What is the slope of the line that passes through the points (-2, -5) and (6, 7)? A.3/2 B.2/3 C.-3/2 D.-2/3
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Answer:
Part A
The rotation of the point (x, y) 180° about the origin gives the point (-x, -y)
Therefore, we have the points, ΔLMN, L(1, 1), and M(2, 2), and N(3, 3)
The coordinates of the vertices of the image, ΔL'M'N' are L'(-1, -1), and M'(-2, -2), and N'(-3, -3)
Please find attached the graph of triangle ΔLMN and ΔL'M'N'
Part B
The lines drawn through L and L' and through M and M' are colinear
Part C
A line is defined by two points, such as <em>L </em>and <em>M.</em> Rotation of a line through 180° about the origin will give an image location on the same path as the extension of the original line.
The third point, <em>N</em>, however, defines a plane, and the rotation of a plane by 180° will give an image which is turned upside down with regards to the preimage
Therefore, a trough in the preimage becomes a peak in the image and the lines drawn through N and N' crosses the colinear lines drawn throgh M and M' and L and L'
Step-by-step explanation:
