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an arbitrary point on L .Then, by the definition, the slope of L is given by




               Example  Find the equation of the line through (– 2, 3) with slope – 4.
               Solution Here m = – 4 and given point (x0 , y0) is (– 2, 3).
               By slope-intercept form formula equation of the given line is y – 3 = – 4 (x + 2) or
               4x + y + 5 = 0, which is the required equation.
               10.3.3 Two-point form
               Let the line L passes through two given points P1 (x1, y1) and P2 (x2, y2).
               Let P (x, y) be a general point on L .The three points P1, P2 and P are collinear, therefore, we
               have slope of P1P = slope of P1P2




               10.3.4 Slope-intercept form
               Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin
               Thus, the point (x, y) on the line with slope m and y-intercept c lies on the line if and
               only ify = mx +c
               Case II Suppose line L with slope m makes x-intercept d. Then equation of L is  y = m(x – d)

               Example  Write the equation of the lines for which tan = , where is the inclination of the line

               and (i) y-intercept is       (ii) x-intercept is 4.



               Solution (i) Here, slope of the line is m = tan =    and y - intercept c =   .Therefore, by slope-

               intercept form ,the equation of the line is 2y-x+3=0 .which is the required equation.


               (ii) Here, we have m = tan =   and d = 4.

               Therefore, by slope-intercept , the equation of the line is 2y-x+4=0
               which is the required equation.
               10.3.5 Intercept - form Suppose a line L makes x-intercept a and y-intercept b on the


               axes. the equation of the line is


               Example  Find the equation of the line, which makes intercepts –3 and 2 on the
               x- and y-axes respectively.
               Solution Here a = –3 and b = 2. By intercept form (5) above, equation of the line is 2x-3y+6=0
               10.3.6 Normal form : the equation of the line having normal distance p from the origin and
               angle which the normal makes with the positive direction of x-axis is given by
               x cos + y sin = p
               Example  Find the equation of the line whose perpendicular distance from the
               origin is 4 units and the angle which the normal makes with positive direction of x-axis is 15°.
               Solution Here, we are given p = 4 and
                      0
               = 15
               By the normal form , the equation of the line is
                        0
                                   0
               X cos 15  + y sin 15  =4
               or  (√     )    (√     )      √
               This is the required equation.
               10.4 General Equation of a Line
               10.4.1 Different forms of Ax + By + C = 0
               (a) Slope-intercept form If B 0, then Ax + By + C = 0 can be written as
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