# Cartesian Coordinates on a Plane

## Position of a Point on a Plane

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. Both x-axis and y-axis have the same unit of length. Sometimes they are simply called the coordinate axes.

The coordinates of any point M on the xy-plane are determined by two real numbers x and y, which are orthogonal projections of the point 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.

The fact that point M has coordinates x and y is denoted as $${M\left({x,y}\right)}.$$ The origin O has coordinates $${\left({0,0}\right).}$$

The coordinate axes divide the plane into four quadrants. The quadrant where both x- and y-coordinates are positive is usually called the $$1\text{st}$$ quadrant. The signs of coordinates of points in other quadrants are shown in the figure below.

## Distance Between Two Points on a Plane

Consider a two-dimensional Cartesian coordinate system Oxy and points $$M_1\left({x_1,y_1}\right)$$ and $$M_2\left({x_2,y_2}\right).$$

Using the Pythagorean theorem, we get the following formula for the distance $$d\left({M_1,M_2}\right)$$ between points $$M_1\left({x_1,y_1}\right)$$ and $$M_2\left({x_2,y_2}\right):$$

$d\left({M_1,M_2}\right) = \sqrt{\left({x_2 - x_1}\right)^2 + \left({y_2 - y_1}\right)^2 }.$

## Centroid of a Triangle

In a triangle with uniform density, the centroid or center of gravity coincides with the point of intersection of the medians. The point of intersection of the medians $$M\left({x_0,y_0}\right),$$ in a triangle has the following coordinates:

$x_0 = \frac{x_1 + x_2 + x_3}{3},\;y_0 = \frac{y_1 + y_2 + y_3}{3},$

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.

## Incenter of a Triangle

The coordinates of the point of intersection of the angle bisectors (incenter) of a triangle are given by the formulas:

$x_0 = \frac{ax_1 + bx_2 + cx_3}{a + b + c},\;y_0 = \frac{ay_1 + by_2 + cy_3}{a + b + c},$

where a = BC, b = AC, c = AB.

## Circumcenter of a Triangle

The point of intersection of the perpendicular bisectors of a triangle is the centre of the circumscribed circle (circumcenter) and has the coordinates

$x_0 = \frac{\left| {\begin{array}{*{20}{c}} {{x_1^2}+{y_1^2}}&{y_1}&1\\ {{x_2^2}+{y_2^2}}&{y_2}&1\\ {{x_3^2}+{y_3^2}}&{y_3}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|},\;y_0 = \frac{\left| {\begin{array}{*{20}{c}} {x_1}&{{x_1^2}+{y_1^2}}&1\\ {x_2}&{{x_2^2}+{y_2^2}}&1\\ {x_3}&{{x_3^2}+{y_3^2}}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|}$

## Orthocenter of a Triangle

The point of intersection of the of the altitudes of a triangle (orthocenter) is given by

$x_0 = \frac{\left| {\begin{array}{*{20}{c}} {y_1}&{{x_2}{x_3}+{y_1^2}}&1\\ {y_2}&{{x_3}{x_1}+{y_2^2}}&1\\ {y_3}&{{x_1}{x_2}+{y_3^2}}&1 \end{array}} \right|}{\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|},\;y_0 = \frac{\left| {\begin{array}{*{20}{c}} {x_1^2}+{y_2}{y_3}&{x_1}&1\\ {x_2^2}+{y_3}{y_1}&{x_2}&1\\ {x_3^2}+{y_1}{y_2}&{x_3}&1 \end{array}} \right|}{\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|}$

## Area of a Triangle

$A = \left({\pm}\right)\frac{1}{2}\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right| = \left({\pm}\right)\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}\\ {{x_3} - {x_1}}&{{y_3} - {y_1}} \end{array}} \right|,$

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.

