1. The document provides examples and explanations of key concepts in geometry including Cartesian coordinates, distance between points, types of triangles, area of triangles and polygons, division of line segments, slope and inclination of lines, and angle between two lines. 2. One example shows that the points (-2, 0), (2, 3) and (5, -1) are the vertices of a right triangle by applying the Pythagorean theorem. 3. Another example finds the area of the triangle with vertices (5, 4), (-2, 1) and (2, -3) to be 20 square units using the area formula.

1. LESSON UNIT 001
Cartesian or Rectangular Coordinates
y
4
II 3 I
(-,+) (+,+)
2
1
x
-4 -3 -2 -1 1 2 3 4
-1
-2 The x-coordinate, or abscissa, of a point P is the
(-,-) (+,-)
directed distance from the y-axis to the point. The
III -3 IV
y-coordinate, or ordinate, of a point P is the
-4 directed distance form the x-axis to the point.
Distance Between Two Points
y
P2 (x2, y2)
d y2- y1
P1 (x1, y1)
x
x1 x2- x1
x2
Where d is the distance between P1 and P2. Using Pythagorean Theorem
d = (x2 - x1)2 + (y2 - y1)2
1. Find the distance between (2, -5) and (-1, -1).
2. Find the value of k so that (3k + 4, 2k-1) is equivalent from (4,-1) and (-2,5).
Common triangles used in Geometry
Isosceles triangle - two sides are equal
Scalene triangle - no sides are equal
Equiangular triangle - all interior angle are equal
Equilateral triangle - three sides are equal
Right triangle - with a right angle 90o
3. Show that the points (4, -3), (1,5) and (-4,-2) are vertices of scalene triangle.
“Anyone who influences others is a leader”

2. 4. Show that the points (5, 4), (-2, 1) and (2, -3) are the vertices of an isosceles triangle.
5. Show that the points (-2, 0), (2, 3) and (5, -1) are the vertices of a right triangle.
6. Show that (-1, 1), (0, -3), (5, 2) and (4, 6) are the vertices of a parallelogram.
Area of a Triangle
y
P1 (x1, y1)
P3 (x3, y3)
x
x1 x 2 x3
1
A= y1 y2 y3
2
1
A= [(x1y2 + x2y3 + x3y1) - (y1x2 + y2x3 + y3x1)]
2
P2 (x2, y2)
7. Find the area of the triangle with vertices at (5, 4), (-2, 1) and (2, -3).
8. Find the area of a parallelogram with vertices at (-1, 1), (0, -3), (5, 2) and (4, 6).
Area of a polygon of n-sides:
x1 x2 x3 ... xn
1
A= y1 y2 y3 ... yn
2
Division of Line Segment
y P2 (x2, y2)
P (x, y)
y2 - y1
y - y1
P1 (x1, y1)
Let P as the point that divides the line segment
x
from P1 to P2 such that: P1P
x - x1 =r
P1 P2
x2 - x1
Using similar triangles:
x - x1 y - y1
=r =r
x2 - x 1 y2 - y1
x = x1 + r (x2 - x1) y = y1 + r (y2 - y1)
1
If midpoint of the line segment from P1 to P2, if r =
2
x1 + x2 y1 + y2
x= y=
2 2
“Reputation is made in a moment. Character is built in a lifetime.”

3. 9. Find the midpoint of the line segment joining (3, -5) and (5, 4).
10. Find the coordinates of P2 if the midpoint of line segment from P1 (-3, 2) to P2 is at (4, -1).
11. Find the coordinates of the points that divide the line segment from (-5, -7) to (1, 2) into three equal parts.
12. Find the vertices of a triangle whose midpoints of the sides are (4, 0), (-2, 3) and (-1, -2).
13. Find the coordinates of the point that is three fourths of the way from (5, 3) to (-2, 1).
Slope and Inclination of a Line
The inclination θ of a line is such that 0o < θ < 180o, or, in radian measure, 0 < θ < π
y y
θ θ
x x
The slope (m) of a line is the tangent of the inclination.
m = tan θ
14. Draw a line through P(2, 2) with inclination 35o.
15. Find the inclination if the slope is -1.
16. Find the inclination if the slope is 3/3.
Let P1 (x1, y1) and P2 (x2, y2) be two given points, and indicate the slope by m.
y
P2 (x2, y2)
y2 - y1
P1 (x1, y1)
θ
x2 - x1 R (x2, y1)
θ RP2 = y2 - y1
x
m = tan θ =
P1R x2 - x1
“A great man shows his greatness by the way he treats little men.”

