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Coordinate geometry
Distance of two points
In the planar Cartesian coordinate system, the distance of the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} (the length of the vector whose initial and terminal points are the two points) is
\latex{AB=\sqrt{(b_{1}-a_{1} )^{2}+(b_{2}-a_{2} )^{2} }.}
Point of division of a segment, centroid of a triangle
I. Midpoint of a segment
In the planar Cartesian coordinate system, the position vectors of the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} are, respectively, \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}}, while that of the midpoint \latex{F(x;y)} of the segment \latex{ AB }is \latex{\overrightarrow{f}} (Figure \latex{ 104 }). The position vector of the segment's midpoint expressed using the position vectors of the endpoints is
\latex{\overrightarrow{f}=\frac{\overrightarrow{a}+\overrightarrow{b} }{2}.}
It follows from the equation concerning position vectors that the coordinates of the midpoint are
\latex{x=\frac{a_{1}+b_{1} }{2}} and \latex{y=\frac{a_{2}+b_{2} }{2}.}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{f}}
\latex{\overrightarrow{b}}
\latex{A(a_{1};a_{2} )}
\latex{F(x;y)}
\latex{B(b_{1};b_{2} )}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 104
II. Trisecting point of a segment
The trisecting points of the segment determined by points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} are \latex{H_{1} (x_{1};y_{2} )} and \latex{H_{2} (x_{2};y_{2} ),} the corresponding position vectors are, respectively, \latex{\overrightarrow{a},\overrightarrow{b},\overrightarrow{h_{1} }} and \latex{\overrightarrow{h_{2} }.} (Figure 105)
The position vectors of the trisecting points, expressed using the position vectors of the endpoints, are
The position vectors of the trisecting points, expressed using the position vectors of the endpoints, are
\latex{\overrightarrow{h_{1} }=\frac{2\times\overrightarrow{a}+\overrightarrow{b} }{3}} and \latex{\overrightarrow{h_{2} }=\frac{2\times\overrightarrow{a}+2\times \overrightarrow{b} }{3},}
and the coordinates of the trisecting points are
\latex{H_{1}\left(\frac{2\times a_{1}+b_{1} }{3};\frac{2\times a_{2}+b_{2} }{3} \right)} and \latex{H_{2}\left(\frac{ a_{1}+2\times b_{1} }{3};\frac{ a_{2}+2\times b_{2} }{3} \right).}
\latex{A(a_{1};a_{2} )}
\latex{B(b_{1};b_{2} )}
\latex{H_{1}(x_{1};y_{1} )}
\latex{H_{2}(x_{2};y_{2} )}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{b}}
\latex{\overrightarrow{h_{1} }}
\latex{\overrightarrow{h_{2} }}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 105
*III.The coordinates of the point of division with ratio \latex{ p : q } of a segment
The position vector of the point \latex{R(x; y)}, which divides the segment determined by \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} in a ratio of \latex{p : q,} is (Figure 106):
\latex{\overrightarrow{r}=\frac{q\times \overrightarrow{a}+p\times \overrightarrow{b} }{p+q},}
from which it follows that the coordinates are
\latex{R\left(\frac{q\times a_{1}+p\times b_{1} }{p+q};\frac{q\times a_{2}+p\times b_{2} }{p+q} \right). }
\latex{A(a_{1};a_{2} )}
\latex{B(b_{1};b_{2} )}
\latex{\overrightarrow{AR}}
\latex{R(x;y)}
\latex{\frac{p}{p+q}\times A}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{r}}
\latex{\overrightarrow{AB}}
\latex{\frac{q}{p+q}\times AB}
\latex{\overrightarrow{b}}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 106
IV. Centroid of a triangle
The position vector \latex{\overrightarrow{s}} of the centroid \latex{S(x; y)} of the triangle determined by the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} and \latex{C(c_{1};c_{2} )}, expressed using the position vectors \latex{\overrightarrow{a},\overrightarrow{b}} and \latex{\overrightarrow{c}} pointing to the vertices, is
\latex{\overrightarrow{s}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} }{3},}
while the coordinates of the centroid are
\latex{S\left(\frac{a_{1}+b_{1}+c_{1} }{3};\frac{a_{2}+b_{2}+c_{2} }{3} \right).}
Equation of curves in the coordinate system
DEFINITION: The equation of a curve in the planar coordinate system is an equation with two variables which is satisfied by the points \latex{P(x; y)} of the curve and there are no other points satisfying it.