$A = \left({\pm}\right) \left[{\left({x_1 - x_2}\right)\left({y_1 + y_2}\right) + \left({x_2 - x_3}\right)\left({y_2 + y_3}\right) + \left({x_3 - x_4}\right)\left({y_3 + y_4}\right) + \left({x_4 - x_1}\right)\left({y_4 + y_1}\right)}\right],$

where $$A\left({x_1,y_1}\right),$$ $$B\left({x_2,y_2}\right),$$ $$C\left({x_3,y_3}\right)$$ and $$D\left({x_4,y_4}\right)$$ are the vertices of the quadrilateral ABCD. The sign in the right side is chosen so that the area of the quadrilateral is nonnegative

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find on the $$y-$$axis the point $$M$$ at a distance $$d = 13$$ from the point $$L\left({5,-4}\right).$$

### Example 2

Determine the centre of gravity of a homogeneous lamina having the form of a triangle with vertices $$A\left({-1,3}\right),$$ $$B\left({3,6}\right)$$ and $$C\left({2,-1}\right).$$

### Example 3

Calculate the area of a triangle whose vertices are points $$A\left({3,2}\right),$$ $$B\left({-2,5}\right),$$ $$C\left({2,-3}\right).$$

### Example 4

Show that points $$A\left({0,2}\right),$$ $$B\left({3,7}\right),$$ $$C\left({8,4}\right)$$ and $$D\left({5,-1}\right)$$ are the vertices of a square.

### Example 5

The area of the triangle is $$10$$ square units and its two vertices are points $$A\left({-2,2}\right)$$ and $$B\left({5,1}\right).$$ Find the coordinates of the third vertex $$C$$ if it is known that it lies on the $$x-$$axis.

### Example 6

Find the center and radius of the circle circumscribed about the triangle with vertices $$A\left({2,8}\right),$$ $$B\left({8,0}\right)$$ and $$C\left({2,0}\right).$$

### Example 1.

Find on the $$y-$$axis the point $$M$$ at a distance $$d = 13$$ from the point $$L\left({5,-4}\right).$$

Solution.

Since the point $$M$$ lies on the $$y-$$axis, the abscissa of the point $$M$$ is zero: $$M = M\left({0,y}\right).$$ The condition $$ML = d = 13$$ can be written as

$ML = d = \sqrt{\left({x_L-x_M}\right)^2 + \left({y_L-y_M}\right)^2}.$

Substitute the known values and find the $$y-$$coordinate of the point $$M:$$

$13 = \sqrt{\left({5-0}\right)^2 + \left({-4-y}\right)^2}, \Rightarrow 13 = \sqrt{25 + \left({4+y}\right)^2}, \Rightarrow \left({4+y}\right)^2 = 169-25=144, \Rightarrow 4+y = \pm12, \Rightarrow y_1 = 8, y_2 = -16.$

So the problem has two solutions: $$M_1\left({0,8}\right)$$ and $$M_2\left({0,-16}\right).$$

### Example 2.

Determine the centre of gravity of a homogeneous lamina having the form of a triangle with vertices $$A\left({-1,3}\right),$$ $$B\left({3,6}\right)$$ and $$C\left({2,-1}\right).$$

Solution.

We will determine the coordinates of the centroid by the formula

$x_0 = \frac{x_1 + x_2 + x_3}{3},\;y_0 = \frac{y_1 + y_2 + y_3}{3}.$

Substituting the coordinates of the vertices of the triangle we get

$x_0 = \frac{-1 + 3 + 2}{3} = \frac{4}{3};$
$y_0 = \frac{3 + 6 + -1}{3} = \frac{8}{3}.$

Therefore, the center of gravity of the triangular lamina (centroid) has coordinates $$\left({\frac{4}{3},\frac{8}{3}}\right).$$

### Example 3.

Calculate the area of a triangle whose vertices are points $$A\left({3,2}\right),$$ $$B\left({-2,5}\right),$$ $$C\left({2,-3}\right).$$

Solution.

