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Alena Robel
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∙ 13y agoAB is 2
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The slope of line ab is 3 what is the slope of line CD if CD is parallel to ab?
3
Can a line segment have 2 midpoints?
No it has to have at least 3 or more
The length of the red line segment is 8 and the length of the blue line segment is 3 How long is the major axis of the ellipse?
4 11 10.8
If the midpoint of a horizontal line segment with a length of 8 is 3 -2 then the coordinates of its endpoints are?
If the midpoint of a horizontal line segment with a length of 8 is (3, -2), then the coordinates of its endpoints are (6, -2) and (0, -4).
What is the midpoint of a line segment with endpoints -1 -1 and 3 3?
Midpoint: (1, 1)
M is the midpoint of line segment AB. A has coordinates (3-7) and M has coordinates (-11). What is the coordinates of B?
B is (-5, 9).
The slope of line ab is 3 what is the slope of line CD if CD is parallel to ab?
3
How do you divide a triangle into 2 triangles and 2 trapezoids?
Let's assume the triangle has points A, B, and C. Method 1 (3 lines) Draw two lines across the triangle parallel to line segment AB. Now you have two trapezoids and one triangle. Draw another line from C to the any point on the closest of the two lines you just drew, splitting the triangle into two more triangles. Method 2 (2 lines) Draw one line across the triangle parallel to line segment AB. Now you have one trapezoid and one triangle. Draw a second line that passes through C and is perpendicular to AB, splitting the trapezoid into two trapezoids and the triangle into 2 triangles. Method 3 (3 lines) Draw one line from point C to any point on line segment AB. Then draw a line parallel to AC and one parallel to BC, but don't let them cross the line you just drew.
How the segment addition postulate can be modeled using a real world object?
the world is an oval so ab make a line so if you dived what you said by 2 it equals 3
A line segment has an endpoint at (3, 2). If the midpoint of the line segment is (6, -2), what are the coordinates of the point at the other end of the line segment?
(9,4)
Given two points A and B in the three dimensional space what is the set of points equidistant from A and B?
A plane is the set of all points in 3-D space equidistant from two points, A and B. If it will help to see it, the set of all points in a plane that are equidistant from points A and B in the plane will be a line. Extend that thinking off the plane and you'll have another plane perpendicular to the original plane, the one with A and B in it. And the question specified that A and B were in 3-D space. Another way to look at is to look at a line segment between A and B. Find the midpoint of that line segment, and then draw a plane perpendicular to the line segment, specifying that that plane also includes the midpoint of the line segment AB. Same thing. The set of all points that make up that plane will be equidistant from A and B. At the risk of running it into the ground, given a line segment AB, if the line segment is bisected by a plane perpendicular to the line segment, it (the plane) will contain the set of all points equidistant from A and B.
How do you find the midpoint of a line segment?
The midpoint of a line segment would be to add up all of the numbers in the segment. Then divide by the number of numbers that were added. For example if a 1, 3, and 8 were added, divide by 3. This will give the midpoint of the line segment.
If line AB contains the points 4 19 and -3 5 what is the slope of a line perpendicular to AB?
-1/2 or -0.50
What is the midpoint of a line segment with the endpoints (-4 -3) and (7 -5)?
The midpoint of the line segment of (-4, -3) and (7, -5) is at (1.5, -4)
If AB contains the points 6 -2 and -3 16 what is the slope of a line perpendicular to AB?
0.5
How do you work out an equation for the perpendicular bisector of the line segment AB when A is at -4 8 and B is at 0 -2?
First find the midpoint of the line segment AB which is: (-2, 3) Then find the slope of AB which is: -5/2 The slope of the perpendicular bisector is the positive reciprocal of -5/2 which is 2/5 Then by using the straight line formula of y-y1 = m(x-x1) form an equation for the perpendicular bisector which works out as:- y-3 = 2/5(x-(-2)) y = 2/5x+4/5+3 y = 2/5x+19/5 => 5y = 2x+19 So the equation for the perpendicular bisector can be expressed in the form of:- 2x-5y+19 = 0
How do you draw a triangle with an area of 5 square units?
== == 1) Draw a line segment AB of 5 units 2) Draw the perpendicular bisector CD of AB such that Cd meerts AB at C. 3) Mark off CE = 2 units on CD 4) Draw the straight line segments AE & BE. ABE is your triangle. Its base (AB) = 5 and height (CE) = 2, so its area = [base x ht] / 2 = 5 sq units
