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 \({\left({0,0}\right).}\)
Quadrants
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 \(M_2\left({x_2,y_2}\right):\)
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:
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:
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
Orthocenter of a Triangle
The point of intersection of the of the altitudes of a triangle (orthocenter) is given by
Area of a Triangle
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.
Area of a Quadrilateral
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
Substitute the known values and find the \(y-\)coordinate of the point \(M:\)
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
Substituting the coordinates of the vertices of the triangle we get
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
Substitute the coordinates of the vertices:
and calculate the determinant:
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:
To make sure that this figure is a square and not a rhombus, we also need to check the lengths of the diagonals:
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:
Expanding the determinant by the third row we get the equation for \(x:\)
So we have two equations.
Case 1.
Case 2.
As you can see the third point \(C\) can have \(2\) positions:
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
Substitute the coordinates of the vertices and calculate \(x_0\) and \(y_0:\)
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: