Questions on Geometry: Points, lines, angles, perimeter answered by real tutors!

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Lines m and n intersect at point A. Line k is perpendicular to both lines m and n at point A. Which statement must be true?
1) Lines m, n and k are in the same plane.
2) Lines m and n are in two different planes.
3) Lines m and n are perpendicular to each other.
4) Line k is perpendicular to the plane containing lines m and n.
The correct answer choice is number 4, but I don't know why it is not 3.
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If you draw 2 intersecting lines on paper, a 3rd line can be perpendicular to both if the 1st 2 lines are perpenicular, or if they're not.
3 might be true, but is not a requirement.
Use a paper on a desk and do it. a pencil held perpendicular to both lines is perpendicular, actually the term is normal, to the paper. This is true regardless of the angle between the lines on the paper.
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PS 3-space is a lot more than 50% harder than 2-space.

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MUST is the key

#3 COULD be true (or not), but #4 MUST be true

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Circle -- a bit tricky to prove using elementary geometry techniques, but we can definitely find an example. For example, suppose that the area of each shape is 1. It can easily be shown that the rectangle with fixed area and the minimal perimeter is a square. The perimeter of the square is 4, and now you can just find the radius of the circle with area 1, then find the circumference.

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to begin with, we need to understand what ordinate and abscissa mean.
the following link addresses that issue:
http://www.blurtit.com/q550444.html
the short answer is:
abscissa is the x value of the coordinate.
ordinate is the y value of the coordinate.
they are giving you the x value of the point P.
that's the abscissa of 1.
you have to find the ordinate of the point P.
that would be the y value of the point P.
coordinates are given in pairs of (x,y).
x is the first value of the pair and points to the x-value of the point.
y is the second value of the pair and points to the y-value of the point.
a diagram of what i think your triangle will look like is shown below:
the triangle is labeled APC.
there is a perpendicular from point P to point D on AC.
point A is (-4,0)
point P is (1,y) ***** y is the value we are looking for.
point C is (4,0)
point D is (1,0)
the abscissa of point P is 1 which makes it the x value of the coordinate pair represented by point P.
you are looking for the ordinate of point P.
the ordinate is the y value of the coordinate pair.
i labeled it "y" until we could find a value for it.
you are given that the distance between point A and C is 8.
you are also given that the abscissa of the point P is 1.
this means that the vertical line to point P lies on the value of x = 1.
this means that there are 5 units from point A to D.
this means that there are 3 units from point D to C.
you need to find the length of the line from point P to D.
this is the altitude of the right triangles formed.
the 2 right triangles formed are APD and CPD.
all you know is that the sum of the distance from point P to A and P to C is equal to 9.
in the diagram i labeled the following:
the length of the line AP is called b.
the length of the line PC is called a.
the length of the line PD is called c.
by using the pythagorean formula, we can find the value of c.
the pythagorean formula states that the hypotenuse squared is equal to the sum of the legs squared.
in triangle APD, b is the hypotenuse and 5 and c are the legs.
in triangle CPD, a is the hypotenuse and 3 and c are the legs.
our formulas that we start with are:
5^2 + c^2 = b^2
3^2 + c^2 = a^2
we solve for c^2 in both formulas to get:
c^2 = b^2 - 5^2
c^2 = a^2 - 3^2
since they are both equal to c^2, then they are both equal to each other, so we get:
b^2 - 5^2 = a^2 - 3^2
we are given that a + b = 9
from that we can solve for b in terms of a to get:
b = 9 - a
we can then substitute for b in the equation of b^2 - 5^2 = a^2 - 3^2 to get:
(9 - a)^2 - 5^2 = a^2 - 3^2
we now solve for a.
we simplify the equation to get:
81 - 18a + a^2 - 5^2 = a^2 - 3^2
we subtract a^2 from both sides of the equation and we add 5^2 to both sides of the equation to get:
81 - 18a + a^2 - a^2 - 5^2 + 5^2 = a^2 - a^2 - 3^2 + 5^2
we combine like terms to get:
81 - 18a = -3^2 + 5^2
we simplify further to get:
81 - 18a = -9 + 25
we combine like terms again to get:
81 - 18a = 16
we subtract 81 from both sides of this equation to get:
- 18a = 16 - 81
we combine like terms to get:
- 18a = - 65
we divide both sides of this equation by -18 to get:
a = 65/18 which simplifies to:
a = 3.611111111
sinca a + b = 9, this means that:
a = 3.611111111
b = 5.388888889
we now have a value for a and b and we can use those values to solve for c using the pythagorean formula again.
the result of that operation is that:
c = 2.010005835
we rounde all answers to the nearest hundredth or whatever rounding is required to get:
a = 3.61
b = 5.39
c = 2.01
the value of c is the length of the line PD which also becomes the y value of the coordinate of P.
the coordinate of P becomes (1,2.01)
the diagram of your triangles formed is shown below:

