# Analytical Geometry

• Jun 20th 2006, 11:56 AM
SaRah<3
Analytical Geometry
Hey guys.
I need lots of help.
I need to find out how to do these questions: Answers (showing your work, like how to get the answer) would be really great. I'm really stuck with this stuff. And its drving me insane because I don't undertsand it.

1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5). Find the area of the triangle.

2. For the line segment DE, one endpoint is D(3,1) and the midpoint is M(0,-4). Find the coordinates of endpoint E.

3.Triangle DEF is a right triangle with vertices D(-2,5), E(-4,1), and (2,3). Verify that the midpoint of the hypotenuse is equidistant from the three vertices of the triangle.

4. Verify that the vertices A(-6,1), B(2,-5), C(6,1), and D(2,4)are the vertices of a trapezoid.

5. Quadrilaterial RSTU has vertices R(3,2), S(0,4), T(-2,1), and U(1,-1). Verify that
a) quadrilateral RSTU is a square
b) the diagonals of quadrilateral RSTU perpendicularly bisect each other and are equal in length.

I know these are alot of questions....but if you guys could try and anser them all i'd really appercaite it.
• Jun 20th 2006, 12:28 PM
earboth
1. problem only
Quote:

Originally Posted by SaRah<3
Hey guys.
I need lots of help.
I need to find out how to do these questions: Answers (showing your work, like how to get the answer) would be really great. I'm really stuck with this stuff. And its drving me insane because I don't undertsand it.
1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5). Find the area of the triangle. ...

Hello,

first you have to calculate the length of both legs of this triangle. Use the Pythagorian Theorem:

$\overline{AC}=\sqrt{(3-1)^2+(3-(-5))^2} = \sqrt{68}$

$\overline{BC}=\sqrt{(-1-1)^2+(4-(-5))^2} = \sqrt{85}$

$\overline{AB}=\sqrt{(3-(-1))^2+(3-4))^2} = \sqrt{17}$

As you can see at once (BC) is the hypotenuse.

Because a right triangle is a half rectangle the area is calculated here by:

$a=\frac{1}{2}\cdot \overline{AC}\cdot \overline{AB}$. So:

$a=\frac{1}{2}\cdot \sqrt{68}\cdot \sqrt{17}= 17$.

Greetings

EB
• Jun 20th 2006, 12:50 PM
SaRah<3
hey thanks alot.

ummm do you think you could help me with the other questions I put in the first post?
Even just a few of the questions would be real great.
• Jun 20th 2006, 12:56 PM
Soroban
Hello, SaRah<3!

Here's #1 . . . in baby steps . . .

Quote:

1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5).
Find the area of the triangle.

A sketch is always a good idea . . .
Code:

    (-1,4)       B o  |           * |             *         *  | *             |  *           * |    *             |      *           *|        *             |          o A             *        * (3,3)       - - - + - - - * - - -             |*    *             |  *             | o             | C (1,-5)

Even a rough sketch makes us suspect that: $AB \perp AC$

We find that: . $m_{AB}\:=\:\frac{4-3}{-1-3}\:=\:\frac{1}{-4}\:=\:-\frac{1}{4}$

. . and that: . $m_{AC}\:=\:\frac{-5-3}{1-3}\:=\:\frac{-8}{-2}\:=\:+4$

Since the slopes are negative reciprocals of each other: $\angle A = 90^o$

The base of the triangle is: . $AC\;=\;\sqrt{(1-3)^2 + (-5-3)^2}\:=\:\sqrt{68}\:=\:2\sqrt{17}$

The height of the triangle is: . $AB\;=\;\sqrt{-1-3)^2 + (4-3)^2}\:=\;\sqrt{17}$

The area of a triangle is: . $A \;=\;\frac{1}{2}(base)(height)$

. . so we have: . $A\;=\;\frac{1}{2}\left(2\sqrt{17}\right)\left( \sqrt{17}\right) \;= \;17$ square units.

Edit: EB's too fast for me . . . and with a shorter solution!
• Jun 20th 2006, 01:11 PM
Soroban
Hello again, SaRah<3!

Here's #4 . . .

Quote:

4. Verify that the vertices A(-6,1), B(2,-5), C(6,1), and D(2,4)are the vertices of a trapezoid.

A trapezoid is a quadrilateral with one pair of parallel sides.

