### Video Transcript

Dilate the square π΄π΅πΆπ· from the origin by a scale factor of a half and state the coordinates of the image.

Dilate is another mathematical word for enlarge. However, a dilation or enlargement can increase or decrease the size of a shape. As the scale factor in this question is one-half, then our image will be one-half the size of the initial square. When answering a question of this type, we need to be careful about the point of dilation, also known as the centre of enlargement. In this case, this is from the origin, which has coordinates zero, zero. All four vertices or corners of the square will be half the distance from the origin that they currently are.

When dealing with a centre at the origin, the easiest way is to multiply all of the coordinates by the scale factor. Point π΄ has coordinates negative six, two. We go along the π₯-axis to negative six and up the π¦-axis to two. Remember, when dealing with coordinates, we go along the corridor and then up the stairs. The π₯-coordinate comes first. Point π΅ has coordinates negative 10, two. We go along the π₯-axis to negative 10 and up the π¦-axis to two. Point πΆ has coordinates negative 10, six. And finally, point π· has coordinates negative six, six.

As the scale factor in this case is one-half, we need to multiply each of the coordinates by one-half. This is the same as dividing by two. Negative six divided by two is equal to negative three and two divided by two is equal to one. This means that the corresponding point in the image π΄ prime has coordinates negative three, one. We repeat this process for point π΅. Negative 10 divided by two is equal to negative five and two divided by two is equal to one. Therefore, π΅ prime has coordinates negative five, one. πΆ prime has coordinates negative five, three. And, finally, π· prime has coordinates negative three, three.

We can plot all four of these points onto the coordinate grid. We go along the π₯-axis to negative three and up the π¦-axis to one. We then go along the π₯-axis to negative five and up the π¦-axis to one. We repeat this for πΆ prime and π· prime. Joining up the four points creates a two-by-two square. As the original square had side length four, we can see that the image is one-half of the size. We can, therefore, say that the image is definitely the correct size.

We can prove the points are in the right place by drawing a line from the origin through the points. The line from the origin to point π΅ passes through the point π΅ prime. Likewise, the line from the origin to point π· passes through π· prime. We could repeat this for points π΄ and πΆ. We can, therefore, say that the distance from π to π· prime is half of the distance from π to π·. We can prove this by counting squares to get from π to π·. We go left six squares and up three squares. To get from π to π· prime, we go left three squares and up three squares. As three is a half of six, this point is correct.

This method is very useful when the centre is not at the origin. We count the number of units from our centre to each of the points and then halve the values for a scale factor of one-half. If the scale factor was one-third, we would find one-third of these values. Likewise, if the scale factor was two or three, we would double or triple, respectively, the distances from the centre to each corner or point of our shape.

Dilating the square π΄π΅πΆπ· from the origin by scale factor of a half gives the four coordinates negative three, one; negative five, one; negative five, three; and negative three, three.