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Question
If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.
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Solution 1
The co-ordinates of the midpoint (xm, ym) between two points (x1, y1) and (x2, y2) is given by,
(xm, ym) = `((x_1+x_2)/2)"," ((y_1+y_2)/2)`
Let the three vertices of the triangle be A(xA, yA), B(xB, yB) and C(xC, yC).
The three midpoints are given. Let these points be `M_(AB) (3,4), M_(BC) (4, 6) and M_(CA) (5, 7)`.
Let us now equate these points using the earlier mentioned formula,
`(3,4) = ((x_A + x_B)/2)"," ((y_A + y_B)/2)`
Equating the individual components we get,
xA + xB = 6
yA + yB = 8
Using the midpoint of another side we have,
`(4,6) = ((x_B + x_C)/2)","((y_B + y_C)/2)`
Equating the individual components we get,
xB + xC = 8
yB + yC = 12
Using the midpoint of the last side we have,
`(5,7) = ((x_A + x_C)/2)","((y_A + y_C)/2)`
Equating the individual components we get,
xA + xC = 10
yA + yC = 14
Adding up all the three equations which have variable ‘x’ alone we have,
xA + xB + xB + xC + xA + xC = 6 + 8 + 10
2(xA + xB + xC) = 24
xA + xB + xC = 12
Substituting xB + xC = 4 in the above equation we have,
xA + xB + xC = 12
xA + 8 = 12
xA = 4
Therefore,
xA + xC = 10
xC = 10 - 4
xC = 6
And
xA + xB = 6
xB = 6 - 4
xB = 2
Adding up all the three equations which have variable ‘y’ alone we have,
yA + yB + yB + yC + yA + yC = 8 + 12 + 14
2(yA + yB + yC) = 34
yA + yB + yC = 17
Substituting yB + yC = 12 in the above equation we have
yA + yB + yC = 17
yA + 12 = 17
yA = 5
Therefore
yA + yC = 14
yC = 14 – 5
yC = 9 And
yA + yB = 8
yC = 14 – 5
yC = 9 And
yA + yB = 8
yB = 8 – 5
yB = 3
Therefore, Co-ordinates of the three vertices of the triangle are A (4, 5), B (2, 3), C (6, 9).
Solution 2
Given mid-points:
M1 = (3, 4)
M2 = (4. 6)
M3 = (5, 7)
1. Set up the Equations
`(x_1+x_2)/2 = 3 => x_1 + x_2 = 6`
`(x_2+x_3)/2 = 4 => x_2 + x_3 = 8`
`(x_3+x_1)/2 = 5 => x_3 + x_1 = 10`
Similarly for the y-coordinates:
`(y_1 + y_2)/2 = 4 => y_1 + y_2 = 8`
`(y_2 + y_3)/2 = 6 => y_2 + y_3 = 12`
`(y_3 + y_1)/2 = 7 => y_3 + y_1 = 14`
Solve for the x-coordinates
(x1 + x2) + (x2 + x3) + (x3 + x1) = 6 + 8 + 10
2(x1 + x2 + x3) = 24
x1 + x2 + x3 = 12
x1 = (x1 + x2 + x3) – (x2 + x3) = 12 – 8 = 4
x2 = (x1 + x2 + x3) – (x3 + x1) = 12 – 10 = 2
x3 = (x1 + x2 + x3) – (x1 + x2) = 12 – 6 = 6
Solve for the y-coordinates
(y1 + y2) + (y2 + y3) + (y3 + y1)
= 8 + 12 + 14
2(y1 + y2 + y3) = 34
y1 + y2 + y3 = 17
y1 = (y1 + y2 + y3) – (y2 + y3) = 17 – 12 = 5
y2 = (y1 + y2 + y3) – (y3 + y1) = 17 – 14 = 3
y3 = (y1 + y2 + y3) – (y1 + y2) = 17 – 8 = 9
The vertices of the triangle are A(4, 5), B(2, 3), and C(6, 9).
