Question

Constant rate of change (question in photo)

Answer 1

Given that the set of coordinates in a circle.

We need to cross out the set of coordinates in the circle that do not belong by using the constant rate of change.

The formula for constant rate of change is given by

$m=\frac{y_2-y_1}{x_2-x_1}$

Constant rate of change for the coordinates (2,7), (4,14) and (6,21):

Substituting the coordinates (2,7) and (4,14) in the formula, we get;

$m=\frac{14-7}{4-2}=\frac{7}{2}=3.5$

Constant rate of change for the coordinates (4,14) and (6,21), we get;

$m=\frac{21-14}{6-4}=\frac{7}{2}=3.5$

Constant rate of change for the coordinates (2,7) and (6,21), we get;

$m=\frac{21-7}{6-2}=\frac{14}{4}=3.5$

Thus, all three coordinates have the same constant rate of change.

Constant rate of change for the coordinates (8,2), (12,3) and (16,5):

Substituting the coordinates (8,2) and (12,3) in the formula, we get;

$m=\frac{3-2}{12-8}=\frac{1}{4}=0.25$

Substituting the coordinates (12,3) and (16,5) in the formula, we get;

$m=\frac{5-3}{16-12}=\frac{2}{4}=0.5$

Substituting the coordinates (8,2) and (16,5) in the formula, we get;

$m=\frac{5-2}{16-8}=\frac{3}{8}=0.375$

Since, the constant rate of change is not the same for all the three coordinates, let us cross out the set of coordinates (8,2), (12,3) and (16,5)

Constant rate of change for the coordinates (1,1), (6,6) and (9,9):

Substituting the coordinates (1,1) and (6,6) in the formula, we get;

$m=\frac{6-1}{6-1}=\frac{5}{5}=1$

Substituting the coordinates (6,6) and (9,9) in the formula, we get;

$m=\frac{9-6}{9-6}=\frac{3}{3}=1$

Substituting the coordinates (1,1) and (9,9) in the formula, we get;

$m=\frac{9-1}{9-1}=\frac{8}{8}=1$

Thus, all three coordinates have the same constant rate of change.

Constant rate of change for the coordinates (5,3), (10,6) and (15,9):

Substituting the coordinates (5,3) and (10,6) in the formula, we get;

$m=\frac{6-3}{10-5}=\frac{3}{5}=0.6$

Substituting the coordinates (10,6) and (15,9) in the formula, we get;

$m=\frac{9-6}{15-10}=\frac{3}{5}=0.6$

Substituting the coordinates (5,3) and (15,9) in the formula, we get;

$m=\frac{9-3}{15-5}=\frac{6}{10}=0.6$

Thus, all three coordinates have the same constant rate of change.

Therefore, the set of coordinates that are crossed out from the circle are (8,2), (12,3) and (16,5)