General Form to Different Form

In the study of straight lines in coordinate geometry, it's often necessary to convert the equation of a line from one form to another to make certain calculations easier or to extract specific information about the line. The general form of the equation of a straight line is given by:

$$ Ax + By + C = 0 $$

where (A), (B), and (C) are real numbers, and (A) and (B) are not both zero.

Conversion to Different Forms

There are several other forms of the equation of a straight line, including the slope-intercept form, point-slope form, and intercept form. Below is a table that summarizes these forms and their key characteristics:

Form	Equation	Important Points
Slope-Intercept Form	( y = mx + c )	(m) is the slope, (c) is the y-intercept.
Point-Slope Form	( y - y_1 = m(x - x_1) )	(m) is the slope, ((x_1, y_1)) is a point on the line.
Intercept Form	( \frac{x}{a} + \frac{y}{b} = 1 )	(a) is the x-intercept, (b) is the y-intercept.
Two-Point Form	( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} )	((x_1, y_1)) and ((x_2, y_2)) are two distinct points on the line.
Normal Form	( x\cos\theta + y\sin\theta = p )	(p) is the length of the perpendicular from the origin to the line, (\theta) is the angle made by the normal with the positive x-axis.

Conversion Process

General Form to Slope-Intercept Form

To convert from the general form to the slope-intercept form, solve for (y):

$$ Ax + By + C = 0 $$ $$ By = -Ax - C $$ $$ y = -\frac{A}{B}x - \frac{C}{B} $$

Here, the slope (m = -\frac{A}{B}) and the y-intercept (c = -\frac{C}{B}).

Example

Convert the general form (2x + 3y - 6 = 0) to the slope-intercept form.

$$ 3y = -2x + 6 $$ $$ y = -\frac{2}{3}x + 2 $$

General Form to Point-Slope Form

To convert to the point-slope form, you need a point ((x_1, y_1)) on the line. If you have such a point, you can use the slope (m = -\frac{A}{B}) obtained from the general form:

$$ y - y_1 = m(x - x_1) $$

Example

Given the general form (2x + 3y - 6 = 0) and a point ((3, 0)) on the line, convert to the point-slope form.

$$ y - 0 = -\frac{2}{3}(x - 3) $$ $$ y = -\frac{2}{3}x + 2 $$

General Form to Intercept Form

To convert to the intercept form, divide the general form by (-C) (assuming (C \neq 0)):

$$ Ax + By + C = 0 $$ $$ \frac{Ax}{-C} + \frac{By}{-C} = -\frac{C}{-C} $$ $$ \frac{x}{-\frac{C}{A}} + \frac{y}{-\frac{C}{B}} = 1 $$

Here, (a = -\frac{C}{A}) and (b = -\frac{C}{B}).

Example

Convert the general form (2x + 3y - 6 = 0) to the intercept form.

$$ \frac{2x}{6} + \frac{3y}{6} = 1 $$ $$ \frac{x}{3} + \frac{y}{2} = 1 $$

General Form to Normal Form

To convert to the normal form, you need to find the perpendicular distance (p) from the origin to the line and the angle (\theta) that the normal makes with the positive x-axis. This involves using the coefficients (A) and (B):

$$ p = \frac{|C|}{\sqrt{A^2 + B^2}} $$ $$ \cos\theta = \frac{A}{\sqrt{A^2 + B^2}} $$ $$ \sin\theta = \frac{B}{\sqrt{A^2 + B^2}} $$

Then, the normal form is:

$$ x\cos\theta + y\sin\theta = p $$

Example

Convert the general form (2x + 3y - 6 = 0) to the normal form.

$$ p = \frac{|-6|}{\sqrt{2^2 + 3^2}} = \frac{6}{\sqrt{13}} $$ $$ \cos\theta = \frac{2}{\sqrt{13}} $$ $$ \sin\theta = \frac{3}{\sqrt{13}} $$

$$ x\left(\frac{2}{\sqrt{13}}\right) + y\left(\frac{3}{\sqrt{13}}\right) = \frac{6}{\sqrt{13}} $$

By understanding these conversions, students can manipulate the equation of a line to suit the needs of various problems in coordinate geometry.