3 Coordinate geometry in the (x, y) plane — printable coverage pack

3.1 · Understand and use the equation of a straight line, including the forms y − y1 = m(x − x1) and ax + by + c = 0; gradient conditions for two straight lines to be parallel or perpendicular; use straight line models in a variety of contexts.

A non-vertical straight line has constant gradient $m$ and may be written as $y-y_1=m(x-x_1)$ through a known point or rearranged into $ax+by+c=0$ .
Parallel lines have equal gradients, while non-vertical perpendicular lines have gradients whose product is $-1$ .
For two known points, first calculate $m=(y_2-y_1)/(x_2-x_1)$ , then substitute one point into a line equation and check the other point.
In a linear model, interpret the gradient and intercept in context and respect practical restrictions. A common error is to round a limiting input upwards when this breaks the stated constraint.

Tier 1 · Easy

1. Find the equation of the line with gradient $-3$ that passes through $(2,4)$ . Write it as $y=mx+c$ .[2 marks]
Answer
- $y=-3x+10$
Method: Use point-gradient form: $y-4=-3(x-2)$ . Expanding gives $y-4=-3x+6$ , so $y=-3x+10$ .

Tier 2 · Standard

1. The line $L$ has equation $2x-5y+7=0$ . Find an equation of the line through $(4,-1)$ that is perpendicular to $L$ . Write your equation as $ax+by+c=0$ with integer coefficients.[3 marks]
Answer
- $5x+2y-18=0$
Method: Rearrange $L$ to $y=\dfrac{2}{5}x+\dfrac{7}{5}$ , so its gradient is $\dfrac{2}{5}$ . A perpendicular line has gradient $-\dfrac{5}{2}$ . Hence $y+1=-\dfrac{5}{2}(x-4)$ , which rearranges to $5x+2y-18=0$ .

Tier 3 · Hard

1. A delivery company models the charge $C$ pounds for a journey of $d$ kilometres by a straight line. Journeys of $8$ km and $14$ km cost $\pounds 19.60$ and $\pounds 29.80$ respectively. Find the model for $C$ in terms of $d$ , interpret its gradient, and find the greatest whole number of kilometres that can be travelled for at most $\pounds 45$ .[5 marks]
Answer
- $C=1.7d+6$
- The model charges $\pounds 1.70$ per kilometre.
- $22$ km
Method: The gradient is $(29.80-19.60)/(14-8)=10.20/6=1.70$ , representing the charge per kilometre. Write $C=1.7d+c$ and use $(8,19.60)$ : $19.60=1.7(8)+c$ , so $c=6$ . For the budget, solve $1.7d+6\le45$ , giving $d\le390/17\approx22.94$ . The greatest permitted whole number is therefore $22$ km; $23$ km would cost $\pounds 45.10$ .

3.2 · Understand and use the coordinate geometry of the circle, including the equation (x − a)² + (y − b)² = r²; complete the square for centre and radius; use circle properties (semicircle angle, chord bisector, tangent-radius).

A circle with centre $(a,b)$ and radius $r$ has equation $(x-a)^2+(y-b)^2=r^2$ .
Complete the square separately in $x$ and $y$ to reveal the centre and radius of a circle given in expanded form.
A radius is perpendicular to the tangent at its endpoint; perpendicular chord bisectors meet at the circumcentre, and an angle subtended by a diameter at the circumference is $90^\circ$ .
When testing tangency, require exactly one intersection or set the perpendicular distance from the centre to the line equal to the radius. A common error is to read the centre as $(a,b)$ from $(x+a)^2+(y+b)^2$ instead of $(-a,-b)$ .

Tier 1 · Easy

1. Find the centre and radius of the circle $x^2+y^2-6x+8y-11=0$ .[3 marks]
Answer
- Centre $(3,-4)$
- Radius $6$
Method: Complete both squares: $x^2-6x=(x-3)^2-9$ and $y^2+8y=(y+4)^2-16$ . The equation becomes $(x-3)^2+(y+4)^2=36$ , so the centre is $(3,-4)$ and the radius is $6$ .

Tier 2 · Standard

1. The point $P(5,3)$ lies on the circle with centre $(2,-1)$ . Determine the tangent at $P$ . Write your equation as $ax+by+c=0$ with integer coefficients.[3 marks]
Answer
- $3x+4y-27=0$
Method: The gradient of the radius from $(2,-1)$ to $(5,3)$ is $4/3$ , so the tangent gradient is $-3/4$ . Hence $y-3=-\dfrac{3}{4}(x-5)$ . Multiplying by $4$ and rearranging gives $3x+4y-27=0$ .

Tier 3 · Hard

1. The points $A(-1,2)$ , $B(5,2)$ and $C(1,6)$ lie on a circle. Use perpendicular bisectors to find the equation of this circumcircle.[5 marks]
Answer
- $(x-2)^2+(y-3)^2=10$
Method: The midpoint of horizontal chord $AB$ is $(2,2)$ , so its perpendicular bisector is $x=2$ . The midpoint of $AC$ is $(0,4)$ and the gradient of $AC$ is $2$ , so its perpendicular bisector is $y-4=-\dfrac{1}{2}x$ . At $x=2$ this gives $y=3$ , so the centre is $(2,3)$ . The squared radius is $(-1-2)^2+(2-3)^2=10$ , hence the circumcircle is $(x-2)^2+(y-3)^2=10$ .