We use the following formula

$A = \left({\pm}\right)\frac{1}{2}\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|.$

Substitute the coordinates of the vertices:

$A = \left({\pm}\right)\frac{1}{2}\left| {\begin{array}{*{20}{c}} 3&2&1\\ -2&5&1\\ 2&-3&1 \end{array}} \right|,$

and calculate the determinant:

$A = \left({\pm}\right)\frac{1}{2}\left\{{3\cdot \left| {\begin{array}{*{20}{c}} 5&1\\ -3&1 \end{array}} \right| - 2\cdot \left| {\begin{array}{*{20}{c}} -2&1\\ 2&1 \end{array}} \right| + 1\cdot \left| {\begin{array}{*{20}{c}} -2&5\\ 2&-3 \end{array}} \right|}\right\} = \left({\pm}\right)\frac{1}{2} \left\{{3\cdot8 - 2\cdot\left({-4}\right) + 1\cdot\left({-4}\right)}\right\} = \left({\pm}\right)\frac{1}{2} \left\{{24 + 8 - 4}\right\} = \left({\pm}\right)\frac{1}{2}\cdot28 = \left({\pm}\right)14.$

We will choose the plus sign so that the area is positive. Hence, $${A = 14}.$$

### Example 4.

Show that points $$A\left({0,2}\right),$$ $$B\left({3,7}\right),$$ $$C\left({8,4}\right)$$ and $$D\left({5,-1}\right)$$ are the vertices of a square.

Solution.

Let us check that all four sides have the same length:

$AB = \sqrt{\left({3 - 0}\right)^2 + \left({7 - 2}\right)^2} = \sqrt{3^2 + 5^2} = \sqrt{9+25} = \sqrt{34};$
$BC = \sqrt{\left({8 - 3}\right)^2 + \left({4 - 7}\right)^2} = \sqrt{5^2 + \left({-3}\right)^2} = \sqrt{25+9} = \sqrt{34};$
$CD = \sqrt{\left({8 - 3}\right)^2 + \left({4 - 7}\right)^2} = \sqrt{5^2 + \left({-3}\right)^2} = \sqrt{25+9} = \sqrt{34};$
$DA = \sqrt{\left({0 - 5}\right)^2 + \left({2 - \left({-1}\right)}\right)^2} = \sqrt{\left({-5}\right)^2 + 3^2} = \sqrt{25+9} = \sqrt{34}.$

To make sure that this figure is a square and not a rhombus, we also need to check the lengths of the diagonals:

$AC = \sqrt{\left({8 - 0}\right)^2 + \left({4 - 2}\right)^2} = \sqrt{8^2 + 2^2} = \sqrt{64+4} = \sqrt{68};$
$BD = \sqrt{\left({5 - 3}\right)^2 + \left({-1 -7}\right)^2} = \sqrt{2^2 + \left({-8}\right)^2} = \sqrt{4+64} = \sqrt{68}.$

Since the given quadrilateral $$ABCD$$ has equal sides and equal diagonals, we conclude that it is a square.

### Example 5.

The area of the triangle is $$10$$ square units and its two vertices are points $$A\left({-2,2}\right)$$ and $$B\left({5,1}\right).$$ Find the coordinates of the third vertex $$C$$ if it is known that it lies on the $$x-$$axis.

Solution.

If point $$C$$ lies on the $$x-$$axis, its $$y-$$coordinate is zero, that is $$C = C\left({x,0}\right).$$ Substitute all the coordinates into the formula for the area of a triangle:

$A = \left({\pm}\right)\frac{1}{2}\left| {\begin{array}{*{20}{c}} -2&2&1\\ 5&1&1\\ x&0&1 \end{array}} \right|.$

Expanding the determinant by the third row we get the equation for $$x:$$

$A = 10 = \left({\pm}\right)\frac{1}{2}\left\{{x\cdot \left| {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right| + 1\cdot \left| {\begin{array}{*{20}{c}} -2&2\\ 5&1 \end{array}} \right|}\right\} = \left({\pm}\right)\frac{1}{2} \left\{{x\cdot1 + 1\cdot\left({-12}\right)}\right\} = \left({\pm}\right) \left\{{\frac{x}{2} - 6}\right\}.$

So we have two equations.

#### Case 1.

$\frac{x}{2} - 6 = 10, \Rightarrow \frac{x}{2} = 16, \Rightarrow x = 32.$

#### Case 2.

$\frac{x}{2} - 6 = -10, \Rightarrow \frac{x}{2} = -4, \Rightarrow x = -8.$

As you can see the third point $$C$$ can have $$2$$ positions:

$C\left({32,0}\right) \text{ or } C\left({-8,0}\right).$

### Example 6.

Find the center and radius of the circle circumscribed about the triangle with vertices $$A\left({2,8}\right),$$ $$B\left({8,0}\right)$$ and $$C\left({2,0}\right).$$

Solution.

