Question 1064005
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Answers:
Area = 15.8 square units
Base = 8.1 units
Altitude = 3.9 units


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Explanation


Use the distance formula to determine the length of side AB
A = (x1,y1) = (5,2) and B = (x2,y2) = (-3,1)
{{{d = sqrt( (x1-x2)^2 + (y1-y2)^2 )}}}


{{{d = sqrt( (5-(-3))^2 + (2-1)^2 )}}}


{{{d = sqrt( (5+3)^2 + (2-1)^2 )}}}


{{{d = sqrt( (8)^2 + (1)^2 )}}}


{{{d = sqrt( 64 + 1 )}}}


{{{d = sqrt( 65 )}}}


{{{d = 8.0623}}} approximately 


{{{d = 8.1}}}


Follow similar steps to find the distance from C = (-2,5) to D = (-1.52,1.18) is roughly 3.85004 which rounds to 3.9


Then,
area of triangle = 0.5*base*height
area = 0.5*AB*CD
area = 0.5*8.1*3.9
area = 15.795
area = 15.8 square units.


There's rounding error going on here. 
The area of triangle ABC should be 15.5 square units exactly. 
This can be determined in the next section below.


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When given the coordinates of the vertex points, we could use the <a href="https://en.wikipedia.org/wiki/Shoelace_formula">shoelace formula</a> to fairly quickly find the area of any polygon. 


Given list of vertex points
(5,2)
(-3,1)
(-2,5)


Make a copy of the first point (5,2) and place it at the bottom of the list
(5,2)
(-3,1)
(-2,5)
(5,2)
This will help form a loop.


Space the x and y coordinates out. Each (x,y) point gets its own row.
{{{
drawing(300,200,-5,5,-5,5,
locate(-3.3333,1.6666,5),locate(-1.3333,1.6666,2),
locate(-3.3333,-0.0001,-3),locate(-1.3333,-0.0001,1),
locate(-3.3333,-1.6668,-2),locate(-1.3333,-1.6668,5),
locate(-3.3333,-3.3335,5),locate(-1.3333,-3.3335,2),
locate(-3.3333, 3.3333,matrix(1,2,"Shoelace","Formula"))
)
}}}


Then draw in the diagonals as indicated below.
{{{
drawing(300,200,-5,5,-5,5,
locate(-3.3333,1.6666,5),locate(-1.3333,1.6666,2),
locate(-3.3333,-0.0001,-3),locate(-1.3333,-0.0001,1),
locate(-3.3333,-1.6668,-2),locate(-1.3333,-1.6668,5),
locate(-3.3333,-3.3335,5),locate(-1.3333,-3.3335,2),
red(line(-2.8333,0.9666,-1.4833,-0.3501)),
red(line(-2.8333,-0.7001,-1.4833,-2.0168)),
red(line(-2.8333,-2.3668,-1.4833,-3.6835)),
blue(line(-2.8333,-3.6835,-1.4833,-2.3668)),
blue(line(-2.8333+0.25,-2.0168,-1.4833,-0.7001)),
blue(line(-2.8333+0.25,-0.3501,-1.4833,0.9666)),
locate(-3.3333, 3.3333,matrix(1,2,"Shoelace","Formula"))
)
}}}
Multiply along the red diagonal pairs and add up those products.
<font color=red>5*1</font>+(<font color=red>-3*5</font>)+(<font color=red>-2*2</font>) = 5-15-4 = <font color=red>-14</font>
Do the same for the blue diagonal pairs.
<font color=blue>5*5</font>+(<font color=blue>-2*1</font>)+(<font color=blue>-3*2</font>) = 25-2-6 = <font color=blue>17</font>

Subtract the results.
<font color=red>red</font> - <font color=blue>blue</font> = <font color=red>-14</font> - <font color=blue>17</font> = -31

We get a negative result. Let's take the absolute value to get |-31| = 31

Lastly, take half of this to get the area = (1/2)*31 = 15.5


The area of the triangle is exactly 15.5 square units.
<a href="https://www.geogebra.org/calculator">GeoGebra</a> can be used to verify.


More practice on the shoelace formula is found <a href="https://www.algebra.com/algebra/homework/Parallelograms/Parallelograms.faq.question.1204418.html">here</a>

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What's another way to see how the area is exactly 15.5?


My post is already quite lengthy so this section will be a brief overview rather than getting into the gritty details.
The distance formula will show that
AB = sqrt(65)


The equation of line AB is y = (1/8)x + 11/8
The equation of line CD is y = -8x - 11
Intersect those two equations to determine point D is at the exact location (-99/65, 77/65)
Note how -99/65 = -1.52 and 77/65 = 1.18 approximately.
So that's where your teacher is getting (-1.52, 1.18) for point D.


Use the distance formula to determine that CD = 31/sqrt(65)


Then lastly,
area = 0.5*base*height
area = 0.5*AB*CD
area = 0.5*sqrt(65)*( 31/sqrt(65) )
area = 0.5*31
area = 15.5
which is exact without any rounding done to it.
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