Three vertices of parallelogram ABCD are A(6,3), B(3,8) and C(-6,10). If point D lies in quadrant II, find the coordinates of D.

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In parallelogram ABCD, AD∥BC, and AD=BC so that slopes of AD and BC are identical to each other.

Unit change of BC, from point C to point B, is -2 (=8-10) units on y-axis and 9 (=3-(-6))units on x-axis.

And the slope is the ratio of unit change in y to unit change in x.   Thus, the slope of BC is: m=-2/9.

Unit change of AD, from point A to point D, is the reverse of BC's: Point A goes 9 units leftwards on x-axis, and 2 units upwards on y-axis to reach point D.

Thus, the coordinates of point D(x,y) is: x=6-9=-3, and y=3+2=5, so point D(-3,5) is in the 2nd quadrant.

CK: The slope of AD: (3-5)/(6-(-3))=-2/9, so m=n   CKD.

Therefore, the coordinates of point D is (-3,5), and point D lies in the 2nd quadrant. 

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