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 \(\normalsize{y=3x1}\) and translate it up by \(\normalsize{2}\) in the \(\normalsize{y}\)direction, we get \(\normalsize{y=3x1+2}\) or just
\[\Large{y=3x+1.}\]If we take the same line \(\normalsize{y=3x1}\) and translate it by \(\normalsize{2}\) in the \(\normalsize{x}\) direction, then the equation changes more subtly to \(\normalsize{y=3(x2)1}\). 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 \({\normalsize y = 3(x2)1 = 3x7}\).
Q1 (E): Which of the following equations represents a translation of \(\normalsize{y=3x1}\) by \(\normalsize{3}\) in the negative \(\normalsize{x}\) direction?
a) \(\normalsize{y=3(x3)1}\)
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 \(\normalsize{yk}\) in an equation represents a translation by \(\normalsize{k}\) in the \(\normalsize{y}\) direction.
Replacing \(\normalsize{x}\) by \(\normalsize{xh}\) 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 \(\normalsize{y=3x1}\) (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 \(\normalsize{y=3x1}\) 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 \(\normalsize{y=3x1}\) 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 \(\normalsize{y=3x1}\) first goes to \(\normalsize{(y3)=3x1}\) and then to \(\normalsize{(y3)=3(x/4)1}\). 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 \(y=3(x4)+5\) or \(y=3x7\).
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