Geometry and Trigonometry

Exam analysis

Most tested topics

7. Straight-line graphs

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 $c=A$ .

8. Coordinate geometry of the circle

Equation	Centre	Radius
$(x-a)^2+(y-b)^2=r^2$	$(a,b)$	$r$
$x^2+y^2+2gx+2fy+c=0$	$(-g,-f)$	$\sqrt{g^2+f^2-c}$

The standard form is the distance formula from a general point $(x,y)$ to the fixed centre $(a,b)$ . To convert a general equation, complete each square using $x^2+2gx=(x+g)^2-g^2$ and the matching identity for $y$ .

If $A(-3,6)$ and $B(7,-8)$ are endpoints of a diameter, the centre is their midpoint $(2,-1)$ . The radius squared is $(-3-2)^2+(6+1)^2=74$ , so the circle is $(x-2)^2+(y+1)^2=74$ .

To classify a point, compare its squared distance from the centre with $r^2$ : smaller means inside, equal means on the circumference, larger means outside. In standard form, the centre signs are opposite to those visible inside the brackets.

Substitute the line equation into the circle equation. The resulting quadratic describes the possible intersection coordinates; back-substitute every real root to obtain the full points.

Discriminant $\Delta=b^2-4ac$	Real roots	Geometry
$\Delta>0$	two	the line cuts a chord
$\Delta=0$	one repeated root	the line is tangent
$\Delta<0$	none	the line does not meet the circle

For $(x-5)^2+(y-2)^2=5$ and $y=2x-3$ , substitution simplifies to $5x^2-30x+45=5(x-3)^2=0$ . There is one repeated root, so the line is tangent; $x=3$ gives the contact point $(3,3)$ .

An axis is just a special line: use $x=0$ for the $y$ -axis and $y=0$ for the $x$ -axis. For $(x-4)^2+(y-2)^2=40$ on the $y$ -axis, $(y-2)^2=24$ , giving $(0,2\pm2\sqrt6)$ .

Do not classify from a sketch alone. A repeated quadratic root is one coordinate value, but the point of contact still needs the other coordinate from the line.

The tangent at a point on a circle is perpendicular to the radius through that point. This right-angle fact supplies the tangent gradient; calculus is not required.

Identify the centre, verify or use the contact point, calculate the radius gradient, take its negative reciprocal for the tangent gradient, then use point-gradient form through the contact point.

A circle has centre $(4,-3)$ and contains $A(3,-1)$ . The radius $CA$ has gradient $(-1+3)/(3-4)=-2$ , so the tangent gradient is $1/2$ . Hence the tangent is $y+1=\tfrac12(x-3)$ .

The tangent passes through the point on the circumference, not through the centre. A horizontal radius gives a vertical tangent; a vertical radius gives a horizontal tangent, so the negative-reciprocal formula is not used with an undefined gradient.

For circles with centre distance $d$ and radii $r_1,r_2$ , compare $d$ with $r_1+r_2$ and $|r_1-r_2|$ . These distances describe when the circles can reach each other externally or internally.

Condition	Relationship
$d>r_1+r_2$	separate outside
$d=r_1+r_2$	touch externally
$\|r_1-r_2\|<d<r_1+r_2$	intersect at two points
$d=\|r_1-r_2\|$	touch internally
$d<\|r_1-r_2\|$	one lies inside the other without touching

To find common intersection points, subtract the two expanded circle equations. The $x^2$ and $y^2$ terms cancel, leaving the straight-line equation of the common chord. Solve that line simultaneously with either original circle.

Centres $(8,5)$ and $(10,6.5)$ are $\sqrt{2^2+1.5^2}=2.5$ apart. With radii $4$ and $1.5$ , $|4-1.5|=2.5$ , so the circles touch internally; no contact coordinate is needed for this conclusion.

Subtracting circle equations gives the common-chord line, not the intersection points by itself. If $d=0$ and $r_1=r_2$ , the equations describe the same circle rather than one of the five distinct-circle cases above.

9. Circular measure

An angle of $\theta$ radians is defined by $\theta=s/r$ , where $s$ is the subtended arc length and $r$ is the radius. Because a full circumference is $2\pi r$ , one full turn is $2\pi$ radians and $180^\circ=\pi$ radians.

Quantity, with $\theta$ in radians	Formula
arc length	$s=r\theta$
sector area	$A=\tfrac12r^2\theta$
sector perimeter	$P=2r+r\theta$
triangle made by two radii	$A_\triangle=\tfrac12r^2\sin\theta$
minor segment area	$A_\text{segment}=\tfrac12r^2(\theta-\sin\theta)$

A sector of radius $24$ has area $432$ . From $\tfrac12(24)^2\theta=432$ , $\theta=3/2$ radians. Its arc is therefore $s=24(3/2)=36$ . This also follows from $A=\tfrac12rs$ , which is obtained by combining the first two formulas.