From a quick sketch (which I'll omit here), we suspect that: $AB \parallel CD$

We find that: . $m_{AB} \:=\:\frac{\text{-}5-1}{2-(\text{-}6)}\:=\;\frac{\text{-}6}{8}\;=\;-\frac{3}{4}$

. . . and that: . $m_{CD}\:=\:\frac{4 - 1}{2 - 6}\:=\:\frac{3}{\text{-}4}\:=\:-\frac{3}{4}$

We were right! . . . $AB \parallel CD$

Therefore, $ABCD$ is a trapezoid.

• Jun 20th 2006, 01:12 PM
earboth
Quote:

Originally Posted by Soroban
[size=3]...
Edit: EB's too fast for me . . . and with a shorter solution!

Hello,

I'm sorry - but I live at least 6 hours east of you. So I am slightly in the lead. (I found this expression in my dictionary and I'm not certain if it is appropriate here).

Best wishes

EB
• Jun 20th 2006, 01:38 PM
SaRah<3
Thanks so much for these guys!
Try and keep them coming if you can :) :D
• Jun 20th 2006, 03:02 PM
Soroban
Hello, SaRah<3!

Don't get discouraged . . .
Think of all the stuff you do know: distance formula, slope formula, etc.
These problems make you apply them in various ways.
You must learn to Think your way through them . . .

Quote:

5. Quadrilaterial RSTU has vertices R(3,2), S(0,4), T(-2,1), and U(1,-1).
Verify that:
a) Quadrilateral RSTU is a square
b) The diagonals of quadrilateral RSTU perpendicularly bisect each other and are equal in length.

$a)$ How do we prove that a quadrilateral is a square?
We must show that the four sides are equal (making it a rhombus)
. . then show that it has one right angle.

To show equals sides, we need the distance formula for find the lengths, right?

$\overline{RS}\:=\:\sqrt{0-3)^2 + (4-2)^2} \:= \:\sqrt{4 + 9}\:=\:\sqrt{13}$
$\overline{ST} \:=\:\sqrt{(\text{-}2-0)^2 + (1-4)^2} \:=\:\sqrt{4 + 9} \:=\:\sqrt{13}$
$\overline{TU} \:=\:\sqrt{(1+2)^2 + (\text{-}1-1)^2} \:=\:\sqrt{9+4}\;=\;\sqrt{13}$
$\overline{UR}\:=\:\sqrt{(3-1)^2 + (2+1)^2} \:=\:\sqrt{4+9}\:=\:\sqrt{13}$
. . All four sides are equal; we have a rhombus - opposite sides are parallel.

The slope of $RS$ is: . $m_{RS}\:=\:4 - 2}{0 - 3} \:=\:-\frac{2}{3}$
The slope of $ST$ is: . $m_{ST} \:=\:\frac{1 - 4}{\text{-}2 - 0} \:=\:+\frac{3}{2}$

. . Hence: $RS \perp ST,\;\angle S = 90^o$

Therefore, quadrilateral $RSTU$ is a square.

$b)$ We must show that the diagonals are perpendicular,
bisect each other, and are equal in length.

The diagonal from $R(3,2)$ to $T(-2,1)$

. . It has slope: . $m_{RT}\:=\:\frac{1-2}{\text{-}2-3}\:=\:\frac{1}{5}$

. . Its midpoint is: . $M_{RT}\:=\:\left(\frac{3-2}{2},\;\frac{2+1}{2}\right)\:=\:\left(\frac{1}{2} ,\,\frac{3}{2}\right)$

. . Its length is: . $\overline{RT}\:=\:\sqrt{(\text{-}2-3)^2+(1-2)^2}\:=\:\sqrt{25+1}\:=\:\sqrt{26}$

The diagonal from $S(0,4)$ to $U(1,-1)$

. . It has slope: . $m_{SU}\:=\:\frac{\text{-}1-4}{1-0}\:=\:-5$

. . Its midpoint is: . $M_{SU}\:=\:\left(\frac{0+1}{2},\,\frac{4-1}{2}\right)\:=\:\left(\frac{1}{2},\,\frac{3}{2} \right)$

. . Its length is: . $\overline{SU}\:=\:\sqrt{(1-0)^2+ (\text{-}1-4)^2}\:=\:\sqrt{1 = 25}\:=\:\sqrt{26}$

Therefore:
. . Their slopes are $\frac{1}{5}$ and $-5$ . . . they are perpendicular.

. . They have the same midpoint . . . they bisect each other.

. . They both have length $\sqrt{26}$ . . . they are equal.

• Jun 20th 2006, 05:00 PM
SaRah<3
Wow I wish I was as smart as you. Math is so difficult for me. Im currently failing my grade 10 course. :(
The exam is tommorow.
Ill be so happy if I pass.