Clearest Point- Writing Assignment Chapter 1
There is one subject area in Chapter 1 that I am very confident in and that is Lines. I understand the different equations for lines like the point-slope form; ( y- y1 = m (x-x1) with m=slope and (x1, y1) being a point on the line), the slope-intercept form; ( y= mx + b, with m=slope and b=the y-intercept) and the standard form; (ax + by = c , a and b≠0 in the same equation).
I also can find lines that are parallel or perpendicular to another line. Parallel means there are two lines with the same slope and perpendicular lines have slopes that are the negative reciprocal of the other line’s slope. I also understand how to graph all these types of lines by hand and on my calculator. I can find the slopes and y-intercepts starting from any equation form with a starting point. For example, if I was given two points on a line, I could use the slope formula, y2-y1 / x2-x1, to find the slope of the line. Then, I could plug this slope and one point into the point-slope form. To graph the line, I would have to simplify the point-slope equation into the slope-intercept equation so that I would know the y-intercept and the slope, making graphing the line easy. All that you have to do is plot the y-intercept, (0,x) up or down on the y-axis of the graph and the slope by using the y-intercept as a reference. I will explain this example: Find a line that passes through (1, -5) and that is perpendicular to the line -7x-5y=18. First, you have to rearrange the equation so that it is in slope-intercept form. You have to get y alone. To do that you add the 7x to the right side of the equation and then divide the entire equation by -5 to get y alone on the left side of the equation. Simplified, you obtain an equation of y= – (7/5)x – (18/5).
The Term Paper on Diophantine Equations
1.INTRODUCTION: The mathematician Diophantus of Alexandria around 250A.D. started some kind of research on some equations involving more than one variables which would take only integer values.These equations are famously known as “DIOPHANTINE EQUATION”,named due to Diophantus.The simplest type of Diophantine equations that we shall consider is the Linear Diophantine equations in two variables: ...
The only part of this you need to worry about is the slope. This slope, – (7/5), needs to be rearranged because it must be changed to find the equation of a perpendicular line. To find the slope of a perpendicular line, you find the negative reciprocal. The negative reciprocal of –(7/5) would be (5/7).
All I did was switch the sign (because the negative of a negative number is positive) and flip the fraction, which is what a reciprocal is. Next, you have to use the point given, (1, -5) by plugging it into an equation (that is in slope-intercept form) to find the b value, which is the y-intercept. (The b is in replacement of the -(18/5) from the first equation found.) You plug 1 in for x and -5 in for y. Using the new perpendicular slope and this point, your equation becomes (-5)=(5/7)(1) + b. Next you must get b alone. To do this you subtract (5/7) from the right side and move it to the right side. Now you are left with (-5)-(5/7)=b. To find the left side, you must find a common denominator to subtract the two values, which is 7. Therefore -5 becomes (-35/7).
(-35/7)-(-5/7) = (-40/7)= b. Now that you know the slope, (5/7) and the y-intercept of the line, (-40/7), you use these two pieces of information to write an equation for a line that passes through point (1, -5) and is perpendicular to the line -7x-5y=-18. The equation (in slope-intercept form) is y=(5/7)x –(40/7).
