![]() ![]() Then enter the relative address for the starting point of the line. This cancels the second point and you are asked to supply the starting point again. You are than asked to "Specify next point." Right click. You are asked to "Specify first point." Click on the current object at your given location (end point, mid point.). Assume that you already have an object on the screen. In this example I will use the 2 Points line command. If you are wanting to start drawing an object a certain distance away from an existing object it is easily accomplished. Scroll towards the bottom of the thread I have linked to below and read my description of the Snap icons. If using an absolute address then a point on a line makes no difference: 0,0 is absolute regardless of any point in the drawing. From there you can use relative addressing to specify a location. As an example, you can turn on Snap to End and the cursor will snap to an endpoint of the closest entity. An entity can be a line, rectangle, circle, etc. The point can be an end point, a middle point, a center point and so on. Snaps are just what the name implies: snap the cursor to a point on an entity. Notice these make the same rectangle, so you could probably just choose one of these for your answer.I'm not sure what you mean when you say, "My issue, right now, is how do I select a specific point from a line, polyline, rectangle, etc." What throws me is the word "from a line." If you mean "on a line" then using a Snap will be the most accurate. Now we have take both of these values, and plug them into our area equation to get a corresponding W. L = / 2(2)įor our last step, we have a ±, which tells us we have two answers: 12 + 3, and 12 - 3. Plug these numbers into the quadratic formula to get: Here, our variable is L instead of x, and our equation is 2L 2 - 48L + 270, so The most direct way would be to use the quadratic formula. Now you can solve a quadratic two ways, either by factoring or using the quadratic formula. This is the most common way to write a quadratic, with the variable squared part first, the variable part, then the number without the variable. I prefer to keep the L 2 part positive, so let's subtract both sides by 48L.ĢL 2 - 48L + 270 = 0. You can solve these by moving everything to one side. Now because we have L 2, L, and a number without L, we have what's called a quadratic. ![]() Because we're dividing 270/L, let's multiply everything by L so there are no fractionsĤ8L = 270 + 2L 2. We can use this in the first equation to get: ![]() Now we know what W is equal to in terms of L. To write this in math terms, let's start with the second equation and solve for W: To be able to solve for a variable, you need to use one equation to solve what a variable is equal to in terms of the other, then use the second equation to rewrite everything in terms of one variable. Now we have two equations with two variables, L and W. To get our area, A you need to multiply a rectangle's length by its width, like below:īecause we know our area is 135, we can also say: The area of a shape is the amount of space contained in its outline. To write our perimeter, P in math terms, it could look like this:īecause our perimeter is equal to 48, we can say: To get the perimeter of a rectangle, you'll add up the length of all its sides. The perimeter of a shape is the length of its outline. Let's use this info to write our two equations. Your question gives us two important pieces of info: the perimeter is equal to 48, and the area is equal to 135. This is the volume of the rectangular shape which corresponds to the dimensions entered for length, width and height. To solve for any number of variables, you'll need at least the same number of independent equations, which means we need to write two equations to be able to find L and W. For problems like this, it might be useful to draw a rectangle and label all its sides, like below:īecause these values are unknown right now, we have two variables, L and W. Rectangles have two dimensions, a length which we will call L, and a width that we will call W.
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