# Transformations of lines

- how a translation in either the \(\normalsize{x}\) or \(\normalsize{y}\) direction affects the equation of a line
- how a dilation affects the equation of a line.

## Translating a line

If we take the line \(\normalsize{y=3x-1}\) and translate it up by \(\normalsize{2}\) in the \(\normalsize{y}\)-direction, we get \(\normalsize{y=3x-1+2}\) or just \[\Large{y=3x+1.}\]If we take the same line \(\normalsize{y=3x-1}\) and translate it by \(\normalsize{2}\) in the \(\normalsize{x}\) direction, then the equation changes more subtly to \(\normalsize{y=3(x-2)-1}\). Can you see why? Suppose the point \(\normalsize{[r,s]}\) lies on the original line. Because \[{\Large s = 3r – 1 = 3( (r+2)-2) – 1}.\]Then \(\normalsize{[r+2,s]}\) will lie on the translated line \({\normalsize y = 3(x-2)-1 = 3x-7}\).Q1(E): Which of the following equations represents a translation of \(\normalsize{y=3x-1}\) by \(\normalsize{3}\) in the negative \(\normalsize{x}\) direction?a) \(\normalsize{y=3(x-3)-1}\)

b) \(\normalsize{y=3(x+3)-1}\)Q2(M): Find the equation of the red line, and hence find the equation of its translate the blue line.

Replacing \(\normalsize{y}\) by \(\normalsize{y-k}\) in an equation represents a translation by \(\normalsize{k}\) in the \(\normalsize{y}\) direction.Replacing \(\normalsize{x}\) by \(\normalsize{x-h}\) in an equation represents a translation by \(\normalsize{h}\) in the \(\normalsize{x}\) direction.

## Scaling a line

Replacing \(\normalsize{y}\) by \(\normalsize{ry}\) in an equation represents a dilation by \(\normalsize{\frac{1}{r}}\) in the \(\normalsize{y}\) direction.Replacing \(\normalsize{x}\) by \(\normalsize{sx}\) in an equation represents a dilation by \(\normalsize{\frac{1}{s}}\) in the \(\normalsize{x}\) direction.

## Combining translation and dilation

## Answers

A1.The translate by \(\normalsize{-3}\) in the \(\normalsize{x}\) direction is b) \(\normalsize{y=3(x+3)-1}\). This simplifies to the equation \(\normalsize{y=3x+8}\).A2.The point \([0,5]\) lies on the red line, and so its equation is of the form \(y=kx+5\) for some number \(k\). Since the point \([-3,-4]\) also lies on this line, this becomes \(y=3x+5\). The blue line is the translate in the \(x\) direction of the red line by \(\normalsize{4}\), and therefore has equation \(y=3(x-4)+5\) or \(y=3x-7\).

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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