I. Equation of a line
DATA DETERMINING THE DIRECTION OF A LINE
DATA DETERMINING THE DIRECTION OF A LINE
DEFINITION: A direction vector of a line is an arbitrary nonzero vector parallel to the line. (Figure 107)
It is denoted by \latex{\overrightarrow{v}(v_{1};v_{2} ).}
\latex{\overrightarrow{v}(v_{1};v_{2} )}
\latex{ x }
\latex{ 0 }
\latex{ y }
\latex{ e }
Figure 107
It follows from the definition that if \latex{\overrightarrow{v}} is a direction vector of the line \latex{ e }, then for any real number \latex{\lambda\neq 0,\lambda\times \overrightarrow{v}} is also a direction vector of the \latex{ e }.
DEFINITION: In the plane, a normal vector of a line is an arbitrary nonzero vector perpendicular to the line. (Figure 108) It is denoted by \latex{\overrightarrow{n}(A;B)}.
\latex{\overrightarrow{n}(A;B)}
\latex{ x }
\latex{ 0 }
\latex{ y }
\latex{ e }
Figure 108
For the angle of inclination \latex{\alpha -\frac{\pi }{2}\lt \alpha \leq \frac{\pi }{2}} holds. \latex{\alpha} is positive (negative), if the \latex{ x }-axis has to be rotated around the origin in a positive (negative) direction with an angle from the said interval so that its image becomes parallel to the line.
DEFINITION: If exists, the tangent of the angle of inclination of a line is called the slope or gradient of the line. The gradient of the line with an angle of inclination
\latex{\alpha} is \latex{m=tg\alpha .}
Since the angle of inclination of the \latex{ y }-axis is \latex{\frac{\pi }{2}} , the lines parallel to the \latex{ y }-axis do not have a gradient.
RELATION AMONG THE DATA DEFINING THE DIRECTION OF A LINE
Let the direction vector of a line \latex{ e } not parallel to the \latex{ y }-axis be \latex{\overrightarrow{v}(v_{1};v_{2} ),} its normal vector be \latex{\overrightarrow{n}(A;B),} its gradient be \latex{ m }, and two of its points are \latex{P_{1}(x_{1};y_{1} )} and \latex{P_{2}(x_{2};y_{2} ).}
The following equalities hold (Figure 110):
\latex{\alpha}
\latex{\beta}
\latex{\alpha \gt 0}
\latex{\beta \lt 0}
\latex{ x }
\latex{ 0 }
\latex{ y }
\latex{ e }
Figure 109
\latex{m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }=\frac{v_{2} }{v_{1} }=-\frac{A}{B}.}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{P_{1}(x_{1};y_{1} )}
\latex{P_{2}(x_{2};y_{2} )}
\latex{x_{2}-x_{1}}
\latex{y_{2}-y_{1}}
\latex{\overrightarrow{v}(v_{1};v_{2} )}
\latex{\overrightarrow{n}(A;B )}
\latex{ x }
\latex{ 0 }
\latex{ y }
\latex{ e }
\latex{ y }
\latex{ x }
\latex{ 0 }
\latex{ e }
Figure 110
CRITERIA FOR TWO LINES BEING PARALLEL OR PERPENDICULAR
Let the data defining the direction of the lines \latex{e_{1}} and \latex{e_{2}} not parallel to the \latex{ y }-axis be \latex{\alpha _{1},m_{1},\overrightarrow{v_{1} },\overrightarrow{n_{1} }} and \latex{\alpha _{2},m_{2},\overrightarrow{v_{2} },\overrightarrow{n_{2} }.} (Figure 111)
Any of the following criteria alone is a necessary and sufficient condition for the lines \latex{e_{1}} and \latex{e_{2}} being parallel:
Any of the following criteria alone is a necessary and sufficient condition for the lines \latex{e_{1}} and \latex{e_{2}} being parallel:
- \latex{\alpha _{1}=\alpha _{2}.}
- \latex{m _{1}=m _{2}.}
- There exists a real number \latex{\lambda} different from \latex{ 0 }, for which \latex{\overrightarrow{v_{1} }=\lambda\times \overrightarrow{v_{2} }.}
