CAIE IGCSE Additional Math Geometry and Trigonometry Questions

Use this geometry-and-trigonometry unit hub to connect coordinate lines and circles with circular measure, exact trigonometric values, graphs, identities and equations.

Exam points

Solve coordinate line and circle problems using gradients, distances, ratios, equations, intersections or tangents.
Use radian measure, arc lengths, sector areas and compound circular regions in exact or numerical form.
Interpret trig values and graphs, prove identities and solve equations over stated degree or radian intervals.

Covered syllabus points

7. Straight-line graphs 8. Coordinate geometry of the circle 9. Circular measure 10. Trigonometry

[Maximum number: 7]

4 The coordinates of points A, B, C and D are as follows.

A(-4,3)\quad B(6,-9)\quad C(15,10)\quad D(14,-1)

The line L has equation y=11 x-75 .

The perpendicular bisector of the line A B meets L at the point E .

Find the area of triangle C D E .

$m_{A B}=\frac{-9-3}{6--4}$ oe isw or $-\frac{6}{5}$ $\left[m_{\perp}=\right] \frac{5}{6}$ or $\frac{-1}{\text { their } m_{A B}}$
FT their $m_{A B}$ providing that
$\frac{\text { difference in } y}{\text { difference in } x}$ attempted
[Midpoint] $\left(\frac{-4+6}{2}, \frac{3-9}{2}\right)$ oe isw or (1,-3)
Finds equation of perpendicular bisector:
$y \text {-their }(-3)=-\frac{1}{\text { their }\left(-\frac{6}{5}\right)}(x-\text { their } 1) \mathrm{oe}$
FT their perpendicular gradient providing gradient is $\frac{-1}{\text { their } m_{A B}}$ and using their midpoint

Their midpoint must not be any of A, B, C or D $11 x-75=\frac{5}{6} x-\frac{23}{6}$
Eliminates one variable
dep on previous $\mathbf{M}$ mark
11 x-75= their perp bisector equation
E(7,2)
[Area C D E= ] 40
Dependent on all previous marks

[Maximum number: 4]

Variables x and y are such that when $\ln y$ is plotted against x, a straight-line graph is obtained.
The line passes through the points $(1,\ln 15)$ and $(2,\ln 75)$ .
Show that $y=Ab^x$ , where A and b are integers to be found.

$\ln y=mx+c$ soi
Correct method to find m, e.g. $m=\frac{\ln 75-\ln 15}{2-1}$ oe, or $-m=\ln 15-\ln 75$ oe, or $m=\ln 5$ oe.
M1 FT their c if necessary.
Correct method to find c, e.g. $\ln 15=(\text{their }\ln 5)(1)+c$ oe, $\ln 75=(\text{their }\ln 5)(2)+c$ oe, or $c=\ln 3$ oe, soi.
M1 FT their m if necessary.
$y=3\times 5^x$ nfww, mark final answer

[Maximum number: 5]

A circle has equation $x^2+y^2-25=0$ .
A second circle has the same radius as the first circle, and the coordinates of its centre are both positive.
The two circles intersect at the points A and B.
The line AB has length 6 and is parallel to the line y=-x.
Find the equation of the second circle in the form $x^2+y^2+ax+by+c=0$ , where a, b and c are constants.

Radius: 5 soi
Centre: $(4\sqrt2,4\sqrt2)$ soi oe
B1 for (5.66,5.66) or better, or for distance between centres is 8 soi.
$\left(x-\frac{8}{\sqrt2}\right)^2+\left(y-\frac{8}{\sqrt2}\right)^2=25$
or equivalent expansion.
Final answer: $x^2+y^2-8\sqrt2x-8\sqrt2y+39=0$
Must be exact.
B1, B2, M1, A1 with FT alternatives as in markscheme.

[Maximum number: 4]

Variables x and y are such that when $\sqrt{y}$ is plotted against $x^{3}$ a straight line graph passing through the points (2,5) and (10,21) is obtained.

Find y in terms of x.

$\sqrt{y}=m x^{3}+c$ soi
$m=\frac{21-5}{10-2}$ oe or 2
c=5-2(2) oe or 1
FT their m

$y=\left(2 x^{3}+1\right)^{2}$ oe, isw
Alternative method

$\sqrt{y}=m x^{3}+c$ soi
21=10 m+c and 5=2 m+c and solves to find
m=2 or c=1
Finds the other unknown
FT their m or c

$y=\left(2 x^{3}+1\right)^{2}$ oe, isw

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Covered syllabus points

Question 1

Question 2

Question 3

Question 4