By Rakesh, Gupta
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The target of geometric numerical integration is the simulation of evolution equations owning geometric homes over lengthy instances. Of specific value are Hamiltonian partial differential equations quite often bobbing up in program fields corresponding to quantum mechanics or wave propagation phenomena.
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Rhombus c. Rectangle d. 2); (4,2); (4,5); (1,5) are the vertices of a a. Square b. Rhombus c. Rectangle d. None of these 29. 30. 31. 32. 33. 34. 35. The points (1,1); (5,1); (5,3); (1,3) are the vertices of a b. Rhombus c. Rectangle d. None of these a. Square The points (1,1); (5,1); (7,5); (3,5) are the vertices of a a. Square b. Rhombus c. Rectangle d. Parallelogram The point which divides the join of A (-2,-4) and B (8,11) in the ratio 2:3 (internally) is b. (2,2) a. (2,3) c. (3,2) d. (1,1) 36.
Now, PL = PR+RL = x tan8 +OM = x tan8 +c y = x(m) +c or y Thus, y B x PR i1PRM tan 8 = - =PRJ -, MR x [ => PR = xtan8 . ,' ill = mx +c; where c is the y-intercept = mx +c is the slope-intercept form of straight line. /48/ Example 3 Solution Find the equation of straight line having inclination 60°& y-intercept 5. e. Example 4 y = J3 x+5 Find the equation of straight line having 1 m = J3' c = 4 Here, m = J3' c = 4 (i) (ii) m = -I, c = -3 Solution 1 (i) The equation of straight line in slope-intercept form is y = mx +c 1 J3 x +4 or y = or J3y = x +4J3 or x - J3y +4 J3 = 0 is required equation.
Trapezium (ABLM) X' + ar. Trapezium (AMNC) - ar. : Area of trapezium = "2 (sum pf parallel sides) x (Distance between the parallel sides)] 1 1 = "2(Y2 + Yl)'(X1-X2) +"2(Yl 1 1 + Y3)'(X3-X1) -"2(Y2 + y:J(x 3- x2) = "2 r Y2 Xl- Y2X2 + X1YI-X2Yl +X3Yl -X1Yl+ X3Y3 - X1Y3 - X3Y2 +x zY2 -X3Y3 + x2y 31 1 = "2 [(XtY2- X2Yl) 1 = "2 [(Xl Y2 -X2Yl) +(X3Yl -X1 Y3) - X,-\Y2 +X 2Y3] + (~Y3- X3 Y2) + (X3 YI- XIY3)] Because area of a ~ is always taken to be positive. Thus we take the above value in magnitude. Hence area of ~ ABC is given by ar.