Drawback 1 :
Place a 2-unit by 6-unit rectangle in a coordinate aircraft.
Drawback 2 :
A proper triangle has legs of 5 items and 12 items. Place the triangle in a coordinate aircraft. Label the coordinates of the vertices and discover the size of the hypotenuse.
Drawback 3 :
Within the diagram proven beneath, ΔMLO ≅ ΔKLO, discover the coordinates of level L.
Drawback 4 :
Within the diagram proven beneath, show that ΔOTU ≅ ΔUVO.
1. Reply :
Select a placement that makes discovering distances straightforward. Listed below are two potential placements.
(i) One vertex is on the origin, and three of the vertices have at the least one coordinate that’s 0.
(ii) One aspect is centered on the origin, and the x-coordinates are opposites.
Be aware :
As soon as a determine has been positioned in a coordinate aircraft, you should use the Distance System or the Midpoint System to measure distances or find factors.
2. Reply :
Primarily based on the knowledge given within the query, one potential placement of the triangle is proven beneath.
Discover that one leg is vertical and the opposite leg is horizontal, which assures that the legs meet at proper angles. Factors on the identical vertical section have the identical x-coordinate, and factors on the identical horizontal section have the identical y-coordinate.
We are able to use the Distance System to seek out the size of the hypotenuse.
Distance System :
d = √[(x2 – x1)2 + (y2 – y1)2]
Substitute (x1, y1) = (0, 0), (x2, y2) = (12, 5).
d = √[(12 – 0)2 + (5 – 0)2]
d = √(122 + 52)
d = √(144 + 25)
d = √169
d = 13
3. Reply :
As a result of the triangles are congruent, it follows that
ML ≅ KL
So, level L should be the midpoint of MK. This implies, we can use the Midpoint System to seek out the coordinates of level L.
Midpoint System :
L(x, y) = [(x1 + x2)/2, (y1 + y2)/2]
Substitute (x1, y1) = Ok(160, 0), (x2, y2) = M(0, 160).
L(x, y) = [(160 + 0)/2, (0 + 160)/2]
L(x, y) = (160/2, 160/2)
L(x, y) = (80, 80)
The coordinates of the purpose L are (80, 80).
4. Reply :
Given : Coordinates of determine OTUV.
To show : ΔOTU ≅ ΔUVO
Coordinate Proof :
Segments OV and TU have the identical size.
Distance System :
d = √(x2 – x1)2 + (y2 – y1)2
To seek out the size of OV, substitute
(x1, y1) = O(0, 0) and (x2, y2) = V(h, 0).
OV = √[(h – 0)2 + (0 – 0)2]
OV = √h2
OV = h
To seek out the size of TU, substitute
(x1, y1) = T(m, okay) and (x2, y2) = U(m + h, okay)
TU = √[(m + h – m)2 + (k – k)2]
TU = √h2
TU = h
Horizontal segments TU and OV, every has a slope of 0. This suggests that TU and OV are parallel. Phase OU intersects TU and OV to kind congruent alternate inside angles ∠TUO ≅ ∠VOU.
As a result of OU ≅ OU, we can apply the SAS Congruence Postulate to conclude that
ΔOTU ≅ ΔUVO
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