$$
Circumcenter: $$O\left( {{x_0},{y_0}} \right)$$
Orthocenter: $$H\left( {{x_0},{y_0}} \right)$$
Distance between two points: $$d$$
Real number: $$\lambda$$
Polar angles: $$\varphi,$$ $${\varphi_1},$$ $${\varphi_2}$$
Polar radii: $$r,$$ $${r_1},$$ $${r_2}$$
Area: $$S$$
1. A two-dimensional Cartesian coordinate system is formed by two mutually perpendicular axes. The axes intersect at the point $$O,$$ which is called the origin. In the right-handed system, one of the axes ($$x$$-axis) is directed to the right, the other $$y$$-axis is directed vertically upwards. The coordinates of any point on the $$xy$$-plane are determined by two real numbers $$x$$ and $$y$$, which are orthogonal projections of the points on the respective axes. The $$x$$-coordinate of the point is called the abscissa of the point, and the $$y$$-coordinate is called its ordinate.
2. The distance between two points $$A\left( {{x_1},{y_1}} \right)$$ and $$B\left( {{x_2},{y_2}} \right)$$ on a plane is determined by the expression
$$d = \left| {AB} \right| =$$ $$\sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}}$$
3. Dividing a line segment in the ratio $$\lambda$$
Let the point $$C\left( {{x_0},{y_0}} \right)$$ divide the segment $$AB$$ in the ratio $$\lambda$$. The coordinates of the point $$C$$ are given by the formulas
$${x_0} = {\large\frac{{{x_1} + \lambda {x_2}}}{{1 + \lambda }}\normalsize},\;$$ $${y_0} = {\large\frac{{{y_1} + \lambda {y_2}}}{{1 + \lambda }}\normalsize},\;$$ $$\lambda = {\large\frac{{AC}}{{CB}}\normalsize},\;$$ $$\lambda \ne – 1,$$
where $${x_1}$$, $${y_1}$$ is the coordinate of the point $$A$$, and $${x_2}$$, $${y_2}$$ is the coordinate of the point $$B$$.
4.  5. The coordinates of the midpoint of the segment are obtained from the previous formula at $$\lambda = 1$$:
$${x_0} = {\large\frac{{{x_1} + {x_2}}}{2}\normalsize},\;$$ $${y_0} = {\large\frac{{{y_1} + {y_2}}}{2}\normalsize},\;$$ $$\lambda = {\large\frac{{AC}}{{CB}}\normalsize} = 1.$$
6. The point of intersection of the medians in a triangle has the following coordinates:
$${x_0} = {\large\frac{{{x_1} + {x_2} + {x_3}}}{3}\normalsize},\;$$ $${y_0} = {\large\frac{{{y_1} + {y_2} + {y_3}}}{3}\normalsize},$$
where $$A\left( {{x_1},{y_1}} \right)$$, $$B\left( {{x_2},{y_2}} \right)$$ and $$C\left( {{x_3},{y_3}} \right)$$ are the vertices of the triangle $$ABC$$. In a triangle with uniform density, the point of intersection of the medians $$M\left( {{x_0},{y_0}} \right)$$ is also the center of gravity or centroid.
7. The coordinates of the point of intersection of the angle bisectors (incenter) of a triangle are given by the relationships:
$${x_0} = {\large\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}}\normalsize},\;$$$${y_0} = {\large\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}\normalsize},$$
where $$a = BC$$, $$b = AC$$, $$c= AB$$.
