chapter1Notes

# chapter1Notes - Chapter 1 Review of Straight Lines We start...

This preview shows pages 1–4. Sign up to view the full content.

Chapter 1 Review of Straight Lines We start with a brief review of properties of straight lines, since these properties are fundamentally important to our understanding of more advanced concepts (tangents, slopes, derivatives). Skills described in this introductory material will be required in many contexts. 1.1 Geometric ideas: lines, slopes, equations Straight lines have some important geometric properties, namely: The slope of a straight line is the same everywhere along its length. Definition: slope of a straight line: y x y x Figure 1.1: The slope of a line (usually given the symbol m ) is the ratio of the change in the y value, ∆ y to the change in the x value, ∆ x . We define the slope of a straight line as follows: Slope = y x v.2005.1 - September 4, 2009 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Math 102 Notes Chapter 1 where ∆ y means “change in the y value” and ∆ x means “change in the x value” between two points. See Figure 1.1 for what this notation represents. Equation of a straight line Using this basic geometric property, we can find the equation of a straight line given any of the following information about the line: The y intercept, b , and the slope, m : y = mx + b. A point ( x 0 , y 0 ) on the line, and the slope, m , of the line: y - y 0 x - x 0 = m Two points on the line, say ( x 1 , y 1 ) and ( x 2 , y 2 ): y - y 1 x - x 1 = y 2 - y 1 x 2 - x 1 Remark: any of these can be rearranged or simplified to produce the standard form y = mx + b , as discussed in the problem set. 1.2 Examples and worked problems The following examples will refresh your memory on how to find the equation of the line that satisfies each of the given conditions. (Note: you should also be able to easily sketch the line in each case.) (a) The line has slope 2 and y intercept 4. (b) The line goes through the points (1,1) and (3,-2). (c) The line has y intercept -1 and x intercept 3. (d) The line has slope -1 and goes through the point (-2,-5). Solutions: v.2005.1 - September 4, 2009 2
Math 102 Notes Chapter 1 (a) We can use the standard form of the equation of a straight line, y = mx + b where m is the slope and b is the y intercept to obtain the equation: y = 2 x + 4 (b) The line goes through the points (1,1) and (3,-2). We use the fact that the slope is the same all along the line. Thus, ( y - y 0 ) ( x - x 0 ) = ( y 1 - y 0 ) ( x 1 - x 0 ) = m. Substituting in the values ( x 0 , y 0 ) = (1 , 1) and ( x 1 , y 1 ) = (3 , - 2), ( y - 1) ( x - 1) = (1 + 2) (1 - 3) = - 3 2 . (Note that this tells us that the slope is m = - 3 / 2.) We find that y - 1 = - 3 2 ( x - 1) = - 3 2 x + 3 2 , y = - 3 2 x + 5 2 . (c) The line has y intercept -1 and x intercept 3, i.e. goes through the points (0,-1) and (3,0). We can use the method in (b) to get y = 1 3 x - 1 Alternately, as a shortcut, we could find the slope, m = y x = 1 3 . (Note that ∆ means “change in the value”, i.e. ∆ y = y 1 - y 0 ). Thus m = 1 / 3 and b = - 1 ( y intercept), leading to the same result. (d) The line has slope -1 and goes through the point (-2,-5). Then, ( y + 5) ( x + 2) = - 1 , so that y + 5 = - 1( x + 2) = - x - 2 , y = - x - 7 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern