CG Coordinate Geometry (2 dimensions only) — printable coverage pack

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

1. Points $A(1,2)$ and $B(5,10)$ lie on a straight line. Calculate its gradient.[2 marks]
Answer
- $m=2$
Method: $m=\frac{10-2}{5-1}=\frac{8}{4}=2$ .

Tier 2 · Standard

1. Work out the gradient of the line $3x-2y=12$ .[2 marks]
Answer
- $m=\frac{3}{2}$
Method: Rearrange to make $y$ the subject: $-2y=12-3x$ , so $y=\frac{3}{2}x-6$ . The coefficient of $x$ is the gradient.

Tier 3 · Hard

1. The gradient of the line through $A(-3,4)$ and $B(p,2p+1)$ is $\frac{3}{2}$ . Determine $p$ .[3 marks]
Answer
- $p=15$
Method: Using the two points, $\frac{2p+1-4}{p-(-3)}=\frac{2p-3}{p+3}$ . Set this equal to $\frac{3}{2}$ : $2(2p-3)=3(p+3)$ , so $4p-6=3p+9$ and $p=15$ .

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. A line has gradient $-3$ . State the gradient of a line perpendicular to it.[1 mark]
Answer
- $\frac{1}{3}$
Method: The perpendicular gradient is the negative reciprocal. The negative reciprocal of $-3$ is $\frac{1}{3}$ , and $-3\times\frac{1}{3}=-1$ .

Tier 2 · Standard

1. Points $A(-2,1)$ , $B(4,3)$ , $C(1,5)$ and $D(3,-1)$ are given. Show that $AB$ is perpendicular to $CD$ .[3 marks]
Answer
- $AB\perp CD$
Method: $m_{AB}=\frac{3-1}{4-(-2)}=\frac{2}{6}=\frac{1}{3}$ and $m_{CD}=\frac{-1-5}{3-1}=\frac{-6}{2}=-3$ . Their product is $\frac{1}{3}\times(-3)=-1$ , so the lines are perpendicular.

Tier 3 · Hard

1. 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.[4 marks]
Answer
- $4x+6y=21$
Method: $L$ is $y=\frac{3}{2}x+\frac{7}{2}$ , so its gradient is $\frac{3}{2}$ and its $y$ -intercept is $(0,\frac{7}{2})$ . The perpendicular gradient is $-\frac{2}{3}$ . Hence $y-\frac{7}{2}=-\frac{2}{3}x$ . Multiplying by $6$ and rearranging gives $4x+6y=21$ .

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

1. Work out the distance between $P(1,2)$ and $Q(4,6)$ .[2 marks]
Answer
- $PQ=5$
Method: The coordinate differences are $4-1=3$ and $6-2=4$ . Therefore $PQ=\sqrt{3^2+4^2}=\sqrt{25}=5$ .

Tier 2 · Standard

1. Work out the exact distance between $R(-2,5)$ and $S(6,-1)$ .[3 marks]
Answer
- $RS=10$
Method: $RS=\sqrt{(6-(-2))^2+(-1-5)^2}=\sqrt{8^2+(-6)^2}=\sqrt{100}=10$ .

Tier 3 · Hard

1. 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.[4 marks]
Answer
- Perimeter $=32$
- The triangle is isosceles.
Method: $AB=\sqrt{6^2+8^2}=10$ , $BC=\sqrt{6^2+(-8)^2}=10$ , and $AC=\sqrt{12^2+0^2}=12$ . The perimeter is $10+10+12=32$ . Since $AB=BC$ , the triangle 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

1. Work out the midpoint of the line segment joining $A(-2,5)$ to $B(6,1)$ .[2 marks]
Answer
- $(2,3)$
Method: Average the coordinates: $x=\frac{-2+6}{2}=2$ and $y=\frac{5+1}{2}=3$ . The midpoint is $(2,3)$ .

Tier 2 · Standard

1. Point $P$ divides the segment from $A(-4,1)$ to $B(11,16)$ in the ratio $AP:PB=2:3$ . Find the coordinates of $P$ .[3 marks]
Answer
- $P=(2,7)$
Method: The vector from $A$ to $B$ is $(15,15)$ . Since $AP$ is $\frac{2}{5}$ of $AB$ , add $\frac{2}{5}(15,15)=(6,6)$ to $A$ . This gives $P=(-4,1)+(6,6)=(2,7)$ .

Tier 3 · Hard

1. 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 $AB$ .[4 marks]
Answer
- $k=6$
- Midpoint of $AB$ is $(0,-1)$
Method: For $AP:PB=3:1$ , $P=\frac{A+3B}{4}$ . Using the $x$ -coordinate, $3=\frac{-6+3k}{4}$ , so $12=-6+3k$ and $k=6$ . Thus $B=(6,-7)$ . The midpoint is $(\frac{-6+6}{2},\frac{5+(-7)}{2})=(0,-1)$ .

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

1. State the gradient and the $y$ -intercept of the line $y=3x-4$ .[2 marks]
Answer
- Gradient $=3$
- $y$ -intercept $=(0,-4)$
Method: Compare $y=3x-4$ with $y=mx+c$ . Therefore $m=3$ and $c=-4$ , so the line crosses the $y$ -axis at $(0,-4)$ .

Tier 2 · Standard

1. Work out the equation of the line with gradient $4$ that passes through $(2,-1)$ . Give your answer as $y=mx+c$ .[3 marks]
Answer
- $y=4x-9$
Method: Use point-gradient form: $y-(-1)=4(x-2)$ . Hence $y+1=4x-8$ , so $y=4x-9$ .

