Sec 3 · Additional Mathematics

Coordinate geometry

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1. Mid-points, gradients and collinear points

Name	Formula	Example: $A(-4, 2)$ , $B(2, 6)$
mid-point	$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$	$(-1, 4)$
gradient	$m = \dfrac{y_2 - y_1}{x_2 - x_1}$	$\dfrac{4}{6}$ , so $\dfrac{2}{3}$
length	$\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$	$\sqrt{52}$ , so $2\sqrt{13}$
collinear $A$ , $B$ , $C$	$m_{AB} = m_{AC}$	$C(5, 8)$ : $m_{AC} = \dfrac{2}{3}$

📖 Worked example

The points $A(0, 5)$ , $B(2, 1)$ and $C(k, -3)$ lie on a straight line. Find the value of $k$ .

Step 1 of 2

Gradient of AB: put the coordinates into the gradient formula

m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}

m_{AB} = \dfrac{1 - 5}{2 - 0}

= -2

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2. Parallel and perpendicular lines

Name	Formula	Example
line through $(x_1, y_1)$	$y - y_1 = m(x - x_1)$	$y - 3 = 2(x - 1)$
parallel	$m_1 = m_2$	$y = 2x + 1$ and $y = 2x - 5$
perpendicular	$m_1 m_2 = -1$	$2 \times \left(-\dfrac{1}{2}\right) = -1$

Perpendicular bisector of AB: through M, perpendicular to AB

\displaystyle m = -\dfrac{3}{2}

A(−4, 2) and B(2, 6) joined, with mid-point M(−1, 4) marked; AM and MB marked equal. The line y = −3/2 x + 5/2 crosses AB at M at a right angle.

e.g.

\displaystyle y - 4 = -\dfrac{3}{2}(x + 1)

\displaystyle y = -\dfrac{3}{2}x + \dfrac{5}{2}

$m_{AB} = \dfrac{2}{3}$ , so the perpendicular gradient is $-\dfrac{3}{2}$

📖 Worked example

Find the equation of the perpendicular bisector of the line joining $A(-4, 2)$ and $B(2, 6)$ .

Step 1 of 3

It passes through the mid-point of AB: (average of the x's, average of the y's)

x = \dfrac{-4 + 2}{2}

= -1

y = \dfrac{2 + 6}{2}

= 4

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3. Areas of polygons

Shoelace: vertices in order round the shape, first repeated at the end

\displaystyle \text{Area} = \dfrac{1}{2}\left| \begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix} \right|

down to the right: $x_1 y_2 + x_2 y_3 + x_3 y_1$
up to the right: $y_1 x_2 + y_2 x_3 + y_3 x_1$
area: half the difference, taken as positive

A(−4, 0), B(2, −3), C(1, 5)

\displaystyle \dfrac{1}{2}\left| \begin{matrix} -4 & 2 & 1 & -4 \\ 0 & -3 & 5 & 0 \end{matrix} \right|

e.g.

\displaystyle \dfrac{1}{2}(22 - (-23))

\displaystyle = 22.5

down: $(-4)(-3) + (2)(5) + (1)(0)$ , which is $22$
up: $(0)(2) + (-3)(1) + (5)(-4)$ , which is $-23$

📖 Worked example

Find the area of triangle $ABC$ with vertices $A(-4, 0)$ , $B(2, -3)$ and $C(1, 5)$ .

Step 1 of 3

List the vertices in order round the shape, and repeat the first at the end

\text{Area} = \dfrac{1}{2}\left| \begin{matrix} -4 & 2 & 1 & -4 \\[8pt] 0 & -3 & 5 & 0 \end{matrix} \right|

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4. Equation of a circle

Form	Equation	Centre	Radius
standard	$(x - a)^{2} + (y - b)^{2} = r^{2}$	$(a, b)$	$r$
general	$x^{2} + y^{2} + 2gx + 2fy + c = 0$	$(-g, -f)$	$\sqrt{g^{2} + f^{2} - c}$

Complete the square in x and in y

\displaystyle x^{2} + y^{2} - 10x + 5 = 0

e.g.

