By W J. Johnston
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Additional resources for An Elementary Treatise on Analytical Geometry, with Numerous Examples
G cCxgYs) Let the vertices be A (x^), B (x C (x y ) C draw y2), 2 3 Through A, B, s II Then 3 to the axes. A ABC = A AHB + A BHC + A CHA = |DHDEB + + iaBHCF inCHKG = i[pGAEF _ oADHK] = J(AE. AG - AD. ) + x y -x y 2 3 3 2 . t )], Analytical 14 But this is the expansion of Geometry x x x3 y3 xx yx x3 y3 yx y2 x 2 area = [23- This formula should be remembered. Cor' — Let the vertex A be area The = at the origin = o o x„ y2 y3 4 x1 then : i(x 2 y3 = -x o, 3 = o, (—2, 3), y1 y2 ). learner should prove this case independently.
10. EQUATION TO JOIN OF TWO POINTS § 64. Suppose we take two points (x x y± ), (x2 y2 ) on a straight line The m of the line = tan# CL ~DL = yi - y Thus the m of a line = the difference between the y's 2 of any two points on the line divided by the difference between their x's. § 65- To find the equation to of two points the join (x^), (x 2 y2 ). Let (x, y) be any other point on the The m of the line It also line. = ^"-^ -x x (fig', § = y-yi y-y, X This is 64) 2 x Xj the required equation.
Find the area of the triangle whose vertices are (5, 6). 2 = area - (1, 2), 121 2 3 1 5 6 1 = I(~ 3) - 2(= -3 + 14-27 = - l6. 7) + l(~ In § 27) 41 we shall account for the Ex. Find the area of the triangle (2, 3), (— 4, 1), = + Thus the area contains 8 square units. negative sign. § 24*. 2 area = 2 3 1 -411 — 121 We infer that the 2 (- 1) -3 (-3) 1 (-7) (— =-2 + three points are collinear. Similarly three points (x^), (x2 y2 ), (x3 y3 ) are collinear if *1 1, 2). 9-7=* o. ] i6 Analytical X Geometry [»r.
An Elementary Treatise on Analytical Geometry, with Numerous Examples by W J. Johnston