By Rakesh, Gupta

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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. (ii) Here, m = -I, c = -3 The equation of straight line in slope-intercept form is y = mx +c or y = (-I)x +(-3) or y = -x-3 or x +y+3 = 0 isthe equation of straight line. Intercept Form of Straight Line Let AB be the straight line which cuts x-axis and y-axis at points Rand Q respectively B Y Let OR = a and OQ = b Let L'QRO = LQPL = a and P(x,y) be an L ______ P(x,y,) arbitrary point on the straight line b In ~QPL, Mi QL x o 14---"'a---+t tan a A LP 3) !

5 5x +4y = 20 5x +4y -20 = 0 is the required equation of line A straight line cuts x-axis and y-axis at points (3,0) and (0,5) respectively. Find the equation of straight line. Q Y Since, the straight line cuts x-axis and Solution y-axis at points (3,0) and (5,0). x-intercept (a) = 3 .. and y-intercept (b) = 5 We know, that equation of straight line in Example 6 intercept form is given by -; + ~ =1 x p /50/ or or or 4) line. =1 3 5 5x + 3y = 1 5x + 3y -1 = 0 is the required equation of line. One Point Form or Slope Point Form Let a given straight line cuts x-axis and y-axis at points Rand S respectively.

Vertices of an isosceles triangle c. vertices of a right angled triangle d. collinear (or not form a triangle) The points (2,-2); (8,4); (5,7) are the vertices of a rectangle, then the fourth vertex will be a. (-1,-1) b. (-1,0) c. (0,-1) d. (7/2,5/2) The points (7,3),(3,0),(0,4),(4,-1) are the vertices of a a. Square b. 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.