ab has endpoints at (2,3) and (2,5). after a dilation with a center at the origin and a scale factor of 15 , what would be the length of the image of ab ?
1/5 units
2/5 units
2 units
10 units
Well, which one would be the closest? 10?
need an answer asap
Yes, 10 units would be the closest choice, although the correct answer should be 30 units.
Well, that's quite a stretch we're talking about! If a dilation with a scale factor of 15 is applied to AB, it means the endpoints of the image will be 15 times farther away from the origin. Since AB lies on the vertical line x=2, the image will also lie on the vertical line x=2.
The difference between the y-coordinates of the endpoints is 5 - 3 = 2 units. So, when we stretch this difference by a factor of 15, we get 2 * 15 = 30 units.
Therefore, the length of the image of AB is 30 units. So, the answer is 30 units.
Or in other words, it's like AB worked out at the gym and got a super long makeover!
To find the length of the image of AB after a dilation with a center at the origin and a scale factor of 15, we can follow these steps:
1. Find the distance between the two endpoints of AB.
- The endpoints of AB are (2,3) and (2,5).
- To find the distance between two points in a Cartesian plane, we can use the distance formula: d = √((x2-x1)^2 + (y2-y1)^2).
- Using the distance formula, the distance between (2,3) and (2,5) is: d = √((2-2)^2 + (5-3)^2) = √((0)^2 + (2)^2) = √(0 + 4) = √4 = 2 units.
2. Apply the dilation to the length of AB.
- A dilation with a scale factor of 15 means that the image of AB will be 15 times larger than the original length.
- Multiply the original length of AB by the scale factor: 2 units * 15 = 30 units.
Therefore, the length of the image of AB after the dilation is 30 units.
both points are on the vertical line x=2, so they are just 2 units apart
so multiply all coordinates (and thus distances) by 15
I suspect a typo, since the result is not one of the choices.