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No. A line can be contained by many, many planes, Picture this, A rectangle with corners - going clockwise - A, B, C and D is the screen of your computer. This is a plane figure. 1 inch away from it a line runs from A1 to C1. The line is parallel to the plane. Now, take a sheet of paper with corners E, F, G and H, and place corner E at corner A of the screen, and place corner F at corner C of the screen. The Line AI is now 'contained' in the plane EFGH. and EFGH is perpendicular to ABCD.
There are one or infinitely many points.
H, I, and F
I am guessing that you are asking for a compass-and-straight-edge construction of a right angle. Let us say that the line DE is given, and we are given a point F not on the line, and we are asked to drop a perpendicular line from F to DE. Set the compass radius large enough so that when you draw a circle centered at F, it crosses DE in 2 points, say A and B. Now draw a circle centered at A with a radius larger than AB, and a circle of the same radius centered at B. The 2 circles intersect in 2 points, one on each side of the line ABDE. Let C be the one on the opposite side of the line from F. Use the straight edge to draw the line CF. That is your perpendicular. It also bisects DE, by the way. It is known as the perpendicular bisector of DE.
It has both
What must be true? In your example, we have 4 intersecting lines. g and b are parallel, and f and h are parallel. g and b are perpendicular to f and h. It might look like tic-tac toe for example
D(power)=1\f ;f=infinity D=0 for plane mirror
No. A line can be contained by many, many planes, Picture this, A rectangle with corners - going clockwise - A, B, C and D is the screen of your computer. This is a plane figure. 1 inch away from it a line runs from A1 to C1. The line is parallel to the plane. Now, take a sheet of paper with corners E, F, G and H, and place corner E at corner A of the screen, and place corner F at corner C of the screen. The Line AI is now 'contained' in the plane EFGH. and EFGH is perpendicular to ABCD.
There are one or infinitely many points.
They are: E T F H and L
H M E W F Z some other i think
H, I, and F
Many letters of the English alphabet have perpendicular lines. Perpendicular lines are like two lines making one corner of a square. So any letter that has two lines joined like the corner of a square has perpendicular lines.These letters have right angles: E, F, f, H, I, L T, t, and sometimes X, x.Keep in mind that the angles of the lines making the letters is effected by the font and whether or not the letter is italicized. For example, E is made of three perpendicular lines, but this 'E' has no perpendicular lines.
I am guessing that you are asking for a compass-and-straight-edge construction of a right angle. Let us say that the line DE is given, and we are given a point F not on the line, and we are asked to drop a perpendicular line from F to DE. Set the compass radius large enough so that when you draw a circle centered at F, it crosses DE in 2 points, say A and B. Now draw a circle centered at A with a radius larger than AB, and a circle of the same radius centered at B. The 2 circles intersect in 2 points, one on each side of the line ABDE. Let C be the one on the opposite side of the line from F. Use the straight edge to draw the line CF. That is your perpendicular. It also bisects DE, by the way. It is known as the perpendicular bisector of DE.
It has both
The answer to this probably depends on (a) the font and (b) wheter the uppercase letter or the lowercase letters are considered. In this particular font, in uppercase B D E F H I K L M N P R and T all have perpendicular segments, G has a short perpendicular segment J has a perpendicular segment which ends in a curve U has two perpendicular segments joined by a curve and in lowercase b d h i k l m n p r and u all have perpendicular segments a f g j and t all have perpendicular segments with curved parts.
Both RC plane are awsome because they are both planes.