Question

10. Find the vertices of the triangle the mid points of whose sides are (3,1), (5,6) and
(-3,2).​

Answer 1

Answer:

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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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 :^)