\displaystyle x^{2} - 10x + y^{2} = -5

\displaystyle (x - 5)^{2} - 25 + y^{2} = -5

\displaystyle (x - 5)^{2} + y^{2} = 20

centre $(5, 0)$ , radius $\sqrt{20}$

The same circle from the general form: g = −5, f = 0, c = 5

\displaystyle r = \sqrt{g^{2} + f^{2} - c}

e.g.

\displaystyle \sqrt{(-5)^{2} + 0^{2} - 5}

\displaystyle = \sqrt{20}

📖 Worked example

Find the coordinates of the centre of the circle $x^2 + y^2 - 10x + 5 = 0$ .

Step 1 of 2

Group the x terms and the y terms, then complete the square in each

x^2 - 10x + y^2 = -5

(x - 5)^2 - 25 + y^2 = -5

(x - 5)^2 + y^2 = 20

Practise this →

5. Circles and lines

Question	Method
inside, on or outside?	compare the distance from the centre with $r$
tangent at $P$	perpendicular to the radius at $P$
where a line meets the circle	substitute the line into the circle
	$b^{2} - 4ac > 0$ : two points
	$b^{2} - 4ac = 0$ : tangent
	$b^{2} - 4ac < 0$ : no points

Tangent to x² + y² = 25 at P(−3, 4)

\displaystyle m_{\text{tangent}} = \dfrac{3}{4}

Circle x² + y² = 25 with centre O. The radius OP goes to P(−3, 4); the tangent at P is at right angles to OP.

e.g.

\displaystyle y - 4 = \dfrac{3}{4}(x + 3)

\displaystyle y = \dfrac{3}{4}x + \dfrac{25}{4}

$m_{OP} = -\dfrac{4}{3}$ ; the tangent is perpendicular to it

📖 Worked example

The point $P(-3, 4)$ lies on the circle $x^2 + y^2 = 25$ . Find the equation of the tangent to the circle at $P$ .

Step 1 of 2

The radius goes from the centre to P; the tangent is perpendicular to it. Put the coordinates into the gradient formula

m_{\text{radius}} = \dfrac{y_2 - y_1}{x_2 - x_1}

m_{\text{radius}} = \dfrac{4 - 0}{-3 - 0}

= -\dfrac{4}{3}

m_{\text{tangent}} = \dfrac{3}{4}

📖 Worked example

Find the number of points at which the line $y = 2x$ meets the circle $x^2 + y^2 = 10$ .

Step 1 of 2

Substitute the line into the circle, expand, and make one side zero

x^2 + (2x)^2 = 10

x^2 + 4x^2 = 10

5x^2 - 10 = 0

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Sec 3 · Additional Mathematics

Coordinate geometry

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1. Mid-points, gradients and collinear points

Name	Formula	Example: $A(-4, 2)$ , $B(2, 6)$
mid-point	$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$	$(-1, 4)$
gradient	$m = \dfrac{y_2 - y_1}{x_2 - x_1}$	$\dfrac{4}{6}$ , so $\dfrac{2}{3}$
length	$\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$	$\sqrt{52}$ , so $2\sqrt{13}$
collinear $A$ , $B$ , $C$	$m_{AB} = m_{AC}$	$C(5, 8)$ : $m_{AC} = \dfrac{2}{3}$

📖 Worked example

The points $A(0, 5)$ , $B(2, 1)$ and $C(k, -3)$ lie on a straight line. Find the value of $k$ .

Step 1 of 2

Gradient of AB: put the coordinates into the gradient formula

m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}

m_{AB} = \dfrac{1 - 5}{2 - 0}

= -2

Practise this →

2. Parallel and perpendicular lines

Name	Formula	Example
line through $(x_1, y_1)$	$y - y_1 = m(x - x_1)$	$y - 3 = 2(x - 1)$
parallel	$m_1 = m_2$	$y = 2x + 1$ and $y = 2x - 5$
perpendicular	$m_1 m_2 = -1$	$2 \times \left(-\dfrac{1}{2}\right) = -1$

Perpendicular bisector of AB: through M, perpendicular to AB

\displaystyle m = -\dfrac{3}{2}

e.g.