- There exists a real number \latex{\mu} different from \latex{ 0 }, for which \latex{\overrightarrow{n_{1} }=\mu\times \overrightarrow{n_{2} }.}
\latex{\alpha _{1}}
\latex{\alpha _{2}}
\latex{\overrightarrow{v_{2} }}
\latex{\overrightarrow{n_{2} }}
\latex{\overrightarrow{n_{1} }}
\latex{\overrightarrow{v_{1} }}
\latex{e_{2}}
\latex{e_{2}}
\latex{e_{1}}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 111
Again, denote the data defining the direction of the lines \latex{e_{1}} and \latex{e_{2}} not parallel to the \latex{ y }-axis by \latex{\alpha _{1},m_{1},\overrightarrow{v_{1} },\overrightarrow{n_{1} }} and \latex{\alpha _{2},m_{2},\overrightarrow{v_{2} },\overrightarrow{n_{2} }} as seen in Figure \latex{ 112 }.
Any of the following criteria alone is a necessary and sufficient condition for the lines \latex{e_{1}} and \latex{e_{2}} being perpendicular:
- \latex{m_{1}\times m_{2}=-1.}
- \latex{\overrightarrow{v_{1} }\times \overrightarrow{v_{2} }=0.}
- \latex{\overrightarrow{n_{1} }\times \overrightarrow{n_{2} }=0.}
\latex{\alpha _{2}}
\latex{\alpha _{1}}
\latex{\overrightarrow{v_{1} }}
\latex{\overrightarrow{v_{2} }}
\latex{e_{2}}
\latex{e_{1}}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 112
DIFFERENT FORMS OF THE EQUATION OF A LINE
The equation of a line in the planar Cartesian coordinate system is a linear equation with two variables of the form
\latex{Ax+By+C=0,}
where at least one of \latex{ A } and \latex{ B } is not equal to \latex{ 0 } \latex{(A^{2}+B^{2}\gt 0 ).}
Normal form:
If the line \latex{ e } is given by a point \latex{P_{0}(x_{0};y_{0} )} and its normal vector \latex{\overrightarrow{n}(A;B)} (Figure 113), then its equation is
\latex{Ax+By=Ax_{0}+By_{0}.}
\latex{\overrightarrow{r}}
\latex{\overrightarrow{r_{0} }}
\latex{\overrightarrow{n}(A;B) }
\latex{P_{0}(x_{0};y_{0} )}
\latex{\overrightarrow{P_{0}P }=\overrightarrow{r}-\overrightarrow{r_{0} }}
\latex{P(x;y)}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 113
Form using a direction vector:
If the line \latex{ e } is given by a point \latex{P_{0}(x_{0};y_{0} )} and its direction vector \latex{\overrightarrow{v}(v_{1};v_{2} )} (Figure 114), then its equation is
\latex{v_{2}x-v_{1}y=v_{2}x_{0}-v_{1}y_{0}.}
Form using the gradient:
If the line \latex{ e } not parallel to the \latex{ y }-axis is given by a point \latex{P_{0}(x_{0};y_{0} )} and its gradient \latex{ m }, then its equation is
\latex{y-y_{0}=m\times (x-x_{0} ).}
If the line \latex{ e } intersects the axis \latex{ y } in the point \latex{P_{0}(0;b)} and its gradient is \latex{ m }, then it follows from the previous form that its equation is
\latex{y=mx+b.}
If \latex{ e }is parallel to the \latex{ y }-axis, then its equation is
\latex{x=x_{0}.}
\latex{\overrightarrow{v}(v_{1};v_{2} )}
\latex{\overrightarrow{n}(v_{2};-v_{1} )}
\latex{P_{0}(x_{0};y_{} )}
\latex{ x }
\latex{ 0 }
\latex{ y }
\latex{ e }
Figure 114
Form using two points of the line:
If the line \latex{ e } is given by its points \latex{P_{1}(x_{1};y_{1} )} and \latex{P_{2}(x_{2};y_{2} )} (Figure 115), then its equation is
\latex{(y_{2}-y_{1} )\times (x-x_{1} )=(x_{2} -x_{1} )\times (y-y_{1} ).}
If the line \latex{ e } is not parallel to any of the axes and it intersects the \latex{ x }-axis in point \latex{A(a;0)} and the \latex{ y }-axis in point \latex{B(0;b)} (Figure 116), then its equation is
\latex{\frac{x}{a}+\frac{y}{b}=1.}
This form is called the intercept form.