Tier 3 · Hard

1. 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.[4 marks]
Answer
- $2x+y=1$
- $x$ -intercept $=(\frac{1}{2},0)$
Method: The gradient is $\frac{-7-5}{4-(-2)}=\frac{-12}{6}=-2$ . Through $A$ , $y-5=-2(x+2)$ , so $y=-2x+1$ and hence $2x+y=1$ . At the $x$ -intercept $y=0$ , giving $2x=1$ and $x=\frac{1}{2}$ .

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

1. On coordinate axes, draw $y=2x+1$ for $-1\leq x\leq2$ .[2 marks]
Answer
- The straight line segment through $(-1,-1)$ , $(0,1)$ , $(1,3)$ and $(2,5)$
Method: Substitute $x=-1,0,1,2$ to obtain $y=-1,1,3,5$ . Plot the points and join them with one straight line segment over the stated interval.

Tier 2 · Standard

1. On coordinate axes, draw $2x+3y=6$ for $-3\leq x\leq3$ .[3 marks]
Answer
- The straight line segment through $(-3,4)$ , $(0,2)$ and $(3,0)$
Method: Rearrange to $y=2-\frac{2}{3}x$ . Substitution gives the convenient points $(-3,4)$ , $(0,2)$ and $(3,0)$ . Plot them and draw one straight line through them across the interval.

Tier 3 · Hard

1. A line passes through $(-2,5)$ and is perpendicular to $y=\frac{1}{2}x-1$ . Draw this line for $-1\leq x\leq3$ .[4 marks]
Answer
- The line segment $y=-2x+1$ from $(-1,3)$ to $(3,-5)$
Method: The given gradient is $\frac{1}{2}$ , so the perpendicular gradient is $-2$ . Through $(-2,5)$ , $y-5=-2(x+2)$ , giving $y=-2x+1$ . Plot points such as $(-1,3)$ , $(0,1)$ and $(3,-5)$ , then draw the straight segment.

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

1. State the centre and radius of the circle $x^2+y^2=49$ .[2 marks]
Answer
- Centre $=(0,0)$
- Radius $=7$
Method: Compare with $x^2+y^2=r^2$ . Here $r^2=49$ , so the centre is $(0,0)$ and the positive radius is $r=7$ .

Tier 2 · Standard

1. Point $P(6,y)$ lies on $x^2+y^2=100$ and $y>0$ . Find $y$ .[2 marks]
Answer
- $y=8$
Method: Substitute $x=6$ : $6^2+y^2=100$ , so $y^2=64$ and $y=\pm8$ . Since $y>0$ , $y=8$ .

Tier 3 · Hard

1. 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 $AOB$ .[4 marks]
Answer
- $\angle AOB=90^\circ$
- Area of triangle $AOB=50$
Method: The gradient of $OA$ is $\frac{8}{6}=\frac{4}{3}$ and the gradient of $OB$ is $\frac{6}{-8}=-\frac{3}{4}$ . Their product is $-1$ , so $OA\perp OB$ and $\angle AOB=90^\circ$ . Both are radii of length $10$ , so the area is $\frac{1}{2}\times10\times10=50$ .

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

1. State the centre and radius of $(x-3)^2+(y+2)^2=25$ .[2 marks]
Answer
- Centre $=(3,-2)$
- Radius $=5$
Method: Write $y+2$ as $y-(-2)$ . Comparing with $(x-a)^2+(y-b)^2=r^2$ gives centre $(3,-2)$ and radius $\sqrt{25}=5$ .

Tier 2 · Standard

1. A circle has centre $(-4,1)$ and passes through $(2,9)$ . Find its equation.[3 marks]
Answer
- $(x+4)^2+(y-1)^2=100$
Method: The squared radius is the squared distance from $(-4,1)$ to $(2,9)$ : $r^2=(2+4)^2+(9-1)^2=6^2+8^2=100$ . Insert the centre and $r^2$ into the circle equation.

Tier 3 · Hard

1. The endpoints of a diameter of a circle are $A(-3,5)$ and $B(7,-1)$ . Find the equation of the circle.[4 marks]
Answer
- $(x-2)^2+(y-2)^2=34$
Method: The centre is the midpoint: $(\frac{-3+7}{2},\frac{5+(-1)}{2})=(2,2)$ . The diameter has squared length $10^2+(-6)^2=136$ , so $r^2=\frac{136}{4}=34$ . Hence the equation is $(x-2)^2+(y-2)^2=34$ .

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

1. The point $(3,4)$ lies on $x^2+y^2=25$ . Find the gradient of the tangent there.[2 marks]
Answer
- Tangent gradient $=-\frac{3}{4}$
Method: The radius from $(0,0)$ to $(3,4)$ has gradient $\frac{4}{3}$ . The tangent is perpendicular, so its gradient is the negative reciprocal, $-\frac{3}{4}$ .

Tier 2 · Standard

1. Work out the equation of the tangent to $x^2+y^2=65$ at $(1,8)$ .[3 marks]
Answer
- $x+8y=65$
Method: The radius gradient is $\frac{8}{1}=8$ , so the tangent gradient is $-\frac{1}{8}$ . Thus $y-8=-\frac{1}{8}(x-1)$ . Multiplying by $8$ and rearranging gives $x+8y=65$ .

Tier 3 · Hard

1. The tangent to $(x-2)^2+(y+3)^2=100$ at $P(8,5)$ meets the $y$ -axis at $Q$ . Find the coordinates of $Q$ .[4 marks]
Answer
- $Q=(0,11)$
Method: The centre is $(2,-3)$ . The radius to $P$ has gradient $\frac{5-(-3)}{8-2}=\frac{8}{6}=\frac{4}{3}$ , so the tangent gradient is $-\frac{3}{4}$ . Its equation is $y-5=-\frac{3}{4}(x-8)$ , or $y=-\frac{3}{4}x+11$ . At the $y$ -axis $x=0$ , so $Q=(0,11)$ .