Quadrilateral ABCD has vertices at A(-2,-1) . B (-4,-7) C(-9,-3) D(-8,-6) . It is reflected across the x-axis. What are the new coordinates? Explain.
@ash2326 .. i asked this earlier but i dont think the awsner i got is correct . can u help ?
i got ; A(-2,1) B(-4,7) C(-9,3) D(-8,6). Because you do the opposite number of the "y " .
Hint: Use graph paper
i am
Okay, next plot the points
i did
Okay, next find out how far each point is from the x axis.
Start with the point closest to the x-axis
a(-2,1) is 2 away from the x axis and one over.
so 2.-1 ?? for a
Whatever the distance each point is from the x-axis, its reflected point will be the same distance. Vertical distance only. the "one over" isn't a vertical distance.
omg ,
so confusing
wait!
Actually, you just need to figure out how to count distance from the x-axis. When they say x-axis, you must first identify what "x-axis" is.
cant you just do ; To reflect across the X axis, simply multiply all the Y coordinates by -1.
Vertical distance means "up and down" distance
Okay, just wait...
wait.
Don't try anything yet, just listen
Sigh
imsorry ...
Okay here's what to do. 1. Put your pencil on the x-axis directly beneath the first point you want to find the distance of. 2. Then draw a line from the x-axis to that point. 3. Count the number of units from the x-axis to that point. 4. The number of units represents the distance from the x-axis to the point 5. The reflected point will be the equal and opposite distance from the x-axis.
I hope that makes sense.
It does!
i got ...
A(-2,1) B(-4,7) C(-9,3) D(-8,6
Yeah, that's right. I realize that you already had it. lol
Lol i was double checking :) thanks!
Good job
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