the point A(-3,6) B(0,7)and C(1,9)are the mid-point of the sides DE,EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallellogram
Answers
Explanation:
there is no use of co- ordinate
Answer:
A (-3,6) is the midpoint of DE , B(0,7) is the midpoint of EF and C(1,9) is the midpoint of FD. Let D = (x1, y1) , E = (x2 , y2) and F = (x3 , y3) .Therefore , A is the midpoint of DE .Similarly , (-3, 6) = {(x1 + x2)/2, (y1 + y2)/2} .x1 + x2 = -6.y1 + y2 = 12 .Similarly, B is the midpoint of EF .so, x2 + x3 = 0.y2 + y3 = 14, and the point C is the midpoint of FD .Therefore, x3 + x1 = 2 ………………..iy1 + y3 = 18 ………………………………….ii now, solving equations I and ii we get the values of x1 and x2 and x3 .x1 = (x1 + x2 + x3) - (x2 + x3) =-2 + 0 = -2 .x2 = (x1 + x2 + x3) - (x1 + x3) = -2 -2 = -4 .x3 = (x1 + x2 + x3) - (x1 + x2)= -2 - 6 = -8 .Similarly, we can find the values of y1,y2 and y3 as…………….. y1 = (y1 + y2 + y3)- (y2 + y3)= 22 - 14 = 8 .y2 = (y1 + y2 + y3) - (y1 + y3) = 22 - 18 = 4 .y3 = (y1 + y2 + y3) - (y1 + y2) = 22 - 12 = 10.Accordingly coordinates of D(-2,8) , E(-4, 4) and F(-8, 10) .We have to establish that ABCD is a parallelogram. Since we know, diagonals of parallelogram intersect each other at midpoint. e.g., midpoint of diagonal AC = midpoint of diagonal BD. So, midpoint of AC = {(-3+1)/2, (6+9)/2} = (-1,15/2).midpoint of BD = {(0-2)/2, (7+8)/2} = (-1,15/2).Hence, ABCD is a parallelogram …. PROVED.hence, it is clear that, ABCD is a parallelogram.