Mathematics 9709 · Pure Mathematics 1

Where a line meets a curve

To find where a line and a curve meet, turn the two equations into one quadratic, then solve it.
The discriminant of that quadratic then tells you: cut, touch, or miss.

The one idea: substitute

A meeting point sits on both graphs. So replace the $y$ in the curve with the line's expression. This leaves one equation in $x$ .

The main idea for the whole topic: set them equal, then solve a quadratic.

Equate the two expressions for $y$ .
Solve the resulting equation for $x$ (usually a quadratic).
Substitute back into the line to get each $y$ .

See the simple shape on two straight lines first

Where does $y=2x+1$ meet $y=x+4$ ? At the crossing both $y$ values are the same. So set the right-hand sides equal:

$2x+1=x+4 \;\Rightarrow\; x=3$

Put that back into either line: $y=3+4=7$ . So the meeting point is $(3,\,7)$ . With a curve, only the middle step changes. It becomes a quadratic, not a one-line solve.

When a curve is involved

Same three steps every time. Set them equal, get one side to $=0$ , solve, then read each $y$ from the line.

The easiest case, $y=x+1$ meets $y=x^2+1$ :

Equate: $x^2+1=x+1$ .
Onto one side, $=0$ : $x^2-x=0$ .
Factor out $x$ : $x(x-1)=0$ , so $x=0$ or $x=1$ .
Read $y$ off $y=x+1$ : points $(0,\,1)$ and $(1,\,2)$ .

Worked example

Standard case: $y=2x+3$ meets $y=x^2$

Equate: $x^2 = 2x+3$ .
Onto one side: $x^2 - 2x - 3 = 0$ .
Factorise: $(x-3)(x+1)=0$ , so $x=3$ or $x=-1$ .
From $y=2x+3$ : $x=3\Rightarrow y=9$ , $x=-1\Rightarrow y=1$ .

Answer

(3,\,9)\quad\text{and}\quad(-1,\,1)

The pale line y = 2x + 3 cuts the curve y = x² at the two solutions (3, 9) and (−1, 1).

Why those factors?

Two numbers that multiply to $-3$ and add to $-2$ are $-3$ and $+1$ , giving $(x-3)(x+1)$ .

Common mistake

Do not stop at the

x

values. Each point needs both coordinates. So always put

x

back in to get

y

, using the line (the easier equation).

Worked example

Exam-level: $y=2x-1$ meets the circle $x^2+y^2=2$

Substitute $y=2x-1$ into the circle: $x^2+(2x-1)^2=2$ .
Expand and tidy to $=0$ : $5x^2 - 4x - 1 = 0$ .
Factorise: $(5x+1)(x-1)=0$ , so $x=-\tfrac15$ or $x=1$ .
From the line: $x=1\Rightarrow y=1$ , $x=-\tfrac15\Rightarrow y=-\tfrac75$ .

Answer

(1,\,1)\quad\text{and}\quad\left(-\tfrac15,\,-\tfrac75\right)

Show the expand + factorise in full

Expand the bracket carefully — $(2x-1)^2=(2x)^2-2(2x)(1)+1^2=4x^2-4x+1$ :

$x^2 + 4x^2 - 4x + 1 = 2$

Collect the $x^2$ terms, then move the $2$ across:

$5x^2 - 4x + 1 - 2 = 0 \;\Rightarrow\; 5x^2 - 4x - 1 = 0$

To factorise: multiply to $5\times(-1)=-5$ , add to $-4$ — that's $-5$ and $+1$ :

$(5x+1)(x-1)=0$

This circle pattern (substitute, expand, solve) is exactly what coordinate-geometry questions ask. They often then ask for a chord length or a midpoint. (9709/12 Mar 2022 Q6; 9709/13 Nov 2022 Q10)

Cut, touch or miss, without solving

Each real root is a meeting point. So the discriminant $b^2-4ac$ of the quadratic tells you how the line and curve sit before you solve.

