Transformations of lines

We now apply translations and dilations to lines, or more precisely to the equations of those lines. What happens to the equation of a line if we translate it by \(\normalsize{2}\) in the \(\normalsize{x}\) direction, or by \(\normalsize{-3}\) in the \(\normalsize{y}\) direction? What happens if we apply a dilation \(\normalsize{[x,y]}\) goes to \(\normalsize{[rx,ry]}\)?

In this step you will learn

• 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. This means that

\[{\Large s = 3r - 1 = 3( (r+2)-2) - 1}.\]

But 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.

Summarizing, we have the following:

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

If we multiply the \(\normalsize{y}\) coordinate of a point \(\normalsize{[x,y]}\) which lies on the line \(\normalsize{y=3x-1}\) (in red) by \(\normalsize{\frac{3}{2}}\), the effect is a dilation in the \(\normalsize{y}\) direction by just this factor. This takes the line \(\normalsize{y=3x-1}\) to \(\normalsize{y=\frac{9}{2}x-\frac{3}{2}}\), which is the blue line below.

If we multiply the \(\normalsize{x}\) coordinate of a point \(\normalsize{[x,y]}\) on the line \(\normalsize{y=3x-1}\) by \(\normalsize{\frac{1}{2}}\), the effect is a dilation in the \(\normalsize{x}\) direction by just this factor.

Summarizing, we have the following:

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

So what happens if we combine a translation by \(\normalsize{3}\) in the \(\normalsize{y}\) direction with a dilation by \(\normalsize{4}\) in the \(\normalsize{x}\) direction? The line \(\normalsize{y=3x-1}\) first goes to \(\normalsize{(y-3)=3x-1}\) and then to \(\normalsize{(y-3)=3(x/4)-1}\). This works out to be \(\normalsize{y=(3/4)x+2}\).

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 \(y=3x-7\).