Download A Comprehensive Text Book Of Applied Mathematics by Rakesh, Gupta PDF

By Rakesh, Gupta

ISBN-10: 8182472253

ISBN-13: 9788182472259

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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.

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