Discriminant	Real roots	Line and curve
$b^2-4ac>0$	two	two points — it cuts
$b^2-4ac=0$	one (repeated)	just touches — a tangent
$b^2-4ac<0$	none	they miss completely

b² − 4ac = 12.0 → two real roots

Drag the slider: two roots come together into one (a tangent), then disappear, as b²−4ac passes through zero.

Worked: show y = x − ¼ is a tangent to y = x²

Set them equal and rearrange:

$x^2-x+\tfrac14=0$

With $a=1,\ b=-1,\ c=\tfrac14$ :

$b^2-4ac=(-1)^2-4(1)\left(\tfrac14\right)=1-1=0$

One repeated root. The line touches exactly once, which is exactly what tangent means.

Your turn— tap to reveal the worked answer (quick check)

A line meets a curve and the resulting quadratic is $x^2-6x+9=0$ . Cut, touch, or miss?

$b^2-4ac=(-6)^2-4(1)(9)=36-36=0$ , so one repeated root.

Answer

\text{touches (a tangent)}

Where the marks go

Common mistake

You cannot factorise and solve until one side is

=0

. Rearrange first, then read off

a,b,c

. Most marks are lost in the substitution algebra:

(2x-1)^2

4x^2-4x+1

, not

4x^2+1

. Give both coordinates of each point. If a diagram marks one point, pick the matching root.

Common mistake

“Tangent” means

b^2-4ac=0

(one repeated root), not

>0

. “Does not meet” needs a strict

b^2-4ac<0

Now you try

Do the three steps yourself: set them equal, solve, then read off both coordinates.

Find where $y=x-1$ meets $y=x^2-4x+3$ , giving both coordinates of each point. (9709-style)

Your turn— tap to reveal the worked answer (set them equal, then = 0)

Equate: $x^2-4x+3 = x-1$ .
Onto one side: $x^2-5x+4=0$ .
Factorise: $(x-1)(x-4)=0$ , so $x=1$ or $x=4$ .
From $y=x-1$ : $x=1\Rightarrow y=0$ , $x=4\Rightarrow y=3$ .

Answer

(1,\,0)\quad\text{and}\quad(4,\,3)

PreviousQuadratic inequalities NextHidden quadratics

Where a line meets a curve

To find where a line and a curve meet, turn the two equations into one quadratic, then solve it.
The discriminant of that quadratic then tells you: cut, touch, or miss.

The one idea: substitute

A meeting point sits on both graphs. So replace the $y$ in the curve with the line's expression. This leaves one equation in $x$ .

The main idea for the whole topic: set them equal, then solve a quadratic.

Equate the two expressions for $y$ .
Solve the resulting equation for $x$ (usually a quadratic).
Substitute back into the line to get each $y$ .

See the simple shape on two straight lines first

Where does $y=2x+1$ meet $y=x+4$ ? At the crossing both $y$ values are the same. So set the right-hand sides equal:

$2x+1=x+4 \;\Rightarrow\; x=3$

Put that back into either line: $y=3+4=7$ . So the meeting point is $(3,\,7)$ . With a curve, only the middle step changes. It becomes a quadratic, not a one-line solve.

When a curve is involved

Same three steps every time. Set them equal, get one side to $=0$ , solve, then read each $y$ from the line.

The easiest case, $y=x+1$ meets $y=x^2+1$ :

Equate: $x^2+1=x+1$ .
Onto one side, $=0$ : $x^2-x=0$ .
Factor out $x$ : $x(x-1)=0$ , so $x=0$ or $x=1$ .
Read $y$ off $y=x+1$ : points $(0,\,1)$ and $(1,\,2)$ .

Worked example

Standard case: $y=2x+3$ meets $y=x^2$

Equate: $x^2 = 2x+3$ .
Onto one side: $x^2 - 2x - 3 = 0$ .
Factorise: $(x-3)(x+1)=0$ , so $x=3$ or $x=-1$ .
From $y=2x+3$ : $x=3\Rightarrow y=9$ , $x=-1\Rightarrow y=1$ .

Answer

(3,\,9)\quad\text{and}\quad(-1,\,1)

The pale line y = 2x + 3 cuts the curve y = x² at the two solutions (3, 9) and (−1, 1).