We determine the center of the circumscribed circle $$P\left({x_0,y_0}\right)$$ by the formulas

$x_0 = \frac{\left| {\begin{array}{*{20}{c}} {{x_1^2}+{y_1^2}}&{y_1}&1\\ {{x_2^2}+{y_2^2}}&{y_2}&1\\ {{x_3^2}+{y_3^2}}&{y_3}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|},\;y_0 = \frac{\left| {\begin{array}{*{20}{c}} {x_1}&{{x_1^2}+{y_1^2}}&1\\ {x_2}&{{x_2^2}+{y_2^2}}&1\\ {x_3}&{{x_3^2}+{y_3^2}}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {x_1}&{y_1}&1\\ {x_2}&{y_2}&1\\ {x_3}&{y_3}&1 \end{array}} \right|}$

Substitute the coordinates of the vertices and calculate $$x_0$$ and $$y_0:$$

$x_0 = \frac{\left| {\begin{array}{*{20}{c}} {{2^2}+{8^2}}&{8}&1\\ {{8^2}+{0^2}}&{0}&1\\ {{2^2}+{0^2}}&{0}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {2}&{8}&1\\ {8}&{0}&1\\ {2}&{0}&1 \end{array}} \right|} = \frac{\left| {\begin{array}{*{20}{c}} {68}&{8}&1\\ {64}&{0}&1\\ {4}&{0}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {2}&{8}&1\\ {8}&{0}&1\\ {2}&{0}&1 \end{array}} \right|} = \frac{\cancel{\left({-8}\right)}\cdot\left| {\begin{array}{*{20}{c}} {64}&{1}\\ {4}&{1} \end{array}} \right|}{2 \cdot\cancel{\left({-8}\right)} \cdot\left| {\begin{array}{*{20}{c}} {8}&{1}\\ {2}&{1} \end{array}} \right|} = \frac{64-4}{2\cdot\left({8-2}\right)} = \frac{60}{12} = 5;$
$y_0 = \frac{\left| {\begin{array}{*{20}{c}} {2}&{{2^2}+{8^2}}&1\\ {8}&{{8^2}+{0^2}}&1\\ {2}&{{2^2}+{0^2}}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {2}&{8}&1\\ {8}&{0}&1\\ {2}&{0}&1 \end{array}} \right|} = \frac{\left| {\begin{array}{*{20}{c}} {2}&{68}&1\\ {8}&{64}&1\\ {2}&{4}&1 \end{array}} \right|}{2\left| {\begin{array}{*{20}{c}} {2}&{8}&1\\ {8}&{0}&1\\ {2}&{0}&1 \end{array}} \right|} = \frac{2\cdot\left| {\begin{array}{*{20}{c}} {64}&{1}\\ {4}&{1} \end{array}} \right| - 68\cdot \left| {\begin{array}{*{20}{c}} {8}&{1}\\ {2}&{1} \end{array}} \right| + 1\cdot \left| {\begin{array}{*{20}{c}} {8}&{64}\\ {2}&{4} \end{array}} \right|}{2 \cdot\left({-8}\right) \cdot\left| {\begin{array}{*{20}{c}} {8}&{1}\\ {2}&{1} \end{array}} \right|} = \frac{2\cdot60 -68\cdot6 + 1\cdot\left({-96}\right)}{2 \cdot\left({-8}\right)\cdot6} = \frac{-384}{-96} = 4.$

Hence, the center of the circumscribed circle have coordinates $$P\left({5,4}\right).$$ Now we can easily determine the radius of the inscribed circle $$r$$ by calculating the distance from point $$P$$ to any vertex:

$r = PA = \sqrt{\left({5 - 2}\right)^2 + \left({4 - 8}\right)^2} = \sqrt{3^2 + \left({-4}\right)^2} = \sqrt{9+16} = \sqrt{25} = 5.$