CAIE A-Level Mathematics AS 1.3 Coordinate Geometry Questions

Practise coordinate geometry with straight lines, circles, tangents and intersections by forming equations from points and graphs.

Exam points

find line equations from gradients, points, midpoints or perpendicular relationships
use circle centres and radii to write equations or find tangent equations
eliminate variables to test intersections with a quadratic discriminant or solve coordinates

Covered syllabus points

1.3.1—Straight lines 1.3.2—Line equation forms 1.3.3—Circle equation 1.3.4—Lines and circles 1.3.5—Graph intersections

[Maximum number: 2]

Figure for Question 1 — CAIE A-Level Mathematics AS

The diagram shows the curve with equation $x=y^{2}+1$ . The points A(5,2) and B(2,-1) lie on the curve.

Find an equation of the line A B.

Gradient of $A B=\frac{2-(-1)}{5-2}$
Expect 1, must be from $\Delta y / \Delta x$ .

Equation of A B is y-2=1(x-5) or y+1=1(x-2)
OE. Expect y=x-3.

[Maximum number: 6]

The equation of a curve is $y=(5-2 x)^{\frac{3}{2}}+5$ for $x<\frac{5}{2}$ .

Point A on the curve has y-coordinate 32. Point B on the curve is such that the gradient of the curve at B is -3 .

Find the equation of the perpendicular bisector of A B. Give your answer in the form a x+b y+c=0, where a, b and c are integers.

$k(5-2 x)^{\frac{1}{2}}=-3$
Equating their $\frac{\mathrm{d} y}{\mathrm{~d} x}$ of the form $k(5-2 x)^{\frac{1}{2}}$ to -3 .

[B is ](2,6)
Gradient $A B=m_{1}=\frac{32-6}{-2-2}$ ,gradient of perpendicular $=-\frac{1}{m_{1}}=\frac{4}{26}$

M1*

For A, y must be 32 .
Clear use of $\frac{\text { difference in y co-ordinates }}{\text { difference in x co-ordinates }}$ for points
A and B, condone inconsistent order, and using
$m_{1} m_{2}=-1$ .
If incorrect values or another complete method used, then working must be clear.

Mid point is $\left(\frac{2-2}{2}, \frac{6+32}{2}\right)=(0,19)$

M1*

Finding the midpoint of A B using A and B. If
incorrect values used then all working must be clear.
For A, y must be 32 .

$y-19=\frac{2}{13}(x-0)$
Finding the equation of the perpendicular bisector using their midpoint and their perpendicular
gradient.

2 x-13 y+247=0 or $\pm$ integer multiples of this.

[Maximum number: 5]

Figure for Question 3 — CAIE A-Level Mathematics AS

The diagram shows the circle with centre C(-4,5) and radius $\sqrt{20}$ units. The circle intersects the y-axis at the points A and B. The size of angle A C B is $\theta$ radians.

(a)

Find the equation of the tangent to the circle at the point (-6,9).

[ 3 ]

Obtain gradient of relevant radius is -2
Using $m_{1} m_{2}=-1$ obtain the gradient of the tangent and use it to form a
straight line equation for a line containing (-6, 9)
$m_{1}$ must be from an attempt to find the gradient of the
radius using the centre and the given point.

Obtain $y=\frac{1}{2} x+12$
OE e.g. $y-9=\frac{1}{2}(x+6)$ .

(b)

Find the equation of the circle in the form $x^{2}+y^{2}+a x+b y+c=0$ .

[ 2 ]

State or imply $(x+4)^{2}+(y-5)^{2}=20$
If $x^{2}+y^{2}-2 g x-2 f y+c=0$ is used correctly with
(-g,-f)=(-4,5) and $c=g^{2}+f^{2}-r^{2}$ then M1.

Obtain $x^{2}+y^{2}+8 x-10 y+21=0$
A1 if above method used.

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Covered syllabus points

Question 1

Question 2

Question 3

Question (a)

Question (b)