Question

Find the vertices of a triangle, the midpoints of whose side are(3,1),(5,6,),(-3,2)

Answer 1

aloha user!!
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we have to find the vertices of a triangle.
for the given midpoints of sides are:

(3,1)
(5,6)
(-3,2)

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[tex] \frac{x1 + x2}{2} = 3 =\ \textgreater \ x1 + x2 = 6 [/tex]

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$\frac{y1 + y2}{2} = 1 =\ \textgreater \ y1 + y2 = 2$

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$\frac{x2 + x3}{2} = 5 =\ \textgreater \ x2 + x3 = 10$

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$\frac{y2 + y3}{2} = 6 =\ \textgreater \ y2 + y3 = 12$

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$\frac{x1 + x3}{2} = -3 =\ \textgreater \ x1 + x3 = -6$

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$\frac{y1 + y3}{2} = 2 =\ \textgreater \ y1 + y3 = 4$

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now,

by elimination:

x1 + x2 = 6
x1 + x3 = -6
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-   -         +
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       x2 - x3 = 12 --------------------(1)


by elimination of (1) and x2  + x3 = 10 { from above }

x2 - x3 = 12
x2 +x3 = 10
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2x2= 22
x2= 11

putting the value of x2 in any of the equation above:

x1 + x2 = 6
x1 + 11= 6
x1 = 6 - 11
x1 = -5

now putting the value of x1 in any of the equation we get:

x1 + x3 = -6
-5 + x3 = -6
x3 = -6 + 5
x3 = -1

again by elimination of y1 + y3 = 4 and y1 + y2 = 2

y1 + y3 = 4
y1 + y2 = 2
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-     -        -
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   y3 - y2 = 2--------------------(2)

now by elimination of (2) and y3 + y2 =12 { from above }

y3 - y2 = 2
y3 + y2 = 12
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2y3 = 14
y3 = 7

now putting the value of y3 in any of the equation we get:

y3 +  y2 = 12
7 + y2 = 12
y2 = 12  - 7
y2 = 5

at last, putting the value of y2 in any of the equation:

y1 + y2 = 2 
y1 + 5 = 2
y1 = -3

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hope it helps :^)

Answer 2

hope it's helped you...