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By Percival Frost Joseph Wolstenholme
The Authors of the subsequent Treatise have endeavoured to offer prior to scholars as entire a view of the topic as definite boundaries have allowed them to do. the need of those boundaries has constructed itself during getting ready the paintings in the course of a interval of 4 years. The research of innumerable papers, through the main celebrated mathematicians of all international locations, has confident the authors that the topic is sort of inexhaustible, and that, to have handled all components of it with whatever coming near near to the fulness with which they've got handled the 1st element, may have swelled their paintings in a worried percentage to what it has already attained.
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Extra info for A treatise on solid geometry
Assume that the endpoints of each line segment have integer coordinates. 18 45. 6 46. Ϫ6 Ϫ12 6 12 Ϫ18 12 47. 12 48. Ϫ12 Ϫ6 12 Ϫ12 Ϫ6 6 Ϫ12 In Problems 49–54, plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area. 49. A = 1 - 2, 52; B = 11, 32; C = 1 - 1, 02 50. A = 1 - 2, 52; B = 112, 32; 51. A = 1 -5, 32; B = 16, 02; C = 15, 52 52. A = 1 - 6, 32; B = 13, -52; 53. A = 14, -32; B = 10, - 32; 54. A = 14, -32; B = 14, 12; C = 14, 22 C = 110, -112 C = 1 - 1, 52 C = 12, 12 In Problems 55–62, find the midpoint of the line segment joining the points P1 and P2 .
E XA MP LE 4 Using Algebra to Solve Geometry Problems Consider the three points A = 1 -2, 12, B = 12, 32, and C = 13, 12. (a) (b) (c) (d) Solution Figure 14 y 3 (a) Figure 14 shows the points A, B, C and the triangle ABC. (b) To find the length of each side of the triangle, we use the distance formula, equation (1). d 1A, B2 = 2 3 2 - 1 -22 4 2 + 13 - 12 2 = 216 + 4 = 220 = 225 B = (2, 3) A = (–2, 1) –3 Plot each point and form the triangle ABC. Find the length of each side of the triangle. Verify that the triangle is a right triangle.
The vertical distance from P3 to P2 is the absolute value of the difference of the y-coordinates, 0 y2 - y1 0 . See Figure 11(b). The distance d 1P1 , P2 2 that we seek is the length of the hypotenuse of the right triangle, so, by the Pythagorean Theorem, it follows that y2 – y1 3 d 1P1 , P2 2 4 2 = 0 x2 - x1 0 2 + 0 y2 - y1 0 2 P1 = (x1, y1) x2 – x 1 = 1x2 - x1 2 2 + 1y2 - y1 2 2 x d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2 Figure 11 y y y2 P2 ϭ (x2, y2) y1 P3 ϭ (x2, y1) P1 ϭ (x1, y1) x2 x1 x P2 ϭ (x2, y2) y2 d(P1, P2) y1 P1 ϭ (x1, y1) ⏐y2 Ϫ y1⏐ ⏐x2 Ϫ x1⏐ x2 x1 (a) P3 ϭ (x2, y1) x (b) Now, if the line joining P1 and P2 is horizontal, then the y-coordinate of P1 equals the y-coordinate of P2; that is, y1 = y2 .
A treatise on solid geometry by Percival Frost Joseph Wolstenholme