The coordinates of the intersection point of two intersecting lines in the plane are the solutions of the system of equations consisting of the equations of the lines.
The coordinates of the intersection point of two intersecting lines in the plane are the solutions of the system of equations consisting of the equations of the lines.
\latex{P_{2}(x_{2};y_{2} )}
\latex{P_{1}(x_{1};y_{1} )}
\latex{\overrightarrow{n}(y_{2}-y_{1};x_{1}-x_{2} )}
\latex{\overrightarrow{v}(x_{2}-x_{1};y_{2}-y_{1} )}
\latex{ x }
\latex{ 0 }
\latex{ e }
\latex{ y }
Figure 115
II. Equation of a circle
The equation of the circle with centre \latex{K(u;v)} and radius \latex{ r } (Figure 117) is
\latex{(x-u)^{2}+(y-v)^{2}=r^{2}.}
The following theorem holds:
\latex{(0;b)}
\latex{(a;0)}
\latex{\frac{x}{a}+\frac{y}{b}=1}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 116
THEOREM: A quadratic equation with two variables is the equation of a circle in the Cartesian coordinate system if and only if it can be rearranged into the following form:
\latex{x^{2}+y^{2}+Ax+By+C=0,}
where \latex{ A, B } and \latex{ C } are real numbers which satisfy the inequality
\latex{A^{2}+B^{2}-4C\gt 0.} The centre of the circle belonging to this equation is \latex{K\left(-\frac{A}{2};-\frac{B}{2} \right),} while its radius is
\latex{r=\frac{\sqrt{A^{2}+B^{2}-4C } }{2}.}
\latex{P(x;y)}
\latex{K(u;v)}
\latex{ x }
\latex{ 0 }
\latex{ y }
\latex{ k }
\latex{ r }
Figure 117
III. Mutual position of a circle and a line
There are three possible ways for a circle and a line to be located in the plane as it can be seen in Figure 118. Knowing the equations of the circle and the line, the number of common points and their coordinates can be computed using algebraic tools.
\latex{e\cap k=\varnothing}
\latex{e\cap k=E}
\latex{e\cap k=\left\{M_{1};M_{2} \right\}}
\latex{ e }
\latex{ k }
\latex{ e }
\latex{ E }
\latex{ e }
\latex{ M_{1} }
\latex{ M_{2} }
Figure 118
The coordinates of the common points of the circle with equation \latex{(x-u)^{2}+(y-v)^{2}=r^{2}} and the line \latex{y=mx+b} are the solutions of the system of equations
\latex{\begin{rcases}(x-u)^{2}+(y-v)^{2}=r^{2} \\ y=mx+b \end{rcases}.}
The number of common points can be determined using the discriminant of the quadratic equation with one variable which can be obtained by substituting the second equation into the first one (by eliminating one variable):
- if the discriminant is positive, then the line intersects the circle in two points;
- if the discriminant is \latex{ 0 }, then the line is tangential to the circle;
- if the discriminant is negative, then the line and the circle do not have any points in common.
