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ABCD is a parallelogram where A is (2,-1), B is (6,2) and C is (11,-2).

(a) State the coordinates of the midpoint AC.
(b) Hence find the coordinates of D.
(c) Find the equations of the diagonals of the parallelogram.

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User PAEz
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(a) The coordinates of the midpoint AC can be found by taking the average of the x-coordinates and the average of the y-coordinates of points A and C. Therefore, the midpoint AC is at:

((2+11)/2, (-1-2)/2) = (6.5,-1.5)

(b) Since ABCD is a parallelogram, the midpoint of BD is also at (6.5,-1.5). Let D have coordinates (x,y). Then the midpoint of BD is:

((6+x)/2, (2+y)/2) = (6.5,-1.5)

Solving for x and y, we get:

x = 7 and y = -5

Therefore, the coordinates of D are (7,-5).

(c) The two diagonals of a parallelogram bisect each other. Since AC is one diagonal and it passes through the midpoint of BD, the equation of AC can be found using the two given points A and C:

Slope of AC = (y2 - y1)/(x2 - x1) = (-2 - (-1))/(11 - 2) = -1/9

Using point-slope form of a line, the equation of AC is:

y - (-1) = (-1/9)*(x - 2)

y = (-1/9)*x + (17/9)

Similarly, the other diagonal BD can be found using the two points B and D:

Slope of BD = (y2 - y1)/(x2 - x1) = (-5 - 2)/(7 - 6) = -7

Using point-slope form of a line, the equation of BD is:

y - 2 = (-7)*(x - 6)

y = -7x + 44

Therefore, the equations of the diagonals of the parallelogram are:

AC: y = (-1/9)*x + (17/9)

BD: y = -7x + 44

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User Alexa Adrian
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