Why those factors?

Two numbers that multiply to $-3$ and add to $-2$ are $-3$ and $+1$ , giving $(x-3)(x+1)$ .

Common mistake

Do not stop at the

x

values. Each point needs both coordinates. So always put

x

back in to get

y

, using the line (the easier equation).

Worked example

Exam-level: $y=2x-1$ meets the circle $x^2+y^2=2$

Substitute $y=2x-1$ into the circle: $x^2+(2x-1)^2=2$ .
Expand and tidy to $=0$ : $5x^2 - 4x - 1 = 0$ .
Factorise: $(5x+1)(x-1)=0$ , so $x=-\tfrac15$ or $x=1$ .
From the line: $x=1\Rightarrow y=1$ , $x=-\tfrac15\Rightarrow y=-\tfrac75$ .

Answer

(1,\,1)\quad\text{and}\quad\left(-\tfrac15,\,-\tfrac75\right)

Show the expand + factorise in full

Expand the bracket carefully — $(2x-1)^2=(2x)^2-2(2x)(1)+1^2=4x^2-4x+1$ :

$x^2 + 4x^2 - 4x + 1 = 2$

Collect the $x^2$ terms, then move the $2$ across:

$5x^2 - 4x + 1 - 2 = 0 \;\Rightarrow\; 5x^2 - 4x - 1 = 0$

To factorise: multiply to $5\times(-1)=-5$ , add to $-4$ — that's $-5$ and $+1$ :

$(5x+1)(x-1)=0$

This circle pattern (substitute, expand, solve) is exactly what coordinate-geometry questions ask. They often then ask for a chord length or a midpoint. (9709/12 Mar 2022 Q6; 9709/13 Nov 2022 Q10)

Cut, touch or miss, without solving

Each real root is a meeting point. So the discriminant $b^2-4ac$ of the quadratic tells you how the line and curve sit before you solve.

Discriminant	Real roots	Line and curve
$b^2-4ac>0$	two	two points — it cuts
$b^2-4ac=0$	one (repeated)	just touches — a tangent
$b^2-4ac<0$	none	they miss completely

b² − 4ac = 12.0 → two real roots

Drag the slider: two roots come together into one (a tangent), then disappear, as b²−4ac passes through zero.

Worked: show y = x − ¼ is a tangent to y = x²

Set them equal and rearrange:

$x^2-x+\tfrac14=0$

With $a=1,\ b=-1,\ c=\tfrac14$ :

$b^2-4ac=(-1)^2-4(1)\left(\tfrac14\right)=1-1=0$

One repeated root. The line touches exactly once, which is exactly what tangent means.

Your turn— tap to reveal the worked answer (quick check)

A line meets a curve and the resulting quadratic is $x^2-6x+9=0$ . Cut, touch, or miss?

$b^2-4ac=(-6)^2-4(1)(9)=36-36=0$ , so one repeated root.

Answer

\text{touches (a tangent)}

Where the marks go

Common mistake

You cannot factorise and solve until one side is

=0

. Rearrange first, then read off

a,b,c

. Most marks are lost in the substitution algebra:

(2x-1)^2

4x^2-4x+1

, not

4x^2+1

. Give both coordinates of each point. If a diagram marks one point, pick the matching root.

Common mistake

“Tangent” means

b^2-4ac=0

(one repeated root), not

>0

. “Does not meet” needs a strict

b^2-4ac<0

Now you try

Do the three steps yourself: set them equal, solve, then read off both coordinates.

Find where $y=x-1$ meets $y=x^2-4x+3$ , giving both coordinates of each point. (9709-style)

Your turn— tap to reveal the worked answer (set them equal, then = 0)

Equate: $x^2-4x+3 = x-1$ .
Onto one side: $x^2-5x+4=0$ .
Factorise: $(x-1)(x-4)=0$ , so $x=1$ or $x=4$ .
From $y=x-1$ : $x=1\Rightarrow y=0$ , $x=4\Rightarrow y=3$ .

Answer

(1,\,0)\quad\text{and}\quad(4,\,3)