*IV. Equation of a parabola
*DEFINITION: A parabola is the set of points in the plane whose distance from a given line \latex{ d } and point \latex{ F } of the plane not incident to \latex{ d } is equal. (Figure 119)
\latex{d\div y=-\frac{p}{2}}
\latex{F\left(0;\frac{p}{2} \right)}
\latex{t\div y}
\latex{P(x;y)}
\latex{y}
\latex{\frac{p}{2}}
\latex{p}
\latex{ x }
\latex{ T }
Figure 119
\latex{ F } is called the focus of the parabola and \latex{ d } is called its directrix. The parabola is reflection symmetric to the line \latex{ t } perpendicular to \latex{ d } through \latex{ F }. \latex{ t } is the axis of the parabola. The midpoint \latex{ T } of the segment of \latex{ t } between \latex{ F } and \latex{ d } is incident to the parabola. \latex{ T } is called the vertex of the parabola. The distance of \latex{ F } and \latex{ d } is the focal parameter of the parabola, it is denoted by \latex{p(p\gt 0).}
If the vertex of a parabola with focal parameter \latex{ p } is the origin and its focus is the point \latex{F\left(0;\frac{p}{2} \right)} (Figure \latex{ 119 }), then its equation is
\latex{y=\frac{1}{2p}\times x^{2}.}
This is usually called the vertical equation of a parabola.
\latex{y=\frac{p}{2}}
\latex{P(x;y)}
\latex{F\left(0;-\frac{p}{2} \right)}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 120
The equation of the parabola with focal parameter \latex{ p } and focus \latex{F\left(0;-\frac{p}{2} \right)} (Figure 120) is
\latex{y=-\frac{1}{2p}\times x^{2}.}
The equation of the 'upward opening' parabola with focal parameter \latex{ p }, vertex \latex{T(u;v)} and axis parallel to the \latex{ y }-axis (Figure 121) is\latex{y-v=\frac{1}{2p}\times (x-u)^{2}.}
\latex{F\left(u;v-\frac{p}{2} \right)}
\latex{y=v-\frac{p}{2}}
\latex{P(x;y)}
\latex{T(u;v)}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 121
Exercises
{{exercise_number}}. In the coordinate system, the vertices of a triangle are \latex{A(2;5),B(4;-2),C(-3;1).} Dilate the triangle to double size with the centre of dilation being the origin. Determine the following about the image:
- the coordinates of its vertices;
- the coordinates of its centroid;
- the equation of its circumscribed circle;
- its circumference;
- its area.
{{exercise_number}}. A point-like object is moving along the circle with equation \latex{x^{2}+y^{2}-4x+6y-12=0} due to a net force acting on the object in the direction of the centre of the circle. At the point \latex{(5;-7)} of the track the force perishes. Compute the equation of the new trajectory. (The object will be following a tangent of the circle.)
{{exercise_number}}. Compute the coordinates of the vertices and the area of the triangle bounded by the coordinate axes and the line whose equation is \latex{3x-4y=12.}
{{exercise_number}}. The circle with equation \latex{x^{2}+y^{2}=9} and the points \latex{A(-2;-3)} and \latex{B(-5;-9)} are given. Find the equations of the tangent lines from the trisecting points of the segment \latex{ AB }to the circle.
{{exercise_number}}. Two of the vertices of the triangle \latex{ ABC } are \latex{A(5;1)} and \latex{B(-3;4)}, while vertex \latex{ C } is incident to the line with equation \latex{y=2x.} Find the set of centroids for every such triangles.
{{exercise_number}}. For which values of the parameter \latex{ a } will be the lines \latex{ f } and \latex{ g } will be
- parallel;
- perpendicular,
if the equation of \latex{ f } is \latex{(a+2)\times x+3y=3} while that of \latex{ g } is \latex{x+ay=1?}
{{exercise_number}}. The equations of the lines of two sides of a square are given: \latex{2x-3y=1,2x-3y=12.} Determine the area of the square.
{{exercise_number}}. Determine the size of the angle between the tangent lines from the origin to the parabola with equation \latex{y=\frac{x^{2} }{2}+4.}
{{exercise_number}}. The parabola with equation \latex{y=x^{2}-4x+5} is intersected by a line parallel to the axis of symmetry of the parabola such that the intersection points and the vertex of the parabola determines a regular triangle. Compute the side lengths and the area of the triangle.
{{exercise_number}}. Plot in the Cartesian coordinate system the set of points \latex{P(x;y)} whose coordinates satisfy the following:
- \latex{|x|\geq |y|;}
- \latex{|x|+|y-1|=3;}
- \latex{(2x+y-3)\times (x^{2}-y+1 )=0.}