Question

ΔABC is dilated by a scale factor of 0.5 with the origin as the center of dilation, resulting in the image ΔA′B′C′. If A = (2, 2), B = (4, 3), and C = (6, 3), what is the length of B'C'¯¯¯¯¯¯¯ ?
3 units
4 units
2 units
1 unit

Answer 1

The answer is 4 units. The distance between the two points of B and C is equals 6-4=2, because the y is the same. According to the question, the BC is equals 0.5*B'C'. So the length of B'C' is 2/0.5=4 units.

Answer 2

Answer:

Fourth option is correct. The length of B'C' is 1 unit.

Step-by-step explanation:

The vertices of triangle ABC are A = (2, 2), B = (4, 3), and C = (6, 3).

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$BC=\sqrt{(6-4)^2+(3-3)^2}=2$

The length of BC is 2 units.

It is given that ΔABC is dilated by a scale factor of 0.5 with the origin as the center of dilation, resulting in the image ΔA′B′C′.

It means the ΔABC and ΔA′B′C′ are similar and corresponding sides of ΔA′B′C′ and ΔABC are proportional.

$k=\frac{B'C'}{BC}$

The scale factor is 0.5.

$0.5=\frac{B'C'}{2}$

$0.5\times 2=B'C'$

$1=B'C'$

Therefore the length of B'C' is 1 unit and fourth option is correct.