For an annular sector with outer radius $8$ , inner radius $5$ and angle $\pi/3$ , subtract sector areas: $A=\tfrac12(8^2-5^2)(\pi/3)=13\pi/2$ . Its boundary contains both arcs and two radial gaps, so $P=8\pi/3+5\pi/3+2(8-5)=13\pi/3+6$ .

For any compound shape, mark each boundary piece before adding the perimeter, then decompose the area into sectors, triangles, rectangles or segments and attach a plus or minus sign to each. Use $2\pi-\theta$ for a major angle and keep exact $\pi$ or radical values until the final step.

The formulas $s=r\theta$ and $A=\tfrac12r^2\theta$ require radians. A sector perimeter includes two radii; a segment area is sector minus triangle, not sector alone. Convert degrees before substitution and include units: length for arcs and squared units for areas.

10. Trigonometry

Sine and cosine locate an angle on the unit circle; the other four functions are ratios or reciprocals of them. This connects all six functions for angles of any magnitude, not only acute angles.

Function	In terms of $\sin\theta$ and $\cos\theta$	Reciprocal pair
$\sin\theta$	$\sin\theta$	$\cosec\theta=1/\sin\theta$
$\cos\theta$	$\cos\theta$	$\sec\theta=1/\cos\theta$
$\tan\theta$	$\sin\theta/\cos\theta$	$\cot\theta=1/\tan\theta=\cos\theta/\sin\theta$

Reduce a large or negative angle to a coterminal angle, find its reference angle, then attach the quadrant sign. In quadrant I all six are positive; in II only sine and cosecant are positive; in III only tangent and cotangent are positive; in IV only cosine and secant are positive.

If $\cos\theta=-\sqrt{5/6}$ and $\theta$ is in quadrant III, then $\sin\theta=-\sqrt{1-5/6}=-\sqrt{1/6}$ . Hence $\tan\theta=1/\sqrt5$ , $\sec\theta=-\sqrt{6/5}$ , $\cosec\theta=-\sqrt6$ and $\cot\theta=\sqrt5$ . The quadrant fixes signs that a square root alone cannot decide.

A reciprocal is undefined when its denominator is zero: secant fails where cosine is zero, and cosecant and cotangent fail where sine is zero. Keep the calculator angle mode consistent with the question's degrees or radians.

In $y=a\sin(bx)+c$ or $y=a\cos(bx)+c$ , $|a|$ is the amplitude, $c$ is the midline, and $b$ controls how quickly the cycle repeats. Tangent has no amplitude because it is unbounded.

Family	Amplitude	Period in radians	Period in degrees
$a\sin(bx)+c$	$\|a\|$	$2\pi/\|b\|$	$360^\circ/\|b\|$
$a\cos(bx)+c$	$\|a\|$	$2\pi/\|b\|$	$360^\circ/\|b\|$
$a\tan(bx)+c$	none	$\pi/\|b\|$	$180^\circ/\|b\|$

Multiplying the function by $a$ scales vertical distances from the midline; multiplying $x$ by $b$ divides every horizontal feature by $|b|$ ; adding $c$ shifts the entire graph vertically. A negative $a$ also reflects the graph in its midline.

For $y=2\cos(x/3)-1$ , the amplitude is $2$ , the midline is $y=-1$ , and the period is $2\pi/(1/3)=6\pi$ . For $y=5\tan(x/4)+1$ , there is no amplitude and the period is $\pi/(1/4)=4\pi$ .

Do not use the sine/cosine period rule for tangent, and do not call the vertical multiplier of tangent an amplitude. State whether the horizontal unit is degrees or radians before calculating a period.

A reliable trig sketch starts from one standard cycle, transforms its horizontal and vertical features, and repeats them only as far as the stated domain requires.

Graph	One-cycle anchors in radians	Shape feature
$y=\sin x$	$(0,0),(\pi/2,1),(\pi,0),(3\pi/2,-1),(2\pi,0)$	crosses the midline at multiples of $\pi$
$y=\cos x$	$(0,1),(\pi/2,0),(\pi,-1),(3\pi/2,0),(2\pi,1)$	begins a cycle at a maximum
$y=\tan x$	$(-\pi/4,-1),(0,0),(\pi/4,1)$	increasing branch between $x=-\pi/2$ and $x=\pi/2$

For $y=a\sin(bx)+c$ or $y=a\cos(bx)+c$ , draw the midline $y=c$ , calculate the period, mark quarter-period anchors and scale their heights by $a$ . For $y=a\tan(bx)+c$ , solve $bx=\pi/2+k\pi$ for every asymptote in the domain, label each $x$ -coordinate, then place the centre crossing where $bx=k\pi$ and $y=c$ .

