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2 years ago

11

Help me please, I’ll give brainliest!!

1 answer:

polet [3.4K]2 years ago

5 0

Answer:

A line that is perpendicular to y=2/5x-5 must have a slope of -5/2, but a line parallel to line y=5/2x+6 must have a slope of 5/2, and these slopes do not match up.

Explanation:

In order for a line to be parallel to another line, they must have the same slope. In order for a line to be perpendicular to another line, their slopes must be opposite reciprocals (meaning that when they are multiplied, the product is -1). From here you can find what slope the line must have to perpendicular or parallel to another line, and then you can see that the answers to not match up to be the same line!

Hope this helped :)

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From the graph: we have the coordinates of RST i.e,

R = (2,1) , S = (2,-2) , T = (-1,-2)

Also, it is given the scale factor $\frac{2}{3}$ and center of dilation C (1,-1)

The mapping rule for the center of dilation applied for the triangle as shown below:

$(x, y) \rightarrow (\frac{2}{3}(x-1)+1, \frac{2}{3}(y+1)-1)$

or

$(x, y) \rightarrow (\frac{2}{3}x -\frac{2}{3}+1 , \frac{2}{3}y+\frac{2}{3}-1)$

or

$(x, y) \rightarrow (\frac{2}{3}x+\frac{1}{3} , \frac{2}{3}y-\frac{1}{3} )$

Now,

for R = (2,1)

the image R' = $(\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(1)-\frac{1}{3} )$ or

$(\frac{4}{3}+\frac{1}{3} , \frac{2}{3}-\frac{1}{3} )$

⇒ R' = $(\frac{5}{3} , \frac{1}{3})$

For S = (2, -2) ,

the image S'= $(\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} )$ or

$(\frac{4}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )$

⇒ S' = $(\frac{5}{3} , -\frac{5}{3})$

and For T = (-1, -2)

The image T' = $(\frac{2}{3}(-1)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} )$ or

$(\frac{-2}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )$

⇒ T' = $(\frac{-1}{3} , \frac{-5}{3})$

Now, label the image of RST on the graph as shown below in the attachment:

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