\displaystyle y - 4 = -\dfrac{3}{2}(x + 1)

\displaystyle y = -\dfrac{3}{2}x + \dfrac{5}{2}

$m_{AB} = \dfrac{2}{3}$ , so the perpendicular gradient is $-\dfrac{3}{2}$

📖 Worked example

Find the equation of the perpendicular bisector of the line joining $A(-4, 2)$ and $B(2, 6)$ .

Step 1 of 3

It passes through the mid-point of AB: (average of the x's, average of the y's)

x = \dfrac{-4 + 2}{2}

= -1

y = \dfrac{2 + 6}{2}

= 4

Practise this →

3. Areas of polygons

Shoelace: vertices in order round the shape, first repeated at the end

\displaystyle \text{Area} = \dfrac{1}{2}\left| \begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix} \right|

down to the right: $x_1 y_2 + x_2 y_3 + x_3 y_1$
up to the right: $y_1 x_2 + y_2 x_3 + y_3 x_1$
area: half the difference, taken as positive

A(−4, 0), B(2, −3), C(1, 5)

\displaystyle \dfrac{1}{2}\left| \begin{matrix} -4 & 2 & 1 & -4 \\ 0 & -3 & 5 & 0 \end{matrix} \right|

e.g.

\displaystyle \dfrac{1}{2}(22 - (-23))

\displaystyle = 22.5

down: $(-4)(-3) + (2)(5) + (1)(0)$ , which is $22$
up: $(0)(2) + (-3)(1) + (5)(-4)$ , which is $-23$

📖 Worked example

Find the area of triangle $ABC$ with vertices $A(-4, 0)$ , $B(2, -3)$ and $C(1, 5)$ .

Step 1 of 3

List the vertices in order round the shape, and repeat the first at the end

\text{Area} = \dfrac{1}{2}\left| \begin{matrix} -4 & 2 & 1 & -4 \\[8pt] 0 & -3 & 5 & 0 \end{matrix} \right|

Practise this →

4. Equation of a circle

Form	Equation	Centre	Radius
standard	$(x - a)^{2} + (y - b)^{2} = r^{2}$	$(a, b)$	$r$
general	$x^{2} + y^{2} + 2gx + 2fy + c = 0$	$(-g, -f)$	$\sqrt{g^{2} + f^{2} - c}$

Complete the square in x and in y

\displaystyle x^{2} + y^{2} - 10x + 5 = 0

e.g.

\displaystyle x^{2} - 10x + y^{2} = -5

\displaystyle (x - 5)^{2} - 25 + y^{2} = -5

\displaystyle (x - 5)^{2} + y^{2} = 20

centre $(5, 0)$ , radius $\sqrt{20}$

The same circle from the general form: g = −5, f = 0, c = 5

\displaystyle r = \sqrt{g^{2} + f^{2} - c}

e.g.

\displaystyle \sqrt{(-5)^{2} + 0^{2} - 5}

\displaystyle = \sqrt{20}

📖 Worked example

Find the coordinates of the centre of the circle $x^2 + y^2 - 10x + 5 = 0$ .

Step 1 of 2

Group the x terms and the y terms, then complete the square in each

x^2 - 10x + y^2 = -5

(x - 5)^2 - 25 + y^2 = -5

(x - 5)^2 + y^2 = 20

Practise this →

5. Circles and lines

Question	Method
inside, on or outside?	compare the distance from the centre with $r$
tangent at $P$	perpendicular to the radius at $P$
where a line meets the circle	substitute the line into the circle
	$b^{2} - 4ac > 0$ : two points
	$b^{2} - 4ac = 0$ : tangent
	$b^{2} - 4ac < 0$ : no points

Tangent to x² + y