Equation of a Line – Normal Form
Normal: A normal to a line is a line segment drawn from a point perpendicular to the given line. Let p be the length of the normal drawn from the origin to a line, which subtends an angle ø with the positive direction of x-axis as follows. Then, we have Cos ø = p/m à m = p/Cos ø And Sin ø = p/n à n = p/Sin ø The equation of line in intercept form is, x/m + y/n =1 x/(p/Cos ø) + y/(p/Sin ø) = 1 x Cos ø/p + y Sin ø/p =1 x Cos ø + y Sin ø = p. This is called the normal form of equation of the given line making the angle ø with the positive direction of x-axis and whose perpendicular distance from the origin is p. Converting the general equation of a line into normal form: The equation of a straight line in general form is, ax+by+c=0 From the above equations, we have Cos ø = p/m and Sin ø = p/n From Trigonometric identity, we have Cos2 ø + Sin2ø=1 p2/m2 + p2/n2=1 Solving this equation gives us, P = mn/√(m2+n2), which is the perpendicular distance from the origin to the line x/m + y/n =1 (i.e., nx + my – mn=0) Thus, we have the distance from origin to the line ax+by+c=0, = |c|/√(a2+b2) Thus, for converting the given line into normal form, divide the equation ax+by+c=0 by √(a2+b2). Thus, [a/ √(a2+b2)] x + [b/√(a2+b2)] + c/√(a2+b2)=0 We offer Study Skills tutoring, click here for more information. SchoolTutoring Academy is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in North-Dakota visit: Tutoring in North-Dakota.