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your second problem starts here
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the center of your circle is at (-3,-2)
your chord is bisected at (3,1)
the length of chord is 2.
you want to find the length of the radius of your circle.
the attached diagram shows what i think is the relationship you are looking for.
the radius of a circle can form the perpendicular bisector of any chord in that circle.
this is one of the theorems in geometry that you should probably have already studied.
this is because the radii of the circle form an isosceles triangle with the chord as shown in the diagram.
that isosceles triangle is ABC in the diagram.
it is an isosceles triangle because the sides of the triangle are the radii of the circle.
since you have an isosceles triangle, then the altitude of that triangle is the perpendicular bisector of the base of that triangle which is the chord.
in your problem, the center of the circle is at (-3,-2) and the bisecting point on the chord is (3,1).
this means that you can draw a line from the center of the circle to that point and the line formed is the perpendicular bisector of the triangle.
that perpendicular bisector forms 2 right triangles which are triangle ADB and ADC.
the base of each of those triangle is 2.
the length of the perpendicular bisector which is the line AD is given by the equation:
AD = sqrt (x^2 + y^2) where x and y are the coordinate points of point A and point D.
the formula becomes:
AD = sqrt (-3 - 3)^2 + -2 - 2)^2 which becomes:
AD = sqrt ((-6)^2 + (-4)^2) which becomes:
AD = sqrt (9 + 16) which becomes:
AD = sqrt(25) which becomes:
AD = 5
the length of AD is equal to 5 as shown in the diagram.
the length of AB is found using the pythagorean formula of:
(AB)^2 = 5^2 + 2^2 which becomes:
(AB)^2 = 25 + 4 which becomes:
(AB)^2 = 29
take square root of both sides of this equation to get:
AB = sqrt(29).
that's your answer.
if you solve for AC, you will find that it's value is also sqrt(29).
diagram for your second problem is shown below:

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There are 4 equal sides, so the percentage is (1/4)*100 = 0.25*100 = 25%

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you need the equation of the line formed by the points (-6,1) and (-1,2).
the equation for this line will be:
y = (1/5)x + 11/5
a graph of this equation looks like this:

now you need the equation of the line perpendiular to this line and passing through its midpoint.
the midpoint of that line would be the point halfway between (-6,1) and (-1,2)
that point would be:
(-7/2,3/2)
the equation of the line perpendicular to the original line and passing through that point would be:
y = -5x -16
a graph of that line superimposed on the original line looks like this:

all points that are equidistant from the points of (-6,1) and (-1,2) will be on this perpendicular line.
all points 5 units from the point (-2,7) will be on the circumference of the circle that surrounds (-2,7) with a radius of 5 units.
the formula for a circle is:
(x-h)^2 + (y-k)^2 = r^2
the point (h,k) is the center of the circle.
that makes (h,k) = (-2,7) which makes h = -2 and k = 7.
the formula for the circle becomes:
(x+2)^2 + (y-7)^2 = r^2
since the radius of the circle has to be 5, then r = 5 and r^2 = 25.
the equation for the circle becomes:
(x+2)^2 + (y-7)^2 = 25
to graph this equation, solve for y to get:
y = +/- sqrt(25-(x+2)^2) + 7
the graph of this equation, superimposed on the graph of the other 2 lines, will look like this:

it appears there are 2 points that will satisfy the equation.
those points are the intersection of the line perpendicular to the original line and the circumference of the circle.
a picture of the final graph with the relationships drawn out is shown below:

the point (-5.34,10.72) is 9.74 units from (-6,1) and (-1,2) and is also 5 units away from (-2,7).
the point (-3.66,2.28) is 2.67 units away from (-6,1) and (-1,2) and is also 5 units away from (-2,7).
both the points (-5.34,10.72) and (-3.66,2.28) are on the circumference of the circle and on the line perpendicular to the original line and intersecting halfway between the points (-6,1) and (-1,2).
the points (-3.66,2.28) and -5.34,10.72) have been rounded to 2 decimal places.
those are the solutions to the problem.
the solution is 2 points rather than just 1.

