Straight lines and transformed data: review — 0606
Use coordinate differences consistently and retain exact distances. Choose a line form that matches the given point and slope. Read model constants from transformed axes rather than original coordinates.
Notes and key methods
For \(\displaystyle A(x_{1},y_{1}), B(x_{2},y_{2})\), the displacement is \(\displaystyle (x_{2}-x_{1},y_{2}-y_{1})\), its length is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), and the midpoint averages corresponding coordinates.
The gradient is \(\displaystyle \frac{y_{2}- y_{1}}{x_{2}- x_{1}}\) when the denominator is nonzero. A vertical line has an undefined gradient. Squared distances can be compared without evaluating square roots.
The point-gradient form is \(\displaystyle y-y_{1}=m(x-x_{1})\). Parallel nonvertical lines have equal gradients. Perpendicular nonvertical lines with finite nonzero gradients satisfy \(\displaystyle m_{1}m_{2}=-1\).
Vertical and horizontal lines form a perpendicular pair without using a reciprocal gradient. To find an intercept, set the other coordinate to zero and solve the resulting equation.
For y=Axⁿ with positive x,y,A, logarithms give \(\log_b y=n\log_b x+\log_b A\). A plot of V=log_b y against U=log_b x therefore has gradient n and intercept log_b A.
For \(\displaystyle y=A b^{kx} a\) plot of log_b y against x has gradient k. Write down the transformed variables explicitly and convert an intercept back to A using the inverse exponential.
Check the conditions
A gradient quotient needs a nonzero horizontal difference.
Use corresponding coordinates in the same order.
The line here has a finite nonzero gradient.
The negative-reciprocal rule needs its nonzero-gradient condition.
Inputs and outputs of a log plot are positive.
Read the axis labels before interpreting its slope.
The gradient is \(\displaystyle \frac{7}{2}\); use the point-gradient form through A.
Answer: \(\displaystyle y - 6 = \frac{7 x}{2} - \frac{35}{2}\)
(c)
Average the y-coordinates.
\(\displaystyle \frac{19}{2}\)
Answer: \(\displaystyle \frac{19}{2}\).
(d)
Divide the vertical difference by the horizontal difference.
\(\displaystyle \frac{7}{2}\)
Answer: \(\displaystyle \frac{7}{2}\).
Example 2 · Straight lines, parallels and perpendiculars
\(\displaystyle A=\)\(\left( 6, \ 11\right)\). A line through A has gradient \(\displaystyle m=2\). Here \(\displaystyle a=6\).
Find an equation of the line through A with gradient m.
Find an equation of its perpendicular through A.
Find the gradient of a perpendicular line.
Find its x-intercept.
Show the reasoning
(a)
\(y - 11 = 2 x - 12\); expand if required.
Answer: \(y = 2 x - 1\)
(b)
The perpendicular gradient is \(\displaystyle - \frac{1}{m}\); use the same point A.
Answer: \(\displaystyle y - 11 = 3 - \frac{x}{2}\)
(c)
Take the negative reciprocal of the nonzero gradient.
\(\displaystyle - \frac{1}{2}\)
Answer: \(\displaystyle - \frac{1}{2}\).
(d)
Set \(\displaystyle y=0\) and solve for x.
\(\displaystyle \frac{1}{2}\)
Answer: \(\displaystyle \frac{1}{2}\).
Example 3 · Straight-line transformations of data
A positive model y=Axⁿ is plotted with \(\displaystyle U=\log _{2}x\) and \(\displaystyle V=\log _{2}y\). The line passes through \(\left( 1, \ 6\right)\) and \(\left( 3, \ 14\right)\).
Show that the stated model becomes a straight line in U and V.
Determine A and n from the transformed points.
Find A.
Find the model output y at \(\displaystyle x=2\).
Show the reasoning
(a)
Take base-2 logarithms of y=Axⁿ and use the power and product laws.
Answer: \(V=nU+\log_2 A\)
(b)
The slope is \(\displaystyle \frac{14- 6}{3- 1}= 4\). The intercept is 2, so \(\displaystyle A=2^2=4\).
Answer: \(4\) and \(4\)
(c)
Undo the base-2 logarithm of A.
\(4\)
Answer: \(4\).
(d)
Substitute into the recovered power law.
\(64\)
Answer: \(64\).
Common mistakes
A midpoint is an average, not a difference.
Do not interchange the horizontal and vertical differences in a gradient.
Perpendicular gradients have product −1, not 1.
The point coordinate \(\displaystyle y_{1}\) is not generally the y-intercept.
The intercept is log_b A, not A.
A power law is not linear in x and log y unless n has a different role.