Equations in Intercept Form
Date: 08/27/98 at 17:58:47 From: cole Subject: Intercept form Show that an equation for a line with nonzero x-intercepts and y-intercepts can be written as: x/a + y/b = 1 where a is the x-intercept and b is the y-intercept. This is called the intercept form of the equation of a line. I do not understand what they want me to do. Do I solve it like this: bx + ay - ab = 0? Or do I substitute with variables? Can you please help?
Date: 08/27/98 at 18:59:01 From: Doctor Jaffee Subject: Re: Intercept form Hi Cole, There are a number of ways to approach this problem. I'll present two of them to you to help you be clearer about the concept. First of all, let's consider the situation geometrically. What is an x-intercept? If you say that it is a point on the x-axis where the line crosses, you are right. But what is the y value at that point? If you say 0, you are right again. So, let's go back to the equation x/a + y/b = 1 and replace the y with a zero. You get x/a = 1, which means x must equal a. Now putting this back into geometric terms, we have "when the line crosses the x-axis (that is, when y = 0), the x number must be a (that is, the x-intercept is a). Likewise, the y-intercept is the point where the line crosses the y-axis, and at that point x must equal 0. If you substitute 0 for x in the original equation you eventually end up with y = b, or in other words, the y-intercept is b. Let's look at a second approach. Suppose we start with the equation Ax + By = C. It is important to note that "A" doesn't mean the same as "a". Capital letters generally mean some different value from lower case letters. If we divide both sides by C, we get (A/C)x + (B/C)y = 1. Suppose we are told that a and b are the x- and y-intercepts, respectively. That means that (a,0) and (0,b) are points on the line. If we substitute a for x and 0 for y we get (A/C)*a = 1. That means that A/C must equal 1/a. Furthermore, if we substitute 0 for x and b for y, we get B/C = 1/b. Therefore x/a + y/b = 1 is the equation. I hope this explanation has helped. Write back if it needs further clarification and I or one of the other Doctors will try to help you out. Also, write back with any other questions you have. - Doctor Jaffee, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum