Some please help I need assistance
1 answer:
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Answer:
see below
Step-by-step explanation:
Every vertex moves twice as far from the center of dilation as it is in the pre-image.
Perhaps the easiest image point to find is the one at lower left. In the pre-image it is 2 units left of the center of dilation, so the image point will be 2×2 = 4 units left of the center of dilation. It will be located at (-6, -2).
Other points on the image can be found either by reference to the center of dilation, or by reference to known image points. For example, the next point clockwise is 1 left and 4 up in the pre-image, so will be 2 left and 8 up from (-6, -2) in the image. That is, the pre-image point (-5, 2) becomes image point (-8, 6). You will note that (-5, 2) is 3 left and 4 up from the center of dilation, and that (-8, 6) is 6 left and 8 up from the center of dilation (twice as far away).
Answer:
Step-by-step explanation:
We want to find an equation of a line that's perpendicular to x=1 that also passes through the point (8,-9).
Note that x=1 is a <em>vertical line </em>since x is 1 no matter what y is.
This means that if our new line is perpendicular to the old, then it must be a <em>horizontal line</em>.
So, since we have a horizontal line, then our equation must be our y-value of our point.
Our y-coordinate of our point (8,-9) is -9.
Therefore, our equation is:
And this is in standard form.
And we're done!