![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.”](https://image.slidesharecdn.com/001basicconcepts-110917104853-phpapp01/85/001-basic-concepts-2-320.jpg)
![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](https://image.slidesharecdn.com/001basicconcepts-110917104853-phpapp01/85/001-basic-concepts-5-320.jpg)
001 basic concepts
- 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 “Leadership is calculated risk-taking.”
- 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.”