HELP ME PLS!!! ITS URGENT AS SOON AS POSSIBLE
Triangle XYZ has vertices X0, 0), Y(10, 0), and Z(0, 6). Triangle MYP has vertices M(5, 0), Y(10, 0), and P(x, y). Find the missing coordinates for P if AMYP ~ AXYZ.

Triangle XYZ, with vertices X(0, 0), Y(10, 0), and Z(0, 6), is a right triangle due to the fact that points X and Z lie on the y-axis, while points X and Y lie on the x-axis, forming a right angle at vertex X.

In similar triangles corresponding angles have the same measure. Therefore, given that triangle MYP is similar to triangle XYZ, triangle MYP is also a right triangle with its right angle at vertex M.

The corresponding sides of similar triangles are always in the same ratio, so:

XY : YZ : XZ = MY : YP : MP

Since point M is the midpoint of XY, the side lengths of triangle MYP are half the length of the corresponding sides of triangle XYZ.

To find the length of MP, we need to halve the length of XZ. Since the x-coordinate of points X and Z is the same (x = 0), we just need to find the difference between their y-coordinates to determine the length of XY, which is 6 units. Therefore, the length of MP is half of this, which is 3 units.

Since MY lies on the x-axis and MP is parallel the y-axis, point P shares its x-coordinate with point M, and its y-coordinate is equal to the length of the segment. Therefore, the coordinates of point P are:

[tex]\LARGE\boxed{\boxed{P(5,3)}}[/tex]

HELP ME PLS!!! ITS URGENT AS SOON AS POSSIBLE Triangle XYZ has vertices X0, 0), Y(10, 0), and Z(0, 6). Triangle MYP has vertices M(5, 0), Y(10, 0), and P(x, y). Find the missing coordinates for P if AMYP ~ AXYZ.

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HELP ME PLS!!! ITS URGENT AS SOON AS POSSIBLE
Triangle XYZ has vertices X0, 0), Y(10, 0), and Z(0, 6). Triangle MYP has vertices M(5, 0), Y(10, 0), and P(x, y). Find the missing coordinates for P if AMYP ~ AXYZ.