General Form into Intercept Form

We will learn the transformation of general form into intercept form.

To reduce the general equation Ax + By + C = 0 into intercept form (\(\frac{x}{A}\) + \(\frac{y}{A}\) = 1):

We have the general equation Ax + By + C = 0.

If A ≠ 0, B ≠ 0, C ≠ 0 then from the given equation we get, 

Ax + By = - C (Subtracting c from both sides)

⇒ \(\frac{Ax}{-C}\) + \(\frac{By}{-C}\) = \(\frac{-C}{-C}\), (Dividing both sides by -C)

⇒ \(\frac{Ax}{-C}\) + \(\frac{By}{-C}\) = 1

⇒ \(\frac{x}{-\frac{C}{A}}\) + \(\frac{y}{-\frac{C}{B}}\) = 1, which is the required intercept form (\(\frac{x}{A}\) + \(\frac{y}{A}\) = 1) of the general form of line Ax + By + C = 0.

Thus, for the straight line Ax + By + C = 0,

Intercept on x-axis = -(\(\frac{C}{A}\))  = - \(\frac{\textrm{Constant term}}{\textrm{Coefficient of x}}\)

Intercept on y-axis = -(\(\frac{C}{B}\))  = - \(\frac{\textrm{Constant term}}{\textrm{Coefficient of y}}\)

Note: From the above discussion we conclude that the intercepts made by a straight line with the co-ordinate axes can be determined by transforming its equation to intercept form. To determine the intercepts on the co-ordinate axes we can also use the following method:

To find the intercept on x-axis (i.e., x-intercept), put y = 0 in the given equation of the straight line line and find the value of x. Similarly To find the intercept on y-axis (i.e., y-intercept), put x = 0 in the given equation of the straight line and find the value of y.

Solved examples on transformation of general equation into intercept form:

1. Transform the equation of the straight line 3x + 2y - 18 = 0 to intercept form and find its x-intercept and y-intercept.

Solution:

The given equation of the straight line 3x + 2y - 18 = 0

First add 18 on both sides.

⇒ 3x + 2y =18

Now divide both sides by 18

⇒ \(\frac{3x}{18}\) + \(\frac{2y}{18}\) = \(\frac{18}{18}\)

⇒ \(\frac{x}{6}\) + \(\frac{y}{9}\)  = 1,

which is the required intercept form of the given straight line 3x + 2y - 18 = 0.

Therefore, x-intercept = 6 and y-intercept = 9.

2. Reduce the equation -5x + 4y = 8 into intercept form and find its intercepts.

Solution:

The given equation of the straight line -7x + 4y = -8.

First divide both sides by -8

⇒ \(\frac{-7x}{-8}\) + \(\frac{4y}{-8}\) = \(\frac{-8x}{-8}\)

⇒ \(\frac{7x}{8}\) + \(\frac{y}{-2}\) = 1

⇒ \(\frac{x}{\frac{8}{7}}\) + \(\frac{y}{-2}\) = 1,

which is the required intercept form of the given straight line -5x + 4y = 8.

Therefore, x-intercept = \(\frac{8}{7}\)  and y-intercept = -2.

● The Straight Line

• Straight Line
• Slope of a Straight Line
• Slope of a Line through Two Given Points
• Collinearity of Three Points
• Equation of a Line Parallel to x-axis
• Equation of a Line Parallel to y-axis
• Slope-intercept Form
• Point-slope Form
• Straight line in Two-point Form
• Straight Line in Intercept Form
• Straight Line in Normal Form
• General Form into Slope-intercept Form
• General Form into Intercept Form
• General Form into Normal Form
  
• Point of Intersection of Two Lines
  
• Concurrency of Three Lines
• Angle between Two Straight Lines
• Condition of Parallelism of Lines
• Equation of a Line Parallel to a Line
• Condition of Perpendicularity of Two Lines
• Equation of a Line Perpendicular to a Line
• Identical Straight Lines
• Position of a Point Relative to a Line
• Distance of a Point from a Straight Line
• Equations of the Bisectors of the Angles between Two Straight Lines
• Bisector of the Angle which Contains the Origin
• Straight Line Formulae
• Problems on Straight Lines
• Word Problems on Straight Lines
• Problems on Slope and Intercept

11 and 12 Grade Math 

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