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# Equation of Strighjt lines

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### Equation of Strighjt lines

1. 1. Analytical Geometry T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
2. 2. General Equation of a Line General equation of first degree in two variables, Ax + By + C= 0 where A, B and C are real constants such that A and B are not zero simultaneously. Graph of the equation Ax + By + C= 0 is always a straight line. Therefore, any equation of the form Ax + By + C= 0, where A and B are not zero simultaneously is called general linear equation or general equation of a line. © iTutor. 2000-2013. All Rights Reserved
3. 3. Different forms of Ax + By + C = 0 The general equation of a line can be reduced into various forms of the equation of a line, by the following procedures:  Slope-intercept form If B ≠ 0, then Ax + By + C= 0 can be written as Where and then from equation (1) the slope of the st. line is and y intercept is If B= 0 then x = which is a vertical line whose slope is undefined and x-intercept is B C x B A y  or y = mx + c ------------- (1) B A m  B C c  B A  A C  B C  A C  © iTutor. 2000-2013. All Rights Reserved
4. 4.  Intercept form If C ≠0, then Ax + By + C = 0 can be written as where We know that equation (1) is intercept form of the equation of a line whose x-intercept is y- intercept is If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0, which is a line passing through the origin and, therefore, has zero intercepts on the axes. 1    B C y A C x or 1 b y a x A C a  B C b And A C  B C  © iTutor. 2000-2013. All Rights Reserved
5. 5.  Normal form Let x cos θ +y sin θ = p be the normal form of the line represented by the equation Ax + By + C = 0 or Ax + By = – C. Thus, both the equations are same and therefore Which gives, and Now, Or p CBA   sincos C Ap cos C Bp sin 1sincos 2 22              C Bp C Ap  22 2 2 BA C p   or 22 BA C p   © iTutor. 2000-2013. All Rights Reserved
6. 6. Therefore and Thus, the normal form of the equation Ax + By + C = 0 is x cos θ + y sin θ = p, where , and Proper choice of signs is made so that p should be positive. 22 cos BA A   22 sin BA B   22 cos BA A   22 sin BA B   22 BA C p   © iTutor. 2000-2013. All Rights Reserved
7. 7. Distance of a Point From a Line The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point P (x1, y1) is d. Draw a perpendicular PM from the point P to the line L. If the line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates of the points are L : Ax + By + C = 0 X Y P(x1, y1)       B C R ,0        0, A C Q        B C R ,0       0, A C Q and © iTutor. 2000-2013. All Rights Reserved
8. 8. Thus, the area of the triangle PQ is given by area (  PQR) , which gives Also, area (  PQR) Or 2 area (ΔPQR) QRPM. 2 1         QR PQR)(trianglearea2 PM  000 2 1 111                    yy B C A C B C x AB C A C y B C x 2 11 2 1  ,. 11 CBxAx AB C  © iTutor. 2000-2013. All Rights Reserved
9. 9. 22 22 00 BA AB C B C A C QR              Substituting the values of area (ΔPQR) and QR in (1), we get or Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) is given by 22 11 BA CByAx PM    22 11 BA CByAx d    22 11 BA CByAx d    © iTutor. 2000-2013. All Rights Reserved
10. 10. Distance between two parallel lines We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form y = mx + c1 ... (1) and y = mx + c2 ... (2) Line (1) will intersect x-axis at the point Distance between two lines is equal to the length of the perpendicular from point A to line (2). X Y d o       0,1 m c A © iTutor. 2000-2013. All Rights Reserved
11. 11. Therefore, distance between the lines (1) and (2) is or Thus, the distance d between two parallel lines y = mx + c1 and y = mx + c2 is given by If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0, then above formula will take the form    2 1 c m c m        2 21 1 m cc d    2 21 1 m cc d    22 21 BA cc d    © iTutor. 2000-2013. All Rights Reserved
12. 12. The End Call us for more Information: www.iTutor.com Visit 1-855-694-8886