Enlargements with Negative Scale Factors

A negative unit factor puts the point the same distance on the opposite side of the centre.

1. Here P=(3,2). With the centre at the origin, multiply each coordinate by the scale factor.
2. (3\times(-1),\,2\times(-1))=(-3,-2): the same distance from O, on the opposite side.

Multiply both coordinates by the scale factor, keeping its sign.

1. Here P=(-2,1). With the centre at the origin, multiply each coordinate by the scale factor -2.
2. x: (-2)\times(-2)=4. y: 1\times(-2)=-2.
3. The image is (4,-2): twice as far from O, on the opposite side.

The image is on the opposite side of the centre at half the distance.

1. Here P=(6,-4). With the centre at the origin, multiply each coordinate by -\dfrac12.
2. 6\times\left(-\dfrac12\right)=-3 and -4\times\left(-\dfrac12\right)=2.
3. The image is (-3,2): half as far from O, on the opposite side.

The centre is the midpoint of the point and its image.

1. Here P=(3,-1) and C=(1,2).
2. The vector from C to P is (2,-3).
3. Reverse (2,-3) to (-2,3) and add it to C. The image is Q=(-1,5).

Separate what the minus sign means from what the magnitude means.

1. The minus sign puts Q on the opposite side of C from P.
2. The magnitude 1.5 makes CQ=1.5CP.

Undo multiplication by the scale factor.

1. The image is Q=(-6,4).
2. Divide both coordinates by -2: P=(3,-2).

Compare corresponding coordinates, including their signs.

1. The points are P=(1,-2) and Q=(-3,6).
2. The x-coordinate is multiplied by -3, and -2\times(-3)=6 confirms the same factor vertically.
3. The scale factor is -3.

Use the midpoint because the distances on either side of the centre are equal.

1. The centre is halfway between P=(-3,2) and Q=(5,-4).
2. The midpoint is \left(\dfrac{-3+5}{2},\dfrac{2-4}{2}\right)=(1,-1).

The centre stays fixed; apply the factor to vectors measured from it.

1. The centre is A=(-2,1), so its image remains (-2,1).
2. The vector from A to B is (2,0); multiplying by -2 gives (-4,0), so B maps to (-6,1).
3. The vector from A to C is (0,2), giving image (-2,-3).

Both coordinates change sign and each distance from the origin doubles.

1. With the centre at the origin, multiply each coordinate of T by the scale factor -2.
2. (1,1)\to(-2,-2), (2,1)\to(-4,-2) and (1,2)\to(-2,-4).
3. These are the vertices of triangle A, which is on the opposite side of the origin and twice the size.
4. Triangle B is the enlargement by +2: the same size as A but on the same side of the origin as T.

Lengths use the magnitude of the scale factor.

1. The original perimeter is 4+5+6=15 cm.
2. Each length is multiplied by 3, so the image perimeter is 15\times3=45 cm.
3. The minus sign changes position relative to the centre, not the sign of a length.

Area uses the square of the linear scale factor.

1. The area scale factor is \left(-\dfrac12\right)^2=\dfrac14.
2. The image area is 36\times\dfrac14=9 cm².

Use the area scale factor to find the length scale factor, then decide whether it is positive or negative.

1. The area scale factor is 54\div24=\dfrac94.
2. Areas are multiplied by the square of the length scale factor, so, ignoring its sign, the length scale factor is \sqrt{\dfrac94}=\dfrac32.
3. Triangle B is on the opposite side of C from triangle A, so the scale factor is negative: -\dfrac32=-1.5.

Reverse the direction and double the centre-to-image vector.

1. Here C=(1,-1) and Q=(-2,2).
2. The vector from C to Q is (-3,3). Dividing by -\dfrac12 gives (6,-6).
3. Therefore P=(1+6,-1-6)=(7,-7).

The point-to-image segment consists of one centre-to-original length and two centre-to-image lengths.

1. The points are P=(1,3) and Q=(-5,-3).
2. The centre lies between them, with CQ=2CP, so it is one third of the way from P to Q.
3. The negative factor puts the image on the opposite side of the centre, twice as far away.

4. The vector PQ=(-6,-6); one third is (-2,-2).
5. The centre is (1-2,3-2)=(-1,1).

Measure each relevant displacement from the centre; area uses the square of the factor.

Part (a)

1. Here K=(-3,2) and A=(1,4).
2. The vector KA=(4,2) becomes (-6,-3).
3. Adding this to K gives the image (-9,-1).

Part (b)

1. The rectangle has width 5-1=4 and height 4-2=2, so its area is 8 square units.
2. The area scale factor is \left(-\dfrac32\right)^2=\dfrac94.
3. The image area is 8\times\dfrac94=18 square units.

Compare the length and direction of AB and A'B'. The centre lies on the straight line through A and A'.

Part (a)

1. Corresponding lengths: from A to B is 2 to the right. From A' to B' is -5.5-(-2.5)=-3, so 3 to the left.
2. Compare both the size and direction of the corresponding side.

3. The scale factor is \dfrac{-3}{2}=-1.5. It is negative because the image points the opposite way.

Part (b)

1. The centre-to-image length is 1.5 times the centre-to-original length, on the opposite side.
2. The centre is therefore 1/(1+1.5)=2/5 of the way from A to A'.
3. The displacement AA'=(-2.5,-5) becomes (-1,-2) after multiplying by 2/5.
4. Lines through corresponding point pairs meet at the centre.

5. Adding to A=(0,1) gives the centre (-1,-1).

With the same centre, combine the two scale factors first. For area, square the length scale factor.

Part (a)

1. With the same centre, each point's position from the origin is multiplied by -2 and then by -2.5, so the scale factors multiply.
2. (-2)\times(-2.5)=5. Two negative scale factors give a positive one: the second enlargement takes the shape back to the original side of the origin.
3. The single enlargement has scale factor 5.

Part (b)

1. Areas are multiplied by the square of the length scale factor.
2. From A to C the length scale factor is 5, so the area scale factor is 5^2=25.
3. Area of C =6\times25=150 cm^2.

Find the negative length scale factor from the area ratio, then apply it to the centre-to-point vector.

1. The magnitude of the scale factor is \sqrt{25/4}=2.5. It is negative because the image is on the opposite side of C.
2. From C=(-1,1) to P=(1,3), the vector is (2,2).
3. Multiplying by -2.5 gives (-5,-5), so the image is (-1-5,1-5)=(-6,-4).

The centre lies on the line through A and A' and also on the blue line.

Part (a)

1. A point, its image and the centre of enlargement always lie on one straight line, so the centre is on line AA'.
2. Gradient of AA': \dfrac{7-(-1)}{-5-3}=\dfrac{8}{-8}=-1. Using A=(3,-1): y+1=-(x-3), so y=-x+2.
3. The centre is also on y=x. Where the lines meet, x=-x+2, so 2x=2 and x=1.
4. Intersect the two lines that must contain the centre.

5. The centre of enlargement is (1,1).

Part (b)

1. From the centre (1,1) to A=(3,-1) is the vector \binom{2}{-2}.
2. From the centre (1,1) to A'=(-5,7) is the vector \binom{-6}{6}.
3. Compare centre-to-point distances along the common line.

4. \binom{-6}{6}=-3\times\binom{2}{-2}, so the scale factor is -3.