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The coordinates of the point of division of a line segment. The coordinates of the centroid of a triangle
The coordinates of the midpoint of a line segment
In year \latex{ 10 } we determined the position vector of the midpoint of a line
segment given in point of an arbitrary reference point with the help
of the position vectors of the two end-points of the line segment. Now
we are going to give the coordinates of the midpoint of a line segment
given in the Cartesian coordinate system with the help of the coordinates
of the end-points.
The line segment defined by the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} is given (Figure 10). Let the midpoint of the line segment be \latex{F(x;y )} . Our aim is to determine the coordinates \latex{ x } and \latex{ y } with the help of the coordinates of the points\latex{ A } and \latex{ B }.
segment given in point of an arbitrary reference point with the help
of the position vectors of the two end-points of the line segment. Now
we are going to give the coordinates of the midpoint of a line segment
given in the Cartesian coordinate system with the help of the coordinates
of the end-points.
The line segment defined by the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} is given (Figure 10). Let the midpoint of the line segment be \latex{F(x;y )} . Our aim is to determine the coordinates \latex{ x } and \latex{ y } with the help of the coordinates of the points\latex{ A } and \latex{ B }.
If the position vectors of the points\latex{ A }, \latex{ B } and \latex{ F } are \latex{\overrightarrow{a},\overrightarrow{b}} and \latex{\overrightarrow{f}} respectively,
then based on the parallelogram rule for the addition of vectors
– by using that the diagonals of a parallelogram bisect each other
– we obtain the following:
then based on the parallelogram rule for the addition of vectors
– by using that the diagonals of a parallelogram bisect each other
– we obtain the following:
\latex{\overrightarrow{f}=\frac{\overrightarrow{a}+\overrightarrow{b} }{2}.}
Based on the relation relating to the coordinates of the sum of vectors
and the scalar multiple of a vector
and the scalar multiple of a vector
\latex{x=\frac{a_{1}+b_{1} }{2}} and \latex{y=\frac{a_{2}+b_{2} }{2},}
i.e. the coordinates of the midpoint are obtained as the arithmetic
mean of the corresponding coordinates of the end-points.
mean of the corresponding coordinates of the end-points.
\latex{A(a_{1};a_{2} )}
\latex{F(x;y)}
\latex{B(b_{1};b_{2} )}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{f}}
\latex{\overrightarrow{b}}
\latex{ x }
\latex{ 0 }
\latex{ y }
Figure 10
The assignment rules can differ; therefore the means can give different values too.
Example 1
The intersection point of the diagonals of a parallelogram is the point \latex{M(2;3);} two adjacent vertices are \latex{A(0;5)} and \latex{B(-1;1).} Let us calculate the coordinates of the other two vertices.
\latex{\overrightarrow{f}=\frac{\overrightarrow{a}+\overrightarrow{b} }{2}}
Solution
\latex{F(\frac{a_{1}+b_{1} }{2};\frac{a_{2}+b_{2} }{2} )}
As the diagonals of the parallelogram bisect each other, \latex{ M } bisects both line segments \latex{ AC } and \latex{ BD } (Figure 11).
Thus
Thus
\latex{2=\frac{0+c_{1} }{2}} and \latex{3=\frac{5+c_{2} }{2}}, which implies \latex{c_{1}=4} and \latex{c_{2}=1.}
Similarly:
\latex{2=\frac{-1+d_{1} }{2}} and \latex{3=\frac{1+d_{2} }{2}}, thus \latex{d_{1}=5} and \latex{d_{2}=5.}
The two missing vertices of the parallelogram are obtained as follows:
\latex{C(4;1),D(5;5).}
\latex{A(0;5)}
\latex{B(-1;1)}
\latex{D(d_{1};d_{2} )}
\latex{C(c_{1};c_{2} )}
\latex{M(2;3)}
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 5 }
x
y
Figure 11
The coordinates of the trisection point of a line segment
Let the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} be given, and let us give the coordinates of the trisection point \latex{H_{1}(x_{1};y_{1} )} of the line segment \latex{ AB } closer to \latex{ A } with the help of the coordinates of the end-points.
If the position vectors of the points \latex{ A }, \latex{ B } and \latex{H_{1}} are \latex{\overrightarrow{a},} \latex{\overrightarrow{b}} and \latex{\overrightarrow{h_{1} }} respectively (Figure 12), then
If the position vectors of the points \latex{ A }, \latex{ B } and \latex{H_{1}} are \latex{\overrightarrow{a},} \latex{\overrightarrow{b}} and \latex{\overrightarrow{h_{1} }} respectively (Figure 12), then
\latex{\overrightarrow{h_{1} }=\overrightarrow{a}+\overrightarrow{AH_{1} }}
\latex{A(a_{1};a_{2} )}
\latex{H_{1}(x_{1};y_{1} )}
\latex{H_{2}(x_{2};y_{2} )}
\latex{B(b_{2};b_{2} )}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{h_{1} }}
\latex{\overrightarrow{h_{2} }}
\latex{\overrightarrow{b }}
\latex{ 0 }
\latex{ y }
\latex{ x }
Figure 12
However \latex{\overrightarrow{AH_{1} }=\frac{1}{3}\times \overrightarrow{AB}=(\overrightarrow{b}\times \overrightarrow{a} )} , thus
\latex{\overrightarrow{h_{1} }=\overrightarrow{a}+\frac{1}{3}\times (\overrightarrow{b}-\overrightarrow{a} )=\frac{3\times \overrightarrow{a}+\overrightarrow{b}-\overrightarrow{a} }{3}=\frac{2\times \overrightarrow{a}+\overrightarrow{b} }{3}=\frac{2}{3}\times \overrightarrow{a}+\frac{1}{3}\times \overrightarrow{b}.}
So in terms of the coordinates
\latex{x_{1}=\frac{2\times a_{1}+b_{1} }{3}=\frac{2}{3}\times a_{1}+\frac{1}{3}\times b_{1};}
\latex{y_{1}=\frac{2\times a_{2}+b_{2} }{3}=\frac{2}{3}\times a_{2}+\frac{1}{3}\times b_{2};}
\latex{y_{1}=\frac{2\times a_{2}+b_{2} }{3}=\frac{2}{3}\times a_{2}+\frac{1}{3}\times b_{2};}
It can similarly be proven that the coordinates of the trisection point \latex{H_{2}(x_{2};y_{2} )} of the line segment \latex{ AB } closer to \latex{ B } are
\latex{x_{2}=\frac{a_{1}+2\times b_{1} }{3}=\frac{1}{3}\times a_{1}+\frac{2}{3}\times b_{1};}
\latex{y_{2}=\frac{a_{2}+2\times b_{2} }{3}=\frac{1}{3}\times a_{2}+\frac{2}{3}\times b_{2}.}
\latex{\overrightarrow{h_{1} }=\frac{2}{3}\times \overrightarrow{a}+\frac{1}{3}\times \overrightarrow{b}}
\latex{\overrightarrow{h_{2} }=\frac{1}{3}\times \overrightarrow{a}+\frac{2}{3}\times \overrightarrow{b}}
\latex{H_{1}(\frac{2\times a_{1}+b_{1} }{3};\frac{2\times a_{2}+b_{2} }{3} )}
\latex{H_{2}(\frac{ a_{1}+2\times b_{1} }{3};\frac{ a_{2}+2\times b_{2} }{3} )}
Example 2
Let us calculate the coordinates of the point \latex{P(x;y)} of the line segment defined by the points \latex{A(4;-1)} and \latex{B(-2;6)} if \latex{AP=2\times PB.}
Solution
The condition implies that \latex{ P } is the trisection point of the line segment \latex{ AB } closer to \latex{ B }. Then by taking the relations above into account
\latex{x=\frac{4+2\times (-2)}{3}=0} and \latex{y=\frac{-1+2\times6}{3}=11}
\latex{ P } lies on the \latex{ y }-axis, and its second coordinate (ordinate) is \latex{ 11 }.