For $y=5\tan(x/4)+1$ on $-\pi\le x\le5\pi$ , the period is $4\pi$ . Its asymptotes satisfy $x=2\pi+4k\pi$ , so the only one strictly inside this domain is $x=2\pi$ . The branch crosses its midline at $(0,1)$ and $(4\pi,1)$ ; label the asymptote before drawing the increasing branches.

A curve must approach but never touch a tangent asymptote. Apply the factor $b$ to the input before locating roots, turning points or asymptotes, and plot only the requested degree or radian domain.

The three Pythagorean identities express the same unit-circle relationship in different function families. Choose the identity containing the function you know and the function you need.

\sin^2 A+\cos^2 A=1,\qquad \sec^2 A=1+\tan^2 A,\qquad \cosec^2 A=1+\cot^2 A

The second identity comes from dividing $\sin^2A+\cos^2A=1$ by $\cos^2A$ ; the third comes from dividing by $\sin^2A$ . Use $\sec A=1/\cos A$ , $\cosec A=1/\sin A$ , $\tan A=\sin A/\cos A$ and $\cot A=\cos A/\sin A$ to move between families.

Suppose $\sec x=\alpha$ and $x$ is in quadrant IV. Then $\cos x=1/\alpha$ and $\tan^2x=\alpha^2-1$ . Tangent is negative in quadrant IV, so $\tan x=-\sqrt{\alpha^2-1}$ and $\sin x=\tan x\cos x=-\sqrt{\alpha^2-1}/\alpha$ .

An identity gives a squared value before it gives a sign. Use the stated domain or quadrant to choose $+$ or $-$ , and note that reciprocal functions require a non-zero denominator.

A trigonometric equation is complete only when every solution in the stated domain has been found and checked. Work with the transformed angle first, then convert back to the original variable.

Rewrite reciprocal functions or use an identity until the equation is in one trig function. 2. Factor or solve the resulting algebraic equation. 3. Translate the original domain into the domain of the transformed angle. 4. Find every reference-angle and quadrant solution in that interval. 5. Convert back, list in order and substitute to reject invalid values.

Solve $\sec^2(3x)+\tan(3x)-3=0$ for $0^\circ\le x\le120^\circ$ . Since $\sec^2(3x)=1+\tan^2(3x)$ , let $t=\tan(3x)$ : $t^2+t-2=0$ , so $t=1$ or $t=-2$ . The angle $3x$ ranges from $0^\circ$ to $360^\circ$ , giving $x=15^\circ,38.9^\circ,75^\circ,98.9^\circ$ (to $1$ d.p.).

For secant or cosecant equations, first use the reciprocal relationship and reject values outside the possible sine/cosine range. For cotangent, convert to tangent when convenient. If the equation contains squares, both signs may produce branches that must be tested over the whole domain.

Do not divide by $\sin x$ , $\cos x$ or another expression until checking whether zero makes the original equation true; division can lose solutions. Also exclude angles where the original reciprocal function is undefined.

To prove a trigonometric identity, transform one side until it has exactly the form of the other. Each line must follow from an identity or valid algebra; checking a few angles is not a proof.

Start with the more complicated side. Replace $\sec$ , $\cosec$ , $\tan$ and $\cot$ by sine and cosine when this reduces the number of function types. Then factor, use a common denominator, or create $\sin^2x+\cos^2x$ . Stop as soon as the required expression appears; do not manipulate both sides toward an unproved middle statement.

For $\dfrac{\cos x}{1-\sin x}+\dfrac{1-\sin x}{\cos x}$ , rationalise the first fraction: $\dfrac{\cos x(1+\sin x)}{1-\sin^2x}=\dfrac{1+\sin x}{\cos x}$ . Adding the second fraction gives $\dfrac{1+\sin x+1-\sin x}{\cos x}=\dfrac2{\cos x}=2\sec x$ .

Look for a structure before expanding: $1-\sin^2x$ suggests $\cos^2x$ ; $\sec^2x-1$ suggests $\tan^2x$ ; a sum of fractions suggests a common denominator. The useful identity is the one that removes, rather than creates, complexity.

A proof is valid only where every expression used is defined. Cancelling a factor assumes it is non-zero, so preserve domain restrictions from denominators and reciprocal functions even when the final expression looks defined more widely.

Geometry and Trigonometry

Exam analysis

Most tested topics

7. Straight-line graphs

Build and use a straight-line equation

Recognise parallel and perpendicular lines

Construct a perpendicular bisector from coordinates

Read relationships through straight-line form

8. Coordinate geometry of the circle

Read and construct circle equations

Classify a line's intersection with a circle

Find a circle tangent without calculus

Compare and intersect two circles

9. Circular measure

Build circular measures from radians

10. Trigonometry

Connect all six trigonometric functions

Read amplitude and period from a trig rule

Sketch transformed trigonometric graphs

Choose and use a trigonometric identity

Solve trigonometric equations over a domain

Prove trigonometric relationships from one side