3.3 · Understand and use the parametric equations of curves and conversion between Cartesian and parametric forms.

Parametric equations express both coordinates in terms of a third variable: $x=f(t)$ and $y=g(t)$ .
To obtain a Cartesian equation, eliminate the parameter by making $t$ the subject of one equation and substituting into the other, or by using a suitable identity.
To parametrise a Cartesian curve, choose expressions that satisfy it identically, such as $x=a\cos t$ and $y=b\sin t$ for $x^2/a^2+y^2/b^2=1$ .
Record any parameter interval because it determines which part of the Cartesian curve is traced. A common error is to eliminate $t$ but lose a domain restriction.

Tier 1 · Easy

1. For $x=2t-1$ and $y=t+4$ , eliminate $t$ to obtain a Cartesian equation.[2 marks]
Answer
- $y=\dfrac{x}{2}+\dfrac{9}{2}$
Method: From $x=2t-1$ , $t=(x+1)/2$ . Substitute into $y=t+4$ to get $y=(x+1)/2+4=x/2+9/2$ .

Tier 2 · Standard

1. The curve $x=2t+1$ , $y=t^2-3$ is defined for $t\ge0$ . Find a Cartesian equation and state the corresponding restriction on $x$ .[3 marks]
Answer
- $y=\dfrac{(x-1)^2}{4}-3$ , with $x\ge1$
Method: Rearrange $x=2t+1$ to $t=(x-1)/2$ . Therefore $y=((x-1)/2)^2-3=(x-1)^2/4-3$ . Since $t\ge0$ , $x=2t+1\ge1$ .

Tier 3 · Hard

1. The curve $x=t+1$ , $y=t^2-2t$ meets the line $y=3x-6$ at two points. Find the exact coordinates of both points.[5 marks]
Answer
- $\left(\dfrac{7+\sqrt{13}}{2},\dfrac{9+3\sqrt{13}}{2}\right)$
- $\left(\dfrac{7-\sqrt{13}}{2},\dfrac{9-3\sqrt{13}}{2}\right)$
Method: At an intersection, $t^2-2t=3(t+1)-6$ , so $t^2-5t+3=0$ . Hence $t=(5\pm\sqrt{13})/2$ . Using $x=t+1$ gives $x=(7\pm\sqrt{13})/2$ , and then $y=3x-6=(9\pm3\sqrt{13})/2$ , with matching signs.

3.4 · Use parametric equations in modelling in a variety of contexts.

In a parametric model, the parameter often represents time and the pair $(x(t),y(t))$ gives the object's position at that time.
Translate a contextual event into equations: meeting a boundary fixes one coordinate, while a collision requires both coordinates of two objects to agree at the same time.
Eliminating the parameter can reveal the path, but the parameter range still controls the physically modelled section of that path.
Check units, permitted times and whether a calculated event occurs within the model's domain. A common error is to find where two paths cross without checking that the objects arrive there simultaneously.

Tier 1 · Easy

1. A particle's position after $t$ seconds is modelled by $x=12t$ , $y=20t-5t^2$ , where distances are in metres. Find its position when $t=1.5$ .[2 marks]
Answer
- $(18,18.75)$ metres
Method: Substitute $t=1.5$ : $x=12(1.5)=18$ and $y=20(1.5)-5(1.5)^2=30-11.25=18.75$ .

Tier 2 · Standard

1. An arch is modelled by $x=6t$ , $y=12t-3t^2$ for $0\le t\le4$ , with $x$ and $y$ in metres. Find a Cartesian equation for the arch and its horizontal span.[4 marks]
Answer
- $y=2x-\dfrac{x^2}{12}$ for $0\le x\le24$
- Span $24$ m
Method: Since $x=6t$ , $t=x/6$ . Substitute into $y$ : $y=12(x/6)-3(x/6)^2=2x-x^2/12$ . The parameter interval gives $0\le x\le24$ . The arch meets ground level at its two endpoints, so its horizontal span is $24-0=24$ m.

Tier 3 · Hard

1. Two particles move for $t\ge0$ seconds. Particle $P$ has position $(3t+2,t^2+2)$ and particle $Q$ has position $(12-2t,t+4)$ . Determine whether they collide and, if they do, find the time and position of the collision.[4 marks]
Answer
- They collide at $t=2$ seconds at $(8,6)$ .
Method: Equal $x$ -coordinates require $3t+2=12-2t$ , so $5t=10$ and $t=2$ . Check the other coordinate at this same time: for $P$ , $y=2^2+2=6$ ; for $Q$ , $y=2+4=6$ . Both positions are therefore $(8,6)$ at $t=2$ , so a collision occurs. The linear $x$ equation has only one solution, so there is no other collision.