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What are the distances from a point on the perpendicular bisector to the endpoints of a segment?
The distance will be length of the line divided by 2 because the perpendicular bisector cuts through the line at its centre and at right angles
How can you construct a perpendicular bisector of a given segment using only two parallel edges of a straightedge with out usign any mesurements on the ruler and compass?
Do I have a compass to use or not ? It's not clear from your question, but since you mentioned it at the end of the question, I'll assume that I do have a compass, and in that case, I only need one straight-edge. 1). Plant the compass on one end of the line segment, open it to more than half the length of the segment, draw a long arc that crosses the segment. 2). Keep the same opening, pick up the compass. 3). Plant the compass on the other end of the segment, draw another long arc that crosses the segment. 4). Sell the compass. 5). The two arcs intersect at two points on opposite sides of the segment. With your straight edge, draw a line between these two points. That line is the perpendicular bisector of the original segment.
What is equidistant from the two end points of a line segment?
the middle point * * * * * In 2 dimensions: also any point on line forming the perpendicular bisector of the line segment. In 3 dimensions: the plane formed by the perpendicular bisector being rotated along the axis of the line segment. In higher dimensions: Hyperplanes being rotated along the same axis.
How do you make square root spiral by paper folding and paper cutting?
First of all draw a line segment that is about 2 cm long between two points P0 and P1. At the one of the outer points, draw another line that is at an angle of 90 degrees from the first line segment. This will cause the new line segment to stand straight on the first segment. Draw another line segment between the not used endpoint of the new line segment, let's call it P2, and the not used endpoint of the first line segment. This will create a triangle. Now on the P2 endpoint, draw another line segment that is again at 90 degrees angle. Repeat the previous steps and you will have created a root spiral.
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 do you draw a line segment of root 12cm?
It is easiest to draw it using two right angled triangles.Draw a line AB that is 2 units long. From B, draw BC which is perpendicular to AB and 2 units long. Join AC. From C, draw CD which is perpendicular to AC (clockwise if BC is clockwise from AB, or anticlockwise if BC is anticlockwise) and make CD 2 uinits long. Then AD is a line segment which is sqrt(12) units long.
What is the perpendicular bisector equation of the line segment whose endpoints are at 2 9 and 9 2?
Endpoints: (2, 9) and (9, 2) Midpoint: (5.5, 5.5) Slope of line segment: -1 Perpendicular slope: 1 Perpendicular bisector equation: y-5.5 = 1(x-5.5) => y = x
What are the distances from a point on the perpendicular bisector to the endpoints of a segment?
The distance will be length of the line divided by 2 because the perpendicular bisector cuts through the line at its centre and at right angles
How can you construct a perpendicular bisector of a given segment using only two parallel edges of a straightedge with out usign any mesurements on the ruler and compass?
Do I have a compass to use or not ? It's not clear from your question, but since you mentioned it at the end of the question, I'll assume that I do have a compass, and in that case, I only need one straight-edge. 1). Plant the compass on one end of the line segment, open it to more than half the length of the segment, draw a long arc that crosses the segment. 2). Keep the same opening, pick up the compass. 3). Plant the compass on the other end of the segment, draw another long arc that crosses the segment. 4). Sell the compass. 5). The two arcs intersect at two points on opposite sides of the segment. With your straight edge, draw a line between these two points. That line is the perpendicular bisector of the original segment.
What is equidistant from the two end points of a line segment?
the middle point * * * * * In 2 dimensions: also any point on line forming the perpendicular bisector of the line segment. In 3 dimensions: the plane formed by the perpendicular bisector being rotated along the axis of the line segment. In higher dimensions: Hyperplanes being rotated along the same axis.
How do you make square root spiral by paper folding and paper cutting?
First of all draw a line segment that is about 2 cm long between two points P0 and P1. At the one of the outer points, draw another line that is at an angle of 90 degrees from the first line segment. This will cause the new line segment to stand straight on the first segment. Draw another line segment between the not used endpoint of the new line segment, let's call it P2, and the not used endpoint of the first line segment. This will create a triangle. Now on the P2 endpoint, draw another line segment that is again at 90 degrees angle. Repeat the previous steps and you will have created a root spiral.
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 do you construct 90 degree angle with compass with construction steps?
Draw a line segment and with compass do the following, 1. Take more than half of line segment in compass. 2. From the left end point draw arc on the upper side and lower side of line segment. 3. Draw arc in the same way from right end point such that the arc should cut the arc from left end point . 4. Now join the points in the upper and lower side of the segment where the arcs drawn from left and right end points intersect. 5.you will get perpendicular bisector [ angle of 90 ]
What is square root spiral?
Select any point in the plane as the origin. From the origin, draw a line of unit length. At its end draw a perpendicular line, also of unit length. This is the first line of the spiral. Join the end of this second line to the origin - this will be a line of length sqrt(2). At the end of this line draw a perpendicular line of unit length. This is the second line of the spiral. Join the end of this new line to the origin - this will be a line of length sqrt(3). At the end of this line draw a perpendicular line of unit length. This is the next line of the spiral. Continue until you have had enough! Or, follow the link.
Which equation is the perpendicular bisector of the line segment with endpoints -2 4 and 6 8?
Endpoints: (-2, 4) and (6, 8) Slope: 1/2 Perpendicular slope: -2 Midpoint: (2, 6) Perpendicular bisector equation: y = -2x+10
What is the perpendicular bisector equation meeting the line segment of -4 8 and 0 -2?
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
