Congruence and Similarity

Lesson

In this diagram, the red shape has been transformed into the larger blue shape. It can be seen that the dimensions of the blue shape are three times those of the red shape and the orientation of the blue shape has been reversed.

The green arrows connect points on the red shape with their corresponding transformed points on the blue shape. The green arrows appear to pass through a common central point.

The central point is called the *centre of enlargement*.

All the distances from the point of enlargement in the blue shape are three times the corresponding distance from the point of enlargement in the red shape but in the opposite direction. We say the scale factor, in this case, is $-3$−3. The change in absolute size has been by a factor of three and the direction has been reversed.

The two shapes are said to be *similar *meaning the shapes are essentially the same in that only the scale has changed. The reversed orientation does not affect the similarity.

Any similarity transformation like the one illustrated above can be constructed geometrically using lines and measurements from a point without any need for a coordinate system. However, we can also use reasoning from coordinate geometry to carry out these transformations without needing to draw any pictures.

Take a centre of enlargement that is two units from the origin of a Cartesian coordinate system in the horizontal direction and three units from the origin in the vertical direction. We label the centre point $X(2,3)$`X`(2,3) where the bracketed numbers are the coordinates of the point $X$`X`.

Another point $Y(4,1)$`Y`(4,1) is to be transformed with respect to the given centre by a factor $-2$−2. What are the coordinates of the transformed point $Y'$`Y`′?

We can consider the horizontal and vertical changes separately. The horizontal distance from $X$`X` to $Y$`Y` is $4-2=2$4−2=2 units. Under the enlargement, this is to be multiplied by $-2$−2. The new horizontal distance from the centre must be $-2\times2=-4$−2×2=−4 units. So, the horizontal coordinate, with respect to the origin, must be $2-4=-2$2−4=−2.

The vertical distance from $X$`X` to $Y$`Y` is $1-3=-2$1−3=−2 units. After multiplying by the scale factor this is $-2\times-2=4$−2×−2=4 units from the centre of enlargement. So, the vertical distance from the origin is $3+4=7$3+4=7 units.

Putting this together, we see that the transformed point $Y'$`Y`′ has coordinates $(-2,7)$(−2,7).

Notice that we performed exactly the same operations on the horizontal and vertical measurements. We could simplify the notation by compressing both calculations into one.

First, we found the distances of the point to be transformed from the centre of enlargement. This could be written $(4,1)-(2,3)=(2,-2)$(4,1)−(2,3)=(2,−2).

Then, we multiplied by the scale factor: $-2\times(2,-2)=(-4,4)$−2×(2,−2)=(−4,4).

Finally, we added back the centre point: $(-4,4)+(2,3)=(-2,7)$(−4,4)+(2,3)=(−2,7).

These three steps will be the same for any problem of a similar kind. So, we can compress the whole thing still further using a more general notation. Suppose the centre of enlargement is the point $(a,b)$(`a`,`b`), the point to be transformed is $(x,y)$(`x`,`y`) and the scale factor is $m$`m`. By combining the three steps, we get

$(x',y')=m\left[(x,y)-(a,b)\right]+(a,b)$(`x`′,`y`′)=`m`[(`x`,`y`)−(`a`,`b`)]+(`a`,`b`)

If the points $A(1,1)$`A`(1,1), $B(4,1)$`B`(4,1) and $C(1,5)$`C`(1,5) are connected by lines, a right-angled triangle is formed. What are the corresponding points $A'$`A`′, $B'$`B`′ and $C'$`C`′ when the triangle is enlarged by a factor $-3$−3 using the point $(0,1)$(0,1) as centre?

The procedure used above has to be applied to each of the three points $A$`A`, $B$`B` and $C$`C`.

$A'(a_1,a_2)$A′(a1,a2) |
$=$= | $-3\times\left[(1,1)-(0,1)\right]+(0,1)$−3×[(1,1)−(0,1)]+(0,1) |

$=$= | $-3(1,0)+(0,1)$−3(1,0)+(0,1) | |

$=$= | $(-3,1)$(−3,1) | |

$B'(b_1,b_2)$B′(b1,b2) |
$=$= | $-3\times\left[(4,1)-(0,1)\right]+(0,1)$−3×[(4,1)−(0,1)]+(0,1) |

$=$= | $-3(4,0)+(0,1)$−3(4,0)+(0,1) | |

$=$= | $(-12,1)$(−12,1) | |

$C'(c_1,c_2)$C′(c1,c2) |
= | $-3\times\left[(1,5)-(0,1)\right]+(0,1)$−3×[(1,5)−(0,1)]+(0,1) |

= | $-3(1,4)+(0,1)$−3(1,4)+(0,1) | |

= | $(-3,-11)$(−3,−11) |

Thus, the vertices of the enlarged triangle are $A'(-3,1)$`A`′(−3,1), $B'(-12,1)$`B`′(−12,1) and $C'(-3,-11)$`C`′(−3,−11).