1. ## 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.

2. ## 1. problem only

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

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.

4. Hello, SaRah<3!

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

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!

5. Hello again, SaRah<3!

Here's #4 . . .

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.

6. 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

7. Thanks so much for these guys!
Try and keep them coming if you can

8. 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 . . .

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.

9. 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.

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### Diagonal of Square RSTU

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