Triangle XYZ with vertices X(3,-1) Y(7,-3)Z(3,-5) undergoes an enlargement represented by the mapping:
(x)--(-0.5 0)(x) +(-3)
(y)--( 0 -0.5)(y) +(4)
Evaluate:
1. The vertices of the image of triangle XYZ.
2. The centre of enlargement
PS. The small brackets should be one big bracket for matrices but I don't have it on the computer.
To find the image of triangle XYZ after the enlargement, we need to apply the given mapping to each of the vertices of the original triangle. Here's how we can do it step by step:
1. The given mapping can be represented by the matrix equation:
⎡x′⎤ ⎡-0.5 0⎤ ⎡x⎤ ⎡-3⎤
⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ + ⎢ ⎥
⎣y′⎦ ⎣ 0 -0.5⎦ ⎣y⎦ ⎣4 ⎦
where (x, y) represents the coordinates of a point in the original triangle and (x′, y′) represents the coordinates of its image after the enlargement.
2. Let's apply the transformation to each vertex of the original triangle:
For vertex X(3,-1):
x′ = (-0.5 * 3) + 0 * (-1) - 3 = -4.5
y′ = 0 * 3 - 0.5 * (-1) + 4 = 4.5
So the image of X is X'(-4.5, 4.5).
For vertex Y(7,-3):
x′ = (-0.5 * 7) + 0 * (-3) - 3 = -6.5
y′ = 0 * 7 - 0.5 * (-3) + 4 = 5.5
So the image of Y is Y'(-6.5, 5.5).
For vertex Z(3,-5):
x′ = (-0.5 * 3) + 0 * (-5) - 3 = -4.5
y′ = 0 * 3 - 0.5 * (-5) + 4 = 6.5
So the image of Z is Z'(-4.5, 6.5).
3. Now we can identify the center of enlargement by examining the coordinates of the corresponding vertices before and after the enlargement.
The center of enlargement is the unique point that is equidistant from the original and image vertices. In this case, the common difference in x-coordinates is 3.5 (7 - 3), and the common difference in y-coordinates is 1.5 (-3 - (-1)).
The center of enlargement can be found as the midpoint between the original vertex Y and its image, Y':
x_center = (7 + (-6.5)) / 2 = 0.25
y_center = (-3 + 5.5) / 2 = 1.25
Therefore, the center of enlargement is C(0.25, 1.25).
To summarize:
1. The vertices of the image of triangle XYZ are X'(-4.5, 4.5), Y'(-6.5, 5.5), and Z'(-4.5, 6.5).
2. The center of enlargement is C(0.25, 1.25).