The endpoints of are A(9, 4) and B(5, –4). The endpoints of its image after a dilation are A'(6, 3) and B'(3, –3). can you please Explain how to find the scale factor.
How did you get this part Reiny?
AC = √(144+16) = √160 = 4√10
A'C = √(81+9) = √90 = 3√10
A'C/AC = 3V10/(4√10) = 3/4
BC = √(64+16) = √80 = 4√5
B'C = √(36+9) = √45 = 3√5
B'C/BC = 3√5/(4√5) = 3/4
Dang! Steve is mega smart! He be answering all of the questions I look up.
How did you do this steve?
CA'/CA = √90/√160 = 3/4
CB'/CB = √45/√80 = 3/4
Lily, I KNOW. Like kids from 10 years ago are hella smart compared to me lmao whatever I just need help
I had to puzzle this out myself. In this case, we need to
(a) find the center of dilation
(b) verify that the dilation of A to A' is the same as that of B to B'
Since dilation is a linear scaling, both A and B are moving toward some point C. Naturally, if the dilation were 0, both points would contract to the same point: the intersection of the lines AA' and BB'
The line containing A and A' is (y-3)/(x-6) = 1/3
The line containing B and B' is (y+3)/(x-3) = -1/2
These lines intersect at (-3,0)
So, now, let's verify that the scale factor is the same along both directions
CA'/CA = √90/√160 = 3/4
CB'/CB = √45/√80 = 3/4
So, the scale factor is 3/4, using (-3,0) as the center of dilation.
Seems like we were on some kind of Vulcan mind-meld.
Even our choice of C for the centre of dilation was the same, spooky!
I had never seen that kind of question, and actually printed myself out a sheet of graph paper, lol
money baby
To find the scale factor of a dilation, we can use the formula:
Scale factor = (distance of image)/(distance of original)
In this case, we can calculate the distance of the original line segment AB and the distance of the image line segment A'B'. Then, we can divide the distance of the image by the distance of the original to find the scale factor.
First, let's find the distance of the original line segment AB.
Using the distance formula:
Distance AB = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the coordinates, we have:
Distance AB = sqrt((5 - 9)^2 + (-4 - 4)^2)
Distance AB = sqrt((-4)^2 + (-8)^2)
Distance AB = sqrt(16 + 64)
Distance AB = sqrt(80)
Now, let's find the distance of the image line segment A'B'.
Using the distance formula:
Distance A'B' = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the coordinates, we have:
Distance A'B' = sqrt((3 - 6)^2 + (-3 - 3)^2)
Distance A'B' = sqrt((-3)^2 + (-6)^2)
Distance A'B' = sqrt(9 + 36)
Distance A'B' = sqrt(45)
Finally, let's calculate the scale factor:
Scale factor = Distance A'B' / Distance AB
Scale factor = sqrt(45) / sqrt(80)
To simplify this, we can divide both the numerator and denominator under the square root by their greatest common factor, which is 5:
Scale factor = sqrt(9) / sqrt(16)
Taking the square root of 9 and 16:
Scale factor = 3/4
Therefore, the scale factor of the dilation is 3/4.
Did you notice that slope AB = slope A'B' = 2 ?
So by extending AA' and BB' until they meet we can find the centre of dilation
Making a neat graph shows that this centre of dilation is C(-3,0)
or
You can find the equation of AA',which was y = (1/3)x + 1, and the equation of BB', which was y = (-1/2)x - 3/2
solving these two to get (-3,0)
AC = √(144+16) = √160 = 4√10
A'C = √(81+9) = √90 = 3√10
A'C/AC = 3V10/(4√10) = 3/4
BC = √(64+16) = √80 = 4√5
B'C = √(36+9) = √45 = 3√5
B'C/BC = 3√5/(4√5) = 3/4
scale factor is 3/4