1. ## linear modelling

ImageShack - Hosting :: math01lt1.jpg - question 6
ImageShack - Hosting :: photomathdy6.jpg - graph

how do i do question 6?
can i do it by hand instead of calculator?

2. First of all, your graph is wrong if you keep to the criteria of your instructions. Points B and E must be in quadrant I and II, respectively. So, why not let the coordinates be B(6, 1) and E(-6, 1), although you could put them anywhere you want in those quadrants. Nothing is said about your pentagon being a regular pentagon. Your other coordinates are ok.

A(0, 6), B(6, 1), C(4, -6), D(-4, -6), E(-6, 1)

You don't need a calculator for this.

2. Find the equations of each line that creates your pentagon.

Let's start with the easy ones: Lines AB and AE. Both have y-intercepts of 6. Now, lets find their slopes:

slope of line AB = $\frac{y_2-y_1}{x_2-x_1} = \frac{1-6}{3-0} = \frac{-5}{3}$

slope of line AE = $\frac{1-6}{-3-0} = \frac{5}{3}$

Using the slope intercept form of a linear equation (y = mx + b), we have the equation of line AB to be $y = \frac{-5}{3}x + 6$

Use a similar technique to find the equation of line AE. You can do it.

Now, let's work on line BC. Find the slope like we did before. The slope is $\frac{7}{2}$

This time, use the point-slope form to find the linear equation for BC.
$y-y_1=m(x-x_1)$
Use either point B or C. Let's use B.

$y-1=\frac{7}{2}(x-6)$

If you want to convert that to slope-intercept form, it would be $y=\frac{7}{2}x-20.$

The equations of the other lines can be found accordingly. Good luck.

The slope of AB was $\frac{-5}{3}$, and the slope of AE was $\frac{5}{3}$. They are not negative reciprocals of each other so the peak of the pentagon does not meet at right angles.

To find the perimeter, you must use the distance formula to determine the length of each segment in you pentagon. Remember the distance formula?

$d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Let me know how you are doing and I'll get back to you.