4. Slope of parallel lines are equal if and only if Line 1 is parallel to Line 2.
m1 = m2
Slopes of perpendicular lines are negative reciprocal with each other.
m1m2 = -1
17. Show that the points (1, -2), (-2, 0) and (5, 4) are the vertices of a right triangle.
18. Show that (-1, 1), (0, -3), (5, 2) and (4, 6) are the vertices of a parallelogram.
Angle Between Two Lines
y
L1
L2
θ
180o - α2 α1 inclination of the line
θ θ + α1 + 180o - α2 = 180o
θ = α2 - α2
α1 α2 tan θ = tan (α2 - α2)
x
tan α2 - tan α1
tan θ =
1 + tan α2 tan α1
m 2 - m1
tan θ =
1 + m 2 m1
Note: Use counter clockwise to measure angle from L1 to L2
where: m1 = slope of L1
m2 = slope of L2
θ = angle from L1 to L2
19. Find the angle of a triangle with vertices at (4, 2), (-3, 0) and (2, -5).
20. Find the value of x if the angle from L1 with the slope of 2x+7 to L2 with slope 3 is 135o.
1-3x
21. Find the value of k if m1 = 2k + 5 and m2 = 1 + 8k and L1 is || L2.
k-2 4k - 1
22. The area of a triangle with vertices (5, 2) (x, 4) and (0, -3) is 12 ½, find x.
23. Find the value of x so that the angle from L1 with slope 3x - 5 to L2 with slope 6x + 2 is 0o.
2x + 7 4x - 3
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5. CARTESIAN COORDINATE
Example 5:
Show that the points (-2, 0), (2, 3) and (5, -1) are the vertices of a right triangle.
|AB| = (5 - 2)2 + (-1 - 3)2
B(2,3)
|AB| = 5 units
C(-2,0)
|BC| = (2 + 2)2 + (3 - 0)2
A(5,1)
|BC| = 5 units
|CA| = (-2 - 5)2 + (0 + 1)2
|CA| = 50 units
The vertices shows a right triangle because it satisfies the Pythagorean theorem.
Example 7:
Find the area of the triangle with vertices at (5, 4), (-2, 1) and (2, -3).
2 5 -2 2
1
P2(5,4)
A= -3 4 1 -3
2
1
A= [(8+5+6) - (-15-8-2)]
P3(-2,1) 2
A = 20 sq. units
P1(2,-3)
Example 10:
Find the coordinates of P2 if the midpoint of line segment from P1 (-3, 2) to P2 is at (4, -1).
x1 + x 2 y1 + y2
x= y=
2 2
-3 + x2 2 + x2
4= -1 =
P1(-3,2) 2 2
x2 = 11 y2 = -4
Mid point (4,-1)
P2(x2,y2)
“Unless you are faithful in small matters, you won't be faithful in large ones.” Luke 16:10a

6. Example 11:
Find the coordinates of the points that divide the line segment from (-5, -7) to (1, 2) into three equal parts.
For the coordinate of P3,
P2(1,2)
P1P3
=r= 1
P1P2 3
P4(x4,y4) x3 = x1 + 1 (x2 - x1) y3 = y1 + 1 (y2 - y1)
3 3
x3 = -5 + 1 (1 + 5) x3 = -7 + 1 (2 + 7)
3 3
P3(x3,y3) x3 = -3 y3 = -4
For the coordinate of P4,
P1P4
P1(-5,-7) =r= 2
P1 P2 3
x3 = x1 + 2 (x2 - x1) y3 = y1 + 2 (y2 - y1)
3 3
x3 = -5 + 2 (1 + 5) x3 = -7 + 2 (2 + 7)
3 3
x4 = -1 y4 = -1
Example 17:
Show that the points (1, -2), (-2, 0) and (5, 4) are the vertices of a right triangle.
-2 - 0 -2
C(5,4) mAB = =
1+2 3
4+2 3
mBC = =
A(-2,0) 5-1 2
The points are vertices of a right triangle because the
B(1,-2)
product of the slope is -1.
Example 19:
Find the angle of a triangle with vertices at (4, 2), (-3, 0) and (2, -5).
A(4,2) 2-0 2
mAB = =
B(-3,0) A 4+3 7
B 0+5
mBC = = -1
-3 - 2
C
2+5 7
mAC = =
C(2,-5) 4-2 2
mAB - mAC mAB - mBC mBC - mAC
tan A = tan B = tan C =
1 + mAB mAC 1 + mAB mBC 1 + mBC mAC
A = 57.94 o
B = 60.94 o
C = 60.94o
“Anyone can steer the ship, but it takes a real leader to chart the course.”