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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