7. Straight-line graphs

Learning objectives

7.1Use straight-line equations• Use the equation of a straight line.7.2Use parallel and perpendicular gradients• Know and use the condition for two lines to be parallel or perpendicular.7.3Solve midpoint, length and bisector problems• Solve problems involving midpoint and length of a line.• Find and use the equation of a perpendicular bisector.7.4Transform relationships to straight-line form• Transform given relationships to and from straight-line form.• Determine unknown constants by calculating the gradient or intercept of the transformed graph.• Includes forms such as y = Ax^n, y = Ab^x, y^2 = Ax^3 + B, e^(2y) = Ax^2 + B and y^3 = A ln x + B.

A non-vertical straight line has constant gradient $m$ and equation $y=mx+c$ . The gradient is the change in $y$ per unit change in $x$ , while $c$ is the $y$ -intercept.

$m=\frac{y_2-y_1}{x_2-x_1},\qquad y-y_1=m(x-x_1)$

From two points, calculate $m$ , substitute either point into the point-gradient form, then rearrange only if another form is requested. To find where two lines meet, solve their equations simultaneously; the resulting ordered pair must satisfy both.

Through $(4,23)$ and $(-8,29)$ , $m=(29-23)/(-8-4)=-1/2$ , so $y-23=-\tfrac12(x-4)$ and $y=-\tfrac12x+25$ . Intersecting with $y=2x+5$ gives $-\tfrac12x+25=2x+5$ , hence $(x,y)=(8,21)$ .

A vertical line has equation $x=k$ and undefined gradient, so it cannot be written as $y=mx+c$ . Keep the order of subtraction consistent in the numerator and denominator when calculating $m$ .

Relationship	Gradient condition
parallel non-vertical lines	$m_1=m_2$
perpendicular non-vertical lines	$m_1m_2=-1$ , so $m_2=-1/m_1$
vertical and horizontal lines	perpendicular to each other

If $AB$ has gradient $-4/3$ , a perpendicular line has gradient $3/4$ . Through $(-2,2)$ its equation is $y-2=\tfrac34(x+2)$ , so $y=\tfrac34x+\tfrac72$ . Substitute the given point to check the intercept.

Perpendicular gradients are negative reciprocals, not merely negatives. Equal gradients show the same direction; compare intercepts if you must distinguish two separate parallel lines from the same line.

$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right),\qquad AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The perpendicular bisector of $AB$ passes through its midpoint and is perpendicular to $AB$ . Find the midpoint, calculate the gradient of $AB$ , take its negative reciprocal, then use point-gradient form through the midpoint. Every point on the resulting line is equidistant from $A$ and $B$ .

For $A(-4,3)$ and $B(6,-9)$ , $M=(1,-3)$ and $m_{AB}=(-9-3)/(6+4)=-6/5$ . Therefore $m_{\perp}=5/6$ and the perpendicular bisector is $y+3=\tfrac56(x-1)$ . Substitution or equal-distance calculations can verify any claimed point.

The perpendicular bisector usually passes through neither endpoint. If $AB$ is horizontal, its perpendicular bisector is vertical through the midpoint; if $AB$ is vertical, the bisector is horizontal. Preserve exact radicals unless a decimal accuracy is requested.

A transformed graph is straight when chosen variables $X=f(x)$ and $Y=g(y)$ satisfy $Y=mX+c$ . Name the plotted variables first; then the graph's gradient and intercept can be matched to the original constants.

Original relationship	Plot $Y$ against $X$	Gradient / intercept
$y=Ax^n$	$Y=\ln y$ , $X=\ln x$	$m=n$ , $c=\ln A$
$y=Ab^x$	$Y=\ln y$ , $X=x$	$m=\ln b$ , $c=\ln A$
$y^2=Ax^3+B$	$Y=y^2$ , $X=x^3$	$m=A$ , $c=B$
$e^{2y}=Ax^2+B$	$Y=e^{2y}$ , $X=x^2$	$m=A$ , $c=B$
$y^3=A\ln x+B$	$Y=y^3$ , $X=\ln x$	$m=A$ , $c=B$

If plotting $\ln y$ against $\ln x$ gives points $(6,5)$ and $(8,9)$ , then $m=2$ and $c=-7$ . Thus $\ln y=2\ln x-7$ , so $y=e^{-7}x^2$ . The gradient becomes the power and the intercept must be exponentiated to recover $A$ .

The method also works from an already transformed graph. If $Y=\sqrt[4]{y}$ is plotted against $X=1/x$ through $(0.5,9)$ and $(3,34)$ , then $Y=10X+4$ . Hence $\sqrt[4]{y}=10/x+4$ and $y=(10/x+4)^4$ .

Gradient and intercept belong to the transformed axes, not automatically to $x$ and $y$ . State logarithm-domain conditions, reverse every transformation, and distinguish $c=\ln A$ from $c=A$ .

Next Topic

—8. Coordinate geometry of the circleRead

7. Straight-line graphs

Learning objectives

Build and use a straight-line equation

Recognise parallel and perpendicular lines

Construct a perpendicular bisector from coordinates

Read relationships through straight-line form

Next Topic