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the abscissa of a point on the fourth quadrant is numerically 3 times its ordinate and is 10 units from (-2,4). find the point.
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Let the point be (x,y)
Equations:
x = 3y
10 = sqrt[(x+2)^2+(y-4)^2]
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Substitute for "x" and solve for "y":
10 = sqrt[(3y+2)^2 + (y-4)^2]
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100 = 9y^2+12y+4 + y^2-8y+16
10y^2+4y+20 = 100
5y^2 + 2y-40 = 0
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y = 2.6355 or y = -3.0355
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The 4th quadrant has y being negative and x being positive
But x = 3y implies that x and y have the same sign.
There is a contradiction here.
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If x = 2.6355, y = 7.9065 (this point would be in the 1st quadrant)
If x = -3.0355, y = -9.1065 (this point would be in the 3rd quadrant)
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Cheers,
Stan H.
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equation is y=4
This line is parallel to x axis
a perpendicular line to this will be parallel to y axis.
(3,7) ----its x coordinate is 3
so the equation will be x=3

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True, just look at Pascal's triangle and you'll find them easily.

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Let the length of a side of the square be given by s.
The perimeter of a square is the sum of the four sides or 4s.
You are told the perimeter is 96. So

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R, V and T and colinear points. You are told a ray can be drawn as RTV.
You are told the length from R to V is 31 and the length from T to V is 14.

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Start with the general form in slope-intercept form:

Find the slope of the given line whose equation is:
rewrite in slope-intercept form:
The slope, .
The slopes of perpendicular lines are negative reciprocals.
The slope of the new line is:
, so...
To find b substitute the x- and y-coordinates of the given point (4, 4) and solve for b.

, finally:
is the final equation.

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Lines do not have a finite length, hence it is impossible to have two lines of unequal length, parallel or otherwise.

Now, if what you really meant was "...two parallel line segments of unequal length," that is another story.

Now the problem is that it is not clear whether you are talking about some existing line segment between the two parallel lines or some hypothetical line segement that you are supposed to construct between the two parallel lines, probably a perpendicular.

I also don't know what other information you have. In either case above you have to either be given or be able to derive the coordinates of the points where the line segment whose length you need to calculate intersects the two parallel lines. Then use the distance formula:



where and are the coordinates of the intersecting points.

John

My calculator said it, I believe it, that settles it

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what?

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Lets name the sides f (first) s (second) and t (third)
We know that
f+s+t=34
s=7f
t=9f
using substitution
f+7f+9f=34
17f=34
f=2
then we can plug in 2 for f and get s=14 and t=18

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That's not an equation. There's no equal sign.

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2(1.5x+14)+2x=148
3x+28+2x=148
5x=120
x=120/5
x=24
1.5x+14=50

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The answer is B. One since there is only one unique normal line to any given plane

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All angles are equal in a square

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in an acute triangle with one angle 47 degrees and then i have to find x and there is and exterior angle of 114 degrees can u help me find x
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The supplement of the 114 degree angle = 180-114 = 66 degrees
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The 3 angle is 180-(66+47) = x
x = 180-113
x = 67 degrees
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Cheers,
Stan H.
============

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A and E are the first two, and X and Y  are the
last two letters of the alphabet which can't be 
drawn without lifting the pencil.
In a pencil or pen drawing, if some straight or curved lines 
meet at a point, or ends at a single point, that point is 
called a "vertex" (plural "vertices"), or "node". 
If the number of straight or curved lines meeting at a vertex
is even, we say the vertex is even, and if that number is odd,
we say the vertex is odd.  If a straight or curved line ends at
a point, we say that point is odd since only one line issues
from that point, and 1 is an odd number.
These capital letters in simplest form: 
B, C, D, I, J, L, M, N, O, P, R, S, U, V, W, Z
can all be drawn without lifting the pencil or retracing.
because each has no more than 2 odd vertices in the simplest
way to.  That's because any pencil or pen drawing which can 
be drawn without lifting the pencil or pen or retracing must 
have 1 beginning and 1 end.  That's 2, and no more!  
These can't be drawn without lifting or retracing:
A, E, F, G, H, K, Q, T, X, Y
Here is an example of one letter, A, that cannot be drawn without 
taking up the pen or pencil or retracing, and one letter, P, which 
can be. The odd vertices are indicated with a red circle and the 
even ones with a green circle.  The A has 4 odd vertices and 1 
even vertex.  So it can't be drawn without lifting the pen or pencil
or retracing, because it has more than 2 odd vertices. However
The P has 1 even vertex and 2 odd vertices, so it can be drawn
without lifting the pen or pencil or retracing.
 
But, again, the answer to your problem is A,E, and X,Y.

Edwin