Check the picture below.
Parallel lines never intersect, therefore, there is one solution
1 answer:
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The image of Triangle ABC have coordinates as
A'(-2,5)
B'(-2,-1)
C'(7,-1)
Step-by-step explanation:
The figure show that Points A(-4,-1),B(-4,-3) and C(-1,-3) forms the Triangle ABC
It is asked to scaled the triangle with scale factor of 3 and centered at P(-5,-4)
For point A(-4,-1) :
Distance of A(-4,-1) from center P(-5,-4) in x-direction is 1 units and y-direction is 3 units
After scale by factor of 3,
Distance of A(-4,-1) from center P(-5,-4) is also multiply by 3
New distance in x-direction is 3 units and 9 units in y-direction from center P(-5,-4)
Therefore, New location A' is ((-5)+3,(-4)+9)=(-2,5)
For point B(-4,-3)
Distance of B(-4,-3) from center P(-5,-4) in x-direction is 1 units and y-direction is 1 units
After scale by factor of 3,
Distance of B(-4,-3) from center P(-5,-4) is also multiply by 3
New distance in x-direction is 3 units and 3 units in y-direction from center P(-5,-4)
Therefore, New location B' is ((-5)+3,(-4)+3)=(-2,-1)
For point C(-1,-3)
Distance of C(-1,-3) from center P(-5,-4) in x-direction is 4 units and y-direction is 1 units
After scale by factor of 1,
Distance of C(-1,-3) from center P(-5,-4) is also multiply by 3
New distance in x-direction is 12 units and 3 units in y-direction from center P(-5,-4)
Therefore, New location C' is ((-5)+12,(-4)+3)=(7,-1)
Therefore, The image of Triangle ABC have coordinates as
A'(-2,5)
B'(-2,-1)
C'(7,-1)
