Gary is using an indirect method to prove that segment DE is not parallel to segment BC in the triangle ABC shown below:
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Coordinates of the given triangle ΔPQR P(1,-1), Q(3,-2), R(3,-4).
a) Let us discuss the rule for rotation of 90° clockwise about the origin.
If given coordinate is (h,k) is rotated 90° clockwise about the origin the position of rotated coordinate will be (k, -h).
x and y coordinates would switch their values and after switching x any values, the y value of the coordinates would reverse in sign.
So, the coordinates of rotated ΔPQR P(1,-1), Q(3,-2), R(3,-4) would become
P(1,-1) --> P'(-1,-1)
Q(3,-2) --> Q'( -2,-3)
R(3,-4) --> R'(-4,-3)
b) If given coordinate is (h,k), the coordinates after reflection across the x-axis would besome (h,-k). (Just change in the sign of y-coordinate).
So, the coordinates of rotated ΔPQR P(1,-1), Q(3,-2), R(3,-4) would become
P(1,-1) --> P"(1,1)
Q(3,-2) --> Q"( 3,2)
R(3,-4) --> R"(3,4)
