Given that the triangle is dilated by factor 3, the image will be found as follows; The object is at: A(-7,-3), B(-3,-2), C(-4,-5) when the image was enlarged the new coordinates will be: A'=3(-7,-3)=(-21,-9) B'=3(-3,-2)=(-9,-6) C'=3(-4,-5)=(-12,-15) since the image is centered at the point (-7,-6), the final point will be at: A"=[(-21+-7),(-9+-6)=(-28,-15) B''=[(-9+-7),(-6+-6)]=(-16,-12) C''=[(-12+-7),(-15+-6)]=(-19,-17) thus the coordinates of the final image are: A''(-28,-15),B''(-16,-12),C''(-19,-17)
If you look at the attached graphic, you will see that the altitude (or height) of an equilateral triangle is: Height = side length * sq root (3) / 2 Area = side^2 * sq root (3) / 4