SOLUTION: Find the lengths of the sides of the triangle and area with the vertices A(-2,-3) B(6,1) & C (-2,-5)

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Question 1004834: Find the lengths of the sides of the triangle and area with the vertices A(-2,-3) B(6,1) & C (-2,-5)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Find the lengths of the sides of the triangle and area with the vertices:
A(-2,-3) B(6,1) & C (-2,-5)
use distance formula:
the length of AB is
Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (-2, -3), we can say (x1, y1) = (-2, -3)
So x%5B1%5D+=+-2, y%5B1%5D+=+-3


Since the second point is (6, 1), we can also say (x2, y2) = (6, 1)
So x%5B2%5D+=+6, y%5B2%5D+=+1


Put this all together to get: x%5B1%5D+=+-2, y%5B1%5D+=+-3, x%5B2%5D+=+6, and y%5B2%5D+=+1

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (-2, -3) and (6, 1)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%28-2+-+6%29%5E2+%2B+%28-3+-+1%29%5E2%29 Plug in x%5B1%5D+=+-2, y%5B1%5D+=+-3, x%5B2%5D+=+6, and y%5B2%5D+=+1


d+=+sqrt%28%28-8%29%5E2+%2B+%28-4%29%5E2%29


d+=+sqrt%2864+%2B+16%29


d+=+sqrt%2880%29


d+=+sqrt%2816%2A5%29


d+=+sqrt%2816%29%2Asqrt%285%29


d+=+4%2Asqrt%285%29


d+=+8.94427190999916

==========================================================

Answer:


The distance between the two points (-2, -3) and (6, 1) is exactly 4%2Asqrt%285%29 units


The approximate distance between the two points is about 8.94427190999916 units



So again,


Exact Distance: 4%2Asqrt%285%29 units


Approximate Distance: 8.94427190999916 units





the length of AC is
Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (-2, -3), we can say (x1, y1) = (-2, -3)
So x%5B1%5D+=+-2, y%5B1%5D+=+-3


Since the second point is (-2, -5), we can also say (x2, y2) = (-2, -5)
So x%5B2%5D+=+-2, y%5B2%5D+=+-5


Put this all together to get: x%5B1%5D+=+-2, y%5B1%5D+=+-3, x%5B2%5D+=+-2, and y%5B2%5D+=+-5

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (-2, -3) and (-2, -5)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%28-2+-+%28-2%29%29%5E2+%2B+%28-3+-+%28-5%29%29%5E2%29 Plug in x%5B1%5D+=+-2, y%5B1%5D+=+-3, x%5B2%5D+=+-2, and y%5B2%5D+=+-5


d+=+sqrt%28%28-2+%2B+2%29%5E2+%2B+%28-3+%2B+5%29%5E2%29


d+=+sqrt%28%280%29%5E2+%2B+%282%29%5E2%29


d+=+sqrt%280+%2B+4%29


d+=+sqrt%284%29


d+=+2

==========================================================

Answer:


The distance between the two points (-2, -3) and (-2, -5) is exactly 2 units





the length of BC is
Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (-2, -5), we can say (x1, y1) = (-2, -5)
So x%5B1%5D+=+-2, y%5B1%5D+=+-5


Since the second point is (6, 1), we can also say (x2, y2) = (6, 1)
So x%5B2%5D+=+6, y%5B2%5D+=+1


Put this all together to get: x%5B1%5D+=+-2, y%5B1%5D+=+-5, x%5B2%5D+=+6, and y%5B2%5D+=+1

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (-2, -5) and (6, 1)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%28-2+-+6%29%5E2+%2B+%28-5+-+1%29%5E2%29 Plug in x%5B1%5D+=+-2, y%5B1%5D+=+-5, x%5B2%5D+=+6, and y%5B2%5D+=+1


d+=+sqrt%28%28-8%29%5E2+%2B+%28-6%29%5E2%29


d+=+sqrt%2864+%2B+36%29


d+=+sqrt%28100%29


d+=+10

==========================================================

Answer:


The distance between the two points (-2, -5) and (6, 1) is exactly 10 units





so, AB=8.9,AC=2, and BC=10