Matrix and Coordinate Geomatry (MATH4201)

Matrix and Coordinate Geomatry

Teacher Name: Mohammad Abu Jabed (MAJ)

All the Lecture sheets and class’s document is here. So, anyone can take any lecture of an

date from here.

1st Lecture: Matrix

Solution Sheet:

After Midterm

1.Coordinate Geometry

2.Two Dimensional Coordinate Geometry 

3. Three Dimensional Coordinate Geometry

***Assignment**


 Sort Note: 
In Math 221 one learns that the center of mass of a collection
of n point masses m1;m2; : : : ;mn located at point P1; P2; : : : ; Pn is the
weighted average

holds for continuous mass distributions. The centroid of a triangle is both the center of mass of three equal mass points at its vertices and also the center of mass of a uniform mass distribution spread over its area. Somewhat surprisingly, the center of mass of a triangle made from three uniform rods is the incenter, not the centroid 129, the point of concurrency in Ceva's theorem canalso be viewed as the center of mass of three (unequal) point masses.

Exercise

 Show that the medians of a triangle divide it into six trianglesof equal area. Hint: It is enough to prove this for an equilateral triangle.
Exercise 3.55. Show that the centroid of a triangle divides each median
into two segments one of which is twice as long as the other.
Exercise 3.56. Points P and Q are selected
on two sides of 4ABC, as shown,
and segments AQ and BP are drawn.
Then QX PY are drawn parallel to BP
and AQ, respectively. Show that XY kAB.

Exercise 3.57. In the gure, vertices B
and C of 4ABC are joined to points P and
Q on the opposite sides, and lines BP and
CQ meet at point X. Suppose that BX =
(2=3)BP and CX = (2=3)CQ. Prove that
BP and CQ are medians of 4ABC.

Exercise 3.58. Show that there is no point P inside 4ABC such that every
line through P cuts the triangle into two pieces of equal area. Hint: Show
that if there were such a point, it would have to lie on each median of the
triangle.
Exercise In the gure, the side BC
of 4ABC is trisected by points R and S.
Similarly, T and U trisect side AC and
V and W trisect side AB. Each vertex
of 4ABC is joined to the two trisection
points on the opposite side, and the intersections
of these trisecting lines determine
4XY Z, as shown. Prove that the sides of
4XY Z are parallel to the sides of 4ABC.

Exercise (Varignon's Theorem.)
PointsW, X, Y and Z are the midpoints of
the sides of quadrangle ABCD as shown,
and P is the intersection of WY with
XZ. Two of the four small quadrangles
are shaded. Show that P is the midpoint
of both WY and XZ and that the shaded
area is exactly half of the area of quadrangle
ABCD. Hint: For the second part,
decompose the whole area into four triangles
so that exactly half the area of each
triangle is shaded.

Exercise Points P and Q are chosen
on two sides of 4ABC, as shown, and lines
BP and QC meet at X. Show that X lies
on the median from vertex A if and only if
QPkBC.
Exercise 3.62. Given 4ABC, let A0 be
the point 1=3 of the way from B to C, as
shown. Similarly, B0 is the point 1=3 of
the way from C to A and C0 lies 1=3 of
the way from A to B. In this way, we have
constructed a new triangle, 4A0B0C0 starting
with an arbitrary triangle. Now apply
the same procedure to 4A0B0C0, thereby
creating 4A00B00C00. Show that the sides
of 4A00B00C00 are parallel to the (appropriate)
sides of 4ABC. What fraction of the
area of 4ABC is the area of 4A00B00C00?

Exercise 3.63. If we draw two medians of a triangle, we see that the interior
of the triangle is divided into four pieces: three triangles and a quadrilateral.
Prove that two of these small triangles have equal areas, and show that the
other small triangle has the same area as the quadrilateral.

Congruence
De nition 4.9. Two gures in the plane are said to be congruent i there
is a Euclidean transformation carrying one onto the other. In particular
two triangles 4ABC and 4A0B0C0 are congruent i there is a Euclidean
transformation T such that T(A) = A0, T(B) = B0, and T(C) = C0. We
write
F = F0
as an abbreviation for the sentence F is congruent to F0
Remark 4.10. We use the word gure rather than the word set because
the order of the points is important. The triangles 4ABC and 4CBA
are di erent, and usually there is no Euclidean transformation T such that
T(A) = C, T(B) = B, and T(C) = A.
Remark 4.11. Congruence is an equivalence relation, i.e. for any gures
1. F = F;
2. if F = F0 and F0 = F00, the F = F00;
3. if F = F0, then F0 = F.
This is an immediate consequence of the fact that the Euclidean transformations
form a group.
4.12. Any two points are congruent. In fact, for any two points A and B
there is a unique translation T(P) = P +(B 􀀀A) such that T(A) = B. Two
directed line segments AB and A0B0 are congruent if and only if they have
the same length, i.e. jABj = jA0B0j. This is an immediate consequence of
the following
Theorem 4.13. Let b = jABj be the distance between distinct points A
and B. Then there are exactly two Euclidean transformations T such that
T(A) = (0; 0) and T(B) = (b; 0).
Proof. Using a translation we may assume w.l.o.g. that A = (0; 0). Let
B = (b1; b2) and de ne a, b, and M by
a =
p b1
b21
+ b22
; b =
p b2
b21
+ b22
; M =

a b
􀀀b a

:
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Then M is orthogonal and MB = (b; 0).
Now suppose that T1(A) = T2(A) = (0; 0) and T1(B) = T2(B) = (b; 0),
Then T(0; 0) = (0; 0) and T(b; 0) = (b; 0) where T = T1 T􀀀1
2 . As T is an
Euclidean transformation xing the origin Theorem 4.2 says that it has form
T(x; y) = ( x + y; x y) where 2 + 2 = 1. From T(b; 0) = (b; 0) we
conclude that = 0 and = 1. We conclude that there are two choices for
T, namely T = I the identity or T = S where S(x; y) = (x;􀀀y) is re
ection
in the x-axis. Hence either T1 = T2 or T1 = S T2.
Corollary 4.14. Every triangle is congruent to a triangle 4ABC where
A = (a; 0), B = (b; 0), and C = (0; 0).
Proof. By Theorem 4.13 We may assume that the triangle has vertices A1 =
(0; 0), B1 = (b1; 0), C1 = (c1; c2). Apply the translation T(P) = P 􀀀V where
V = (c1; 0).
Theorem 4.15 (SSS). Two triangles 4ABC and 4A0B0C0 are congruent if
and only if the corresponding sides are equal:
4ABC = 4A0B0C0 () jABj = jA0B0j; jBCj = jB0C0j; jCAj = jC0A0j:
Proof. \Only if" is immediate since Euclidean transformations preserve distance.
Hence assume that the corresponding sides are equal. By Theorem
4.13 we may assume that A = A0 = (0; 0) and B = B0 = (c; 0). Let
a = jBCj and b = jACj then both C and C0 lie on the intersection of the two
circles
x2 + y2 = b2; (x 􀀀 c)2 + y2 = a2:
But these two circles intersect in the two points (x0; y0). (Speci cally,
x0 = (c2 􀀀 a2)=2c and y0 =
p
b2 􀀀 x20
.) Thus either C = C0 or the re
ection
S(x; y) = (x;􀀀y) carries C to C0. Either way 4ABC = 4A0B0C0.
4.4 Similarity Transformations
De nition 4.16. A similarity transformation is an a ne transformation
T which preserves ratios of distances, i.e. there is a positive constant such
that
jA0B0j = jABj
whenever A0 = T(A) and B0 = T(B).
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De nition 4.17. Two gures in the plane are said to be similar i there
is a similarity transformation carrying one onto the other. In particular two
triangles 4ABC and 4A0B0C0 are similar i there is a Euclidean transformation
T such that T(A) = A0, T(B) = B0, and T(C) = C0.
Theorem 4.18. The similarity transformations form a group and similarity
is an equivalence relation.
Proof. As for Euclidean transformations and congruence.
Theorem 4.19. An a ne transformation T(P) = MP + V is a similarity
transformation if and only M has one of the two forms
M =

a b
􀀀b a

or M =

a b
b 􀀀a

where a2 + b2 > 0.
Proof. If T(P) = MP + V is a similarity transformation, then 􀀀1T(P) =
􀀀1MP + 􀀀1V is Euclidean.
4.20. We summarize De nitions 4.6 and 4.16 and Theorem 3.25:
Euclidean transformations preserve distance.
Similarity transformations preserve ratios of distances.
A ne transformations preserve ratios of collinear distances.
4.5 Rotations
De nition 4.21. An a ne transformation T(P) = MP +V is called orientation
preserving i det(M) > 0 and orientation reversing i det(M) <
0.
Remark 4.22. A Euclidean transformation has form T(P) = MP + V
where M has one of the two forms in Theorem 4.2. The rst is orientation
preserving and the second is orientation reversing.
De nition 4.23. A xed point of a transformation T is a point P which
is not moved by T, i.e. T(P) = P.
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Theorem 4.24. An orientation preserving Euclidean transformation which
is not a translation has a unique xed point O; such a transformation is called
a rotation about O.
Proof. Assume that T is an orientation preserving Euclidean Transformation
which is not a translation. Then T(P) = MP + V where
M =

a b
􀀀b a

; V 􀀀

p
q

:
and a2 + b2 = 1, a < 1. To nd the xed point O we solve MO + V = O for
O = (x; y). These equations are
(1 􀀀 a)x 􀀀 by = p; by + (1 􀀀 a)y = q:
The determinant of the matrix of coe cients is (1 􀀀 a)2 + b2 = 2(1 􀀀 a) > 0
so there is a unique solution.
4.25. Recall from Math 222 that rotation through angle is described by
the orthogonal matrix
R =

cos 􀀀sin
sin cos

:
The rotation R carries the point (1; 0) to the point (cos ; sin ). The matrices
R satisfy the identities
R0 = I; R R = R + ; R􀀀1
= R􀀀 :
The second identity is an immediate consequences of the trigonometric addition
formulas:
cos( + ) = cos cos 􀀀sin sin ; sin( + ) = sin cos +cos sin :
De nition 4.26. A matrix of form
R =

c 􀀀s
s c

; c2 + s2 = 1
rotation matrix; the corresponding Euclidean transformation is then a rotation
about the origin (0; 0).
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4.6 Review
Next we review how angles are treated in Math 222. We use calculus notation.
For the proofs see any calculus textbook. We will not use the material in
this section in the formal development, but it motivates