8. The point of intersection of the perpendicular bisectors of a triangle is the centre of the circumscribed circle (circumcenter) and has the coordinates
$${x_0} = {\large\frac{{\left| {\begin{array}{*{20}{c}} {x_1^2 + y_1^2} & {{y_1}} & 1\\ {x_2^2 + y_2^2} & {{y_2}} & 1\\ {x_3^2 + y_3^2} & {{y_3}} & 1 \end{array}} \right|}}{{2\left| {\begin{array}{*{20}{c}} {{x_1}} & {{y_1}} & 1\\ {{x_2}} & {{y_2}} & 1\\ {{x_3}} & {{y_3}} & 1 \end{array}} \right|}}\normalsize},\;$$ $${y_0} = {\large\frac{{\left| {\begin{array}{*{20}{c}} {{x_1}} & {x_1^2 + y_1^2} & 1\\ {{x_2}} & {x_2^2 + y_2^2} & 1\\ {{x_3}} & {x_3^2 + y_3^2} & 1 \end{array}} \right|}}{{2\left| {\begin{array}{*{20}{c}} {{x_1}} & {{y_1}} & 1\\ {{x_2}} & {{y_2}} & 1\\ {{x_3}} & {{y_3}} & 1 \end{array}} \right|}}\normalsize}$$
9. The point of intersection of the altitudes (orthocenter) of a triangle
$${x_0} = {\large\frac{{\left| {\begin{array}{*{20}{c}} {{y_1}} & {{x_2}{x_3} + y_1^2} & 1\\ {{y_2}} & {{x_3}{x_1} + y_2^2} & 1\\ {{y_3}} & {{x_1}{x_2} + y_3^2} & 1 \end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}} {{x_1}} & {{y_1}} & 1\\ {{x_2}} & {{y_2}} & 1\\ {{x_3}} & {{y_3}} & 1 \end{array}} \right|}}\normalsize},\;$$ $${y_0} = {\large\frac{{\left| {\begin{array}{*{20}{c}} {x_1^2 + {y_2}{y_3}} & {{x_1}} & 1\\ {x_2^2 + {y_3}{y_1}} & {{x_2}} & 1\\ {x_3^2 + {y_1}{y_2}} & {{x_3}} & 1 \end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}} {{x_1}} & {{y_1}} & 1\\ {{x_2}} & {{y_2}} & 1\\ {{x_3}} & {{y_3}} & 1 \end{array}} \right|}}\normalsize}$$
10. Area of a triangle
$$S =$$ $$\left( \pm \right){\large\frac{1}{2}\normalsize}\left| {\begin{array}{*{20}{c}} {{x_1}} & {{y_1}} & 1\\ {{x_2}} & {{y_2}} & 1\\ {{x_3}} & {{y_3}} & 1 \end{array}} \right| =$$ $$\left( \pm \right){\large\frac{1}{2}\normalsize}\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}} & {{y_2} – {y_1}}\\ {{x_3} – {x_1}} & {{y_3} – {y_1}} \end{array}} \right|,$$
where $$A\left( {{x_1},{y_1}} \right)$$, $$B\left( {{x_2},{y_2}} \right)$$ and $$C\left( {{x_3},{y_3}} \right)$$ are the vertices of the triangle $$ABC$$, and the sign in the right side is chosen so that the area of the triangle is nonnegative.
$$S =$$ $$\left( \pm \right){\large\frac{1}{2}\normalsize}\big[ {\left( {{x_1} – {x_2}} \right)\left( {{y_1} + {y_2}} \right) }$$ $$+\;{ \left( {{x_2} – {x_3}} \right)\left( {{y_2} + {y_3}} \right) }$$ $$+\;{ \left( {{x_3} – {x_0}} \right)\left( {{y_3} + {y_0}} \right) }$$ $$+\;{ \left( {{x_0} – {x_1}} \right)\left( {{y_0} + {y_1}} \right)} \big],$$
where $$A\left( {{x_1},{y_1}} \right)$$, $$B\left( {{x_2},{y_2}} \right)$$, $$C\left( {{x_3},{y_3}} \right)$$, $$D\left( {{x_0},{y_0}} \right)$$ are the vertices of the quadrilateral $$ABCD$$. The sign in the right side is chosen so that the area of the quadrilateral is nonnegative.
$$d = \left| {AB} \right| =$$ $$\big[ {r_1^2 + r_2^2 }$$ $$-\;{ 2{r_1}{r_2}\cos \left( {{\varphi_2} – {\varphi_1}} \right)}\big]^{\large{\frac{1}{2}}\normalsize}$$
$$x = r\cos \varphi ,\;$$ $$y = r\sin \varphi$$
$$r = \sqrt {{x^2} + {y^2}} ,\;$$ $$\tan \varphi = {\large\frac{y}{x}\normalsize}.$$