The coordinates of the point of division dividing
a line segment in a given ratio (extra-curricular topic)
a line segment in a given ratio (extra-curricular topic)
Now we are looking for the coordinates of the point \latex{R(x;y)} of the
line segment defined by the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} for which \latex{AR:RB=p\div q.} The method used for determining the coordinates of the trisection point can also be applied in this general case.
Let the position vectors of the points \latex{ A }, \latex{ B } and \latex{ R } be \latex{\overrightarrow{a} ,\overrightarrow{b}} and \latex{\overrightarrow{r}} respectively (Figure 13). Since \latex{\frac{AR}{RB}=\frac{p}{q},}
line segment defined by the points \latex{A(a_{1};a_{2} )} and \latex{B(b_{1};b_{2} )} for which \latex{AR:RB=p\div q.} The method used for determining the coordinates of the trisection point can also be applied in this general case.
Let the position vectors of the points \latex{ A }, \latex{ B } and \latex{ R } be \latex{\overrightarrow{a} ,\overrightarrow{b}} and \latex{\overrightarrow{r}} respectively (Figure 13). Since \latex{\frac{AR}{RB}=\frac{p}{q},}
\latex{\frac{AR}{AB}=\frac{AR}{AR+RB}=\frac{1}{1+\frac{RB}{AR} }=\frac{1}{1+\frac{q}{p} }=\frac{p}{p+q}.}
\latex{A(a_{1};a_{2} )}
\latex{\overrightarrow{AR}}
\latex{\frac{p}{p+q}\times AB}
\latex{\overrightarrow{a}}
\latex{R(x;y)}
\latex{\overrightarrow{r}}
\latex{\overrightarrow{AB}}
\latex{\frac{q}{p+q}\times AB}
\latex{\overrightarrow{b}}
\latex{B(b_{1};b_{2} )}
\latex{ 0 }
\latex{ y }
\latex{ x }
Figure 13
\latex{\overrightarrow{r}=\frac{\overrightarrow{qa}+\overrightarrow{pb} }{p+q}}
By using this the position vector \latex{\overrightarrow{r}} of the point \latex{ R } is obtained as follows:
\latex{\overrightarrow{r}=\overrightarrow{a}+\overrightarrow{AR}=\overrightarrow{a}+\frac{p}{p+q}\times \overrightarrow{AB}=\overrightarrow{a}+\frac{p}{p+q}\times (\overrightarrow{b}-\overrightarrow{a} )=}
\latex{=\frac{(p+q)\times \overrightarrow{a}+p\times \overrightarrow{b}-p\times \overrightarrow{a} }{p+q}=\frac{q\times \overrightarrow{a}+p\times \overrightarrow{b} }{p+q} =\frac{q}{p+q}\times \overrightarrow{a}+\frac{p}{p+q}\times \overrightarrow{b}.}
Thus the coordinates of the point \latex{ R } are
\latex{x=\frac{q\times a_{1}+p\times b_{1} }{p+q}=\frac{q}{p+q}\times a_{1}+\frac{p}{p+q}\times b_{1};}
and
\latex{y=\frac{q\times a_{2}+p\times b_{2} }{p+q}=\frac{q}{p+q}\times a_{2}+\frac{p}{p+q}\times b_{2}.}
Notes:
- If \latex{ p = q = 1 }, then the coordinates of the midpoint are obtained.If \latex{ p = 1, q = 2 } or \latex{ p = 2 }, \latex{ q = 1 }, then the coordinates of the trisection points are obtained from the general relation above.
- A direct consequence of the results above is that the position vector \latex{\overrightarrow{r}} of an arbitrary point \latex{ R } of the line segment \latex{ AB } can be expressed with the help of the vectors \latex{\overrightarrow{a} } and \latex{\overrightarrow{b} } pointing to the end-points (as the linear combination of \latex{\overrightarrow{a} } and \latex{\overrightarrow{b} } ) in the form \latex{\overrightarrow{r}=\alpha \times \overrightarrow{a}+\beta \times \overrightarrow{b}} where \latex{\alpha} and \latex{\beta} are non-negative real
numbers for which \latex{\alpha +\beta =1}.
Then if \latex{ R } is an interior point:
\latex{\frac{AR}{RB}=\frac{\frac{AR}{AB} }{\frac{RB}{AB} }=\frac{\beta }{\alpha }.}
If \latex{R=A}, then \latex{\beta =0,\alpha =1;} if \latex{R=B,} then \latex{\alpha =0,\beta =1.}
\latex{R(\frac{qa_{1}+pb_{1} }{p+q};\frac{qa_{2}+pb_{2} }{p+q} )}
Example 3
Let us calculate the coordinates of the point \latex{P(x;y)} line segment defined by the points \latex{A(-4;2)} and \latex{B(3;-6)} for which \latex{AP:PB=2:3}.
Solution
The condition implies that \latex{AP=\frac{2}{5}\times AB} and \latex{PB=\frac{3}{5}\times AB}. Thus
\latex{x=\frac{3\times (-4)+2\times 3}{5}=-\frac{6}{5}} and \latex{y=\frac{3\times2+2\times (-6)}{5}=-\frac{6}{5}}
So the condition of the example is satisfied by the point \latex{P(-\frac{6}{5};-\frac{6}{5} ).}
The coordinates of the centroid of a triangle
In year \latex{ 10 } we gave the position vector of the centroid of a triangle in terms of an arbitrary reference point with the help of the position vectors pointing to the vertices. By brushing up the process used there now we are going to give the coordinates of the centroid in the coordinate system in terms of the coordinates of the vertices.
Let the position vectors of the vertices \latex{A(a_{1};a_{2} ),B(b_{1}b_{2} )} and \latex{C(c_{1}c_{2} )} of the triangle \latex{ ABC } be \latex{\overrightarrow{a},\overrightarrow{b}} and \latex{\overrightarrow{c}} respectively (Figure 14).
Let the position vectors of the vertices \latex{A(a_{1};a_{2} ),B(b_{1}b_{2} )} and \latex{C(c_{1}c_{2} )} of the triangle \latex{ ABC } be \latex{\overrightarrow{a},\overrightarrow{b}} and \latex{\overrightarrow{c}} respectively (Figure 14).
\latex{B(b_{1};b_{2} )}
\latex{\overrightarrow{b}}
\latex{\overrightarrow{f}}
\latex{\overrightarrow{c}}
\latex{\overrightarrow{a}}
\latex{S(x;y)}
\latex{A(a_{1};a_{2} )}
\latex{C(c_{1};c_{2} )}
\latex{ 0 }
\latex{ y }
\latex{ x }
Figure 14
If F is the midpoint of the line segment \latex{ AB }, then the position vector \latex{\overrightarrow{f}} of the point \latex{ F } is
\latex{\overrightarrow{f}=\frac{\overrightarrow{a}+\overrightarrow{b} }{2}.}
If the position vector of the trisection point \latex{ S } of the line segment \latex{ CF } closer to \latex{ F } is \latex{\overrightarrow{s}}, then
\latex{\overrightarrow{s}=\frac{2\times \overrightarrow{f}+\overrightarrow{c} }{3}=\frac{2\times \frac{\overrightarrow{a}+\overrightarrow{b} }{2}+c }{3}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} }{3}.}
As the expression above is irrespective of the side the midpoint of which we start with, the trisection point of all three medians further away from the corresponding vertex is the same point \latex{ S }. With this we have also proven that the medians of the triangle intersect each other at one point. (This proof can also be found in the year \latex{ 10 } book.) The expression in terms of the vectors above implies that the coordinates
of the centroid \latex{S(x;y)} are
of the centroid \latex{S(x;y)} are
\latex{x=\frac{a_{1}+b_{1}+c_{1} }{3}} and \latex{y=\frac{a_{2}+b_{2}+c_{2} }{3}},
i.e. the coordinates of the centroid of the triangle can be obtained as the arithmetic mean of the corresponding coordinates of the vertices.
\latex{\overrightarrow{s}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} }{3}}
Example 4
Two vertices of a triangle are \latex{A(-2;-3),B(1;8)}. Let us calculate the coordinates of the third vertex, if the centroid is \latex{S(4;1)}.
\latex{S(\frac{a_{1}+b_{1}+c_{1} }{3} ;\frac{a_{2}+b_{2}+c_{3} }{3} )}
Solution
If the third vertex of the triangle is \latex{C(x;y)} , then
\latex{4=\frac{-2+1+x}{3},}
which implies \latex{x=13}, and
\latex{1=\frac{-3+8+x}{3},}
which implies \latex{y=-2.}
So the third vertex of the triangle is \latex{C(13;-2).}
Exercises
{{exercise_number}}. Calculate the coordinates of the midpoint of the line segment \latex{ AB }, if
- \latex{A(0;1),B(3;2);}
- \latex{A(4;1),B(-1;6);}
- \latex{A(-2;-5),B(7;-10);}
- \latex{A(8;-7),B(-4;-5);}
{{exercise_number}}. We extend the line segment \latex{ AB } by itself in both directions. Calculate the coordinates of the end-points, if \latex{ A } and \latex{ B } are the points given in the previous exercise.
{{exercise_number}}. Calculate the coordinates of the trisection points of the line segment defined by the points \latex{ A } and \latex{ B } given in exercise \latex{ 1 }.
{{exercise_number}}. We extend the line segment \latex{ AB } by two times \latex{ AB } in both directions. Calculate the coordinates of the end-points, if \latex{ A } and \latex{ B } are the points given in exercise \latex{ 1 }.
{{exercise_number}}. Give the coordinates of the point \latex{ P } of the line segment defined by the points \latex{A(-6;3)} and \latex{B(5;-4)} for which \latex{AP:PB=}
- \latex{1:2;}
- \latex{1:3;}
- \latex{3:2;}
- \latex{3:5;}
- \latex{5:2;}
- \latex{\frac{3}{4}:\frac{5}{6}.}
{{exercise_number}}. Three vertices of a parallelogram are \latex{(-2;2),(2;-3)} and \latex{(-5;-4).} Determine the coordinates of the fourth vertex and the coordinates of the intersection point of the diagonals.
{{exercise_number}}. The vertices of the quadrilateral \latex{ ABCD } are \latex{A(-6;-2),B(-5;-1),C(6;4)} and \latex{D(3;6).} Calculate the coordinates of the midpoint in the case of both midlines of the quadrilateral. What do you experience? Try and generalise.
{{exercise_number}}. Calculate the coordinates of the centroid of the triangle \latex{ ABC }, if
- \latex{A(0;2),B(2;5),S(2;1);}
- \latex{A(-3;1),B(2;6),S(3;-1);}
- \latex{A(2;-3),B(-3;3),S(4;-7).}
{{exercise_number}}. The vertices \latex{ A } and \latex{ B }, and the centroid \latex{ S } of a triangle are given. Calculate the coordinates of the third vertex, if
- \latex{A(0;0),B(2;5),S(2;1);}
- \latex{A(-3;1),B(2;6),S(3;-1);}
- \latex{A(5;1-2),B(-3;3),S(4;-7).}
{{exercise_number}}. The midpoints of the sides of a triangle are \latex{(-2;2),(5;1),(3;4).}
- Calculate the coordinates of the vertices of the triangle.
- Calculate the coordinates of the centroid of the original triangle and the triangle defined by the midpoints of the sides. What do you experience?
{{exercise_number}}. The position vector of the vertex \latex{ A } of the triangle \latex{ ABC } is \latex{\overrightarrow{a}(-2;3),\overrightarrow{AB}=7i-2j} and \latex{CB=3i-6j.} Calculate the coordinates of the vertices and the centroid of the triangle.
