Notes On General Equation of a Line - CBSE Class 11 Maths
An equation of the form Ax + By + C = 0, where A, B and C are real constants, and A and B are never equal to zero simultaneously, is called the general equation of a line. Reducing General Equation to Slope-Intercept Form: Consider the general equation of a line such that the constant B is not equal to zero. Ax + By + C = 0 where B ≠ 0 ⇒ By = -Ax - C Dividing both sides by B: By/B= (-A/B)x -( C/B) ⇒ y = (-A/B)x - (C/B) y = (-A/B)x+ ( - C/B) .... eq (1) y = mx + c ….Slope-Intercept Form Equation represents the slope-intercept form of the equation of a line, with slope (m) is equal to (-A/B) and y-intercept is equal to (- C/B). If B = 0: ⇒ Ax + 0y + C = 0  ⇒ x = - C /A ⇒ x = - C /A This equation represents a vertical line with undefined slope and the x-intercept is equal to - C /A. Reducing General Equation into Intercept Form: Consider the general equation of a line such that the constant C is not equal to zero. Ax + By + C = 0 Þ Ax + By = -C Dividing both sides by -C: (A/-C) x + (B/-C) y = -C/-C ⇒ (A/-C) x +( B/-C) y = 1 x/(-C/A) + y/(-C/B) = 1 Thus, the equation represents the intercept form of the equation of a line, where the x-intercept is -C/A and the y-intercept is -C/B. If C = 0: ⇒ Ax + By + 0 = 0 ⇒ Ax + By = 0 The equation represents a straight line passing through the origin cutting no intercept on any axis. Reducing General Equation into Normal Form: Consider the general equation of a line. Ax + By + C = 0 Ax + By = -C This equation is similar to the normal form of the equation of a line (x cos ω + y sin ω = p). Equating the like terms in equation 1 and the normal form of equation of a line, ⇒ A/cos ω = B/sin ω = -C/p ⇒ cos ω = -Ap/C and sin ω = -Bp/C ⇒ sin2 ω + cos2 ω = (-Bp/C)2 + (-Ap/C)2 = 1 ⇒ B2p2/C2 + A2p2/C2 = 1 ⇒p2/C2(A2 + B2) = 1 ⇒ p2 = C2 / (A2 + B2) Taking square root on both sides, we get: p = ± C / √(A2 + B2) ∴ cos ω = -Ap/C = -A/C x ± C / √(A2 + B2) ⇒ cos ω = ± A/√(A2 + B2) ∴ sin ω = -Bp/C = -B/C x ± C / √(A2 + B2) ⇒ sin ω = ± B/√(A2 + B2) ∴ Ax + By + C = 0 in normal form x cos ω + y sin ω = p is: x (± A/√(A2 + B2)) + y (± B/√(A2 + B2)) = ± C / √(A2 + B2) x (± A/√(A2 + B2)) + y (± B/√(A2 + B2)) = C / √(A2 + B2) The constant term on the right hand side of the equation is always positive as it is the length of the perpendicular from the origin to the line.

#### Summary

An equation of the form Ax + By + C = 0, where A, B and C are real constants, and A and B are never equal to zero simultaneously, is called the general equation of a line. Reducing General Equation to Slope-Intercept Form: Consider the general equation of a line such that the constant B is not equal to zero. Ax + By + C = 0 where B ≠ 0 ⇒ By = -Ax - C Dividing both sides by B: By/B= (-A/B)x -( C/B) ⇒ y = (-A/B)x - (C/B) y = (-A/B)x+ ( - C/B) .... eq (1) y = mx + c ….Slope-Intercept Form Equation represents the slope-intercept form of the equation of a line, with slope (m) is equal to (-A/B) and y-intercept is equal to (- C/B). If B = 0: ⇒ Ax + 0y + C = 0  ⇒ x = - C /A ⇒ x = - C /A This equation represents a vertical line with undefined slope and the x-intercept is equal to - C /A. Reducing General Equation into Intercept Form: Consider the general equation of a line such that the constant C is not equal to zero. Ax + By + C = 0 Þ Ax + By = -C Dividing both sides by -C: (A/-C) x + (B/-C) y = -C/-C ⇒ (A/-C) x +( B/-C) y = 1 x/(-C/A) + y/(-C/B) = 1 Thus, the equation represents the intercept form of the equation of a line, where the x-intercept is -C/A and the y-intercept is -C/B. If C = 0: ⇒ Ax + By + 0 = 0 ⇒ Ax + By = 0 The equation represents a straight line passing through the origin cutting no intercept on any axis. Reducing General Equation into Normal Form: Consider the general equation of a line. Ax + By + C = 0 Ax + By = -C This equation is similar to the normal form of the equation of a line (x cos ω + y sin ω = p). Equating the like terms in equation 1 and the normal form of equation of a line, ⇒ A/cos ω = B/sin ω = -C/p ⇒ cos ω = -Ap/C and sin ω = -Bp/C ⇒ sin2 ω + cos2 ω = (-Bp/C)2 + (-Ap/C)2 = 1 ⇒ B2p2/C2 + A2p2/C2 = 1 ⇒p2/C2(A2 + B2) = 1 ⇒ p2 = C2 / (A2 + B2) Taking square root on both sides, we get: p = ± C / √(A2 + B2) ∴ cos ω = -Ap/C = -A/C x ± C / √(A2 + B2) ⇒ cos ω = ± A/√(A2 + B2) ∴ sin ω = -Bp/C = -B/C x ± C / √(A2 + B2) ⇒ sin ω = ± B/√(A2 + B2) ∴ Ax + By + C = 0 in normal form x cos ω + y sin ω = p is: x (± A/√(A2 + B2)) + y (± B/√(A2 + B2)) = ± C / √(A2 + B2) x (± A/√(A2 + B2)) + y (± B/√(A2 + B2)) = C / √(A2 + B2) The constant term on the right hand side of the equation is always positive as it is the length of the perpendicular from the origin to the line.

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