CG Coordinate Geometry (2 dimensions only) — AQA 8365 spec coverage

CG1

Know and use the definition of a gradient

The gradient measures vertical change per unit horizontal change: $m=\frac{y_2-y_1}{x_2-x_1}$ .
Choose the same order for both subtractions; reversing both orders gives the same gradient.
A positive gradient rises from left to right, a negative gradient falls, and a horizontal line has gradient $0$ .
A common error is to divide the change in $x$ by the change in $y$ .

Tier 1 · Easy

2 marks

ORIGINAL

Points $A(1,2)$ and $B(5,10)$ lie on a straight line. Calculate its gradient.

Tier 2 · Standard

2 marks

ORIGINAL

Work out the gradient of the line $3x-2y=12$ .

Tier 3 · Hard

3 marks

ORIGINAL

The gradient of the line through $A(-3,4)$ and $B(p,2p+1)$ is $\frac{3}{2}$ . Determine $p$ .

CG2

Know the relationship between the gradients of parallel and perpendicular lines

Distinct non-vertical parallel lines have equal gradients; distinct vertical lines are parallel and have undefined gradient.
For two non-vertical perpendicular lines, $m_1m_2=-1$ , so one gradient is the negative reciprocal of the other.
To prove a relationship, calculate both gradients and explicitly compare them.
A common error is to use only the reciprocal and forget the change of sign for perpendicular lines.

Tier 1 · Easy

1 mark

ORIGINAL

A line has gradient $-3$ . State the gradient of a line perpendicular to it.

Tier 2 · Standard

3 marks

ORIGINAL

Points $A(-2,1)$ , $B(4,3)$ , $C(1,5)$ and $D(3,-1)$ are given. Show that $AB$ is perpendicular to $CD$ .

Tier 3 · Hard

4 marks

ORIGINAL

The line $L$ has equation $2y=3x+7$ . A line perpendicular to $L$ passes through the point where $L$ meets the $y$ -axis. Express the new line as $ax+by=c$ using integer coefficients.

CG3

Use Pythagoras' theorem to calculate the distance between two points

For points $(x_1,y_1)$ and $(x_2,y_2)$ , the distance is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .
The coordinate differences are the horizontal and vertical side lengths of a right-angled triangle.
Keep exact surd answers unless a question asks for a decimal or a stated degree of accuracy.
A common error is to forget to square a negative coordinate difference.

Tier 1 · Easy

2 marks

ORIGINAL

Work out the distance between $P(1,2)$ and $Q(4,6)$ .

Tier 2 · Standard

3 marks

ORIGINAL

Work out the exact distance between $R(-2,5)$ and $S(6,-1)$ .

Tier 3 · Hard

4 marks

ORIGINAL

The vertices of a triangle are $A(-4,1)$ , $B(2,9)$ and $C(8,1)$ . Calculate its perimeter and state whether it is isosceles.

CG4

Use ratio to find the coordinates of a point on a line given the coordinates of two other points, including the midpoint

A midpoint is found by averaging corresponding coordinates: $M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ .
If $AP:PB=m:n$ , move the fraction $\frac{m}{m+n}$ of the way from $A$ to $B$ .
Apply the same ratio separately to the $x$ -coordinates and the $y$ -coordinates.
A common error is to weight a point by the segment next to it instead of the opposite segment.

Tier 1 · Easy

2 marks

ORIGINAL

Work out the midpoint of the line segment joining $A(-2,5)$ to $B(6,1)$ .

Tier 2 · Standard

3 marks

ORIGINAL

Point $P$ divides the segment from $A(-4,1)$ to $B(11,16)$ in the ratio $AP:PB=2:3$ . Find the coordinates of $P$ .

Tier 3 · Hard

4 marks

ORIGINAL

Point $P(3,-4)$ divides $A(-6,5)$ to $B(k,-7)$ in the ratio $AP:PB=3:1$ . Find $k$ and then find the midpoint of $AB$ .

CG5

The equation of a straight line: y = mx + c and y - y1 = m(x - x1) and other forms, including interpretation of the gradient and y-intercept

In $y=mx+c$ , $m$ is the gradient and $(0,c)$ is the $y$ -intercept.
A line of gradient $m$ through $(x_1,y_1)$ can be written as $y-y_1=m(x-x_1)$ .
When two points are given, first calculate the gradient and then substitute one point into a line equation.
Check a final equation by substituting a known point; a common error is a sign mistake when expanding brackets.

Tier 1 · Easy

2 marks

ORIGINAL

State the gradient and the $y$ -intercept of the line $y=3x-4$ .

Tier 2 · Standard

3 marks

ORIGINAL

Work out the equation of the line with gradient $4$ that passes through $(2,-1)$ . Give your answer as $y=mx+c$ .

Tier 3 · Hard

4 marks

ORIGINAL

A straight line passes through $A(-2,5)$ and $B(4,-7)$ . Find its equation in the form $ax+by=c$ and find its $x$ -intercept.

CG6

Draw a straight line from given information

Two distinct points determine a straight line, although a third point is useful as a check.
For $y=mx+c$ , plot $(0,c)$ and use the gradient to find another point.
For an equation such as $ax+by=c$ , rearrange it or calculate convenient coordinate pairs.
Use a ruler and extend the line across the full stated interval; do not join plotted points with separate short segments.

Tier 1 · Easy

2 marks

ORIGINAL

On coordinate axes, draw $y=2x+1$ for $-1\leq x\leq2$ .

Tier 2 · Standard

3 marks

ORIGINAL

On coordinate axes, draw $2x+3y=6$ for $-3\leq x\leq3$ .

Tier 3 · Hard

4 marks

ORIGINAL

A line passes through $(-2,5)$ and is perpendicular to $y=\frac{1}{2}x-1$ . Draw this line for $-1\leq x\leq3$ .

CG7

Understand that x^2 + y^2 = r^2 is the equation of a circle with centre (0, 0) and radius r; application of circle geometry facts

The circle $x^2+y^2=r^2$ has centre $(0,0)$ and radius $r$ .
A point lies on the circle when its coordinates satisfy the equation exactly.
Every radius has length $r$ , and a radius is perpendicular to the tangent at its endpoint.
A common error is to quote $r^2$ as the radius instead of taking its positive square root.

Tier 1 · Easy

2 marks

ORIGINAL

State the centre and radius of the circle $x^2+y^2=49$ .

Tier 2 · Standard

2 marks

ORIGINAL

Point $P(6,y)$ lies on $x^2+y^2=100$ and $y>0$ . Find $y$ .

Tier 3 · Hard

4 marks

ORIGINAL

On $x^2+y^2=100$ , take $A(6,8)$ and $B(-8,6)$ , with centre $O$ . Show that angle $AOB$ is $90^\circ$ , and calculate the area of triangle $AOB$ .

CG8

Understand that (x - a)^2 + (y - b)^2 = r^2 is the equation of a circle with centre (a, b) and radius r

The circle $(x-a)^2+(y-b)^2=r^2$ has centre $(a,b)$ and radius $r$ .
The signs inside the brackets are opposite to the signs of the centre coordinates.
Find $r^2$ by substituting a known point or by using the squared distance from the centre.
If the endpoints of a diameter are known, their midpoint is the centre and half their distance is the radius.

Tier 1 · Easy

2 marks

ORIGINAL

State the centre and radius of $(x-3)^2+(y+2)^2=25$ .

Tier 2 · Standard

3 marks

ORIGINAL

A circle has centre $(-4,1)$ and passes through $(2,9)$ . Find its equation.

Tier 3 · Hard

4 marks

ORIGINAL

The endpoints of a diameter of a circle are $A(-3,5)$ and $B(7,-1)$ . Find the equation of the circle.

CG9

The equation of a tangent at a point on a circle

The tangent to a circle is perpendicular to the radius at the point of contact.
For a non-axis-aligned radius, find its gradient and take the negative reciprocal to obtain the tangent gradient.
A vertical radius has a horizontal tangent, while a horizontal radius has a vertical tangent.
Use the point of contact in $y-y_1=m(x-x_1)$ to form the tangent equation.
A common error is to use the radius gradient unchanged, producing the radius line rather than the tangent.

Tier 1 · Easy

2 marks

ORIGINAL

The point $(3,4)$ lies on $x^2+y^2=25$ . Find the gradient of the tangent there.

Tier 2 · Standard

3 marks

ORIGINAL

Work out the equation of the tangent to $x^2+y^2=65$ at $(1,8)$ .

Tier 3 · Hard

4 marks

ORIGINAL

The tangent to $(x-2)^2+(y+3)^2=100$ at $P(8,5)$ meets the $y$ -axis at $Q$ . Find the coordinates of $Q$ .