1) The coordinate of Point X is (5,-6) in Triangle XYZ.
What is the coordinates of X when Triangle XYZ is rotated about the origin 180 degrees and reflected over the x-axis?
(i have diffculties with these type of questions...so can someone explain to me how to do it?)
2) What is the intersection point of perpendicular bisectors of the sides of a triangle called?
3) Help me solve this proof?
Given-Quad ABCD is a parallelogram
Prove- Pt E is the mdpt of the diagonals, BC and AD.
Thans in Advance ;)
What is the coordinates of X when Triangle XYZ is rotated about the origin 180 degrees and reflected over the x-axis?
(i have diffculties with these type of questions...so can someone explain to me how to do it?)
2) What is the intersection point of perpendicular bisectors of the sides of a triangle called?
3) Help me solve this proof?
Given-Quad ABCD is a parallelogram
Prove- Pt E is the mdpt of the diagonals, BC and AD.
Thans in Advance ;)
-
1) Rotating by 180 degrees makes the x and y coordinates the negative of what they started as this gives (-5,6).
Reflecting in the x axis doesn't change the x coordinate but again makes the y the negative of what it was before so (-5,6) becomes (-5,-6).
You can always practice what happens with a piece of tracing paper
2) see http://demonstrations.wolfram.com/ThePer…
3) I assume that Pt E is the intersection of the two diagonals and the diagonals are actually AC and BD. Just show that triangles AED and CEB are congruent because of the fact the AD = CB and the angles and A & C and B & D equal because they are alternate angles.
Therefor AE = CE and E is the mid-point of AC and BE = DE and E is the midpoint of BD
Reflecting in the x axis doesn't change the x coordinate but again makes the y the negative of what it was before so (-5,6) becomes (-5,-6).
You can always practice what happens with a piece of tracing paper
2) see http://demonstrations.wolfram.com/ThePer…
3) I assume that Pt E is the intersection of the two diagonals and the diagonals are actually AC and BD. Just show that triangles AED and CEB are congruent because of the fact the AD = CB and the angles and A & C and B & D equal because they are alternate angles.
Therefor AE = CE and E is the mid-point of AC and BE = DE and E is the midpoint of BD