Sketching the line for a given law

When a law is written as \(Y=mX+c\), the graph of \(Y\) against \(X\) is a straight line. You can describe and sketch it without plotting any data: its gradient is \(m\), it meets the vertical axis at \((0,\,c)\) and, when \(m\ne0\), it meets the horizontal axis where \(X=-\dfrac{c}{m}\).

Why: every value of \(X\) gives exactly one value \(Y=mX+c\), and each increase of \(1\) in \(X\) adds the same amount \(m\) to \(Y\). Taking logarithms of \(y=2x^{1.5}\) gives \(\lg y=1.5\lg x+\lg 2\), so the curve becomes a line once \(\lg y\) and \(\lg x\) are on the axes.

Method

1. Rearrange the law so that the quantity on the vertical axis is alone on one side, in the form \(Y=mX+c\).
2. Read off the gradient \(m\) and the vertical intercept \(c\). Both must be constants.
3. Sketch: mark \((0,\,c)\), slope the line up if \(m>0\) and down if \(m<0\), and mark the horizontal intercept \(\left(-\dfrac{c}{m},\,0\right)\) if it is asked for.
4. For a point on the line, substitute its known coordinate into \(Y=mX+c\).
5. If different quantities are plotted, rearrange the law again for the new axes; the gradient and intercept usually change.

Example

The variables \(x\) and \(y\) are related by \(y^2=4x+9\). Plotting \(Y=y^2\) against \(X=x\) gives \(Y=4X+9\): gradient \(4\), vertical intercept \(9\). The point on the line with \(X=4\) has \(Y=4(4)+9=25\).

Dividing the same law by \(x\) gives \(\dfrac{y^2}{x}=9\left(\dfrac{1}{x}\right)+4\). Plotting \(\dfrac{y^2}{x}\) against \(\dfrac{1}{x}\) gives a line with gradient \(9\) and intercept \(4\): the two constants swap roles.