Triangle ABC has vertices at A(-5,2), B(-4,-6) and C(4,3). It is translated 1 unit left and 2 units up. What are the vertices of A'B'C'?*

Step-by-step explanation:

move 1 unit left means subtracting 1 from x-coordinate

move 2 units up means adding 2 to y-coordinate

A' = (-5-1, 2+2) = (-6, 4)

B' = (-4-1, -6+2) = (-5, -4)

C' = (4-1, 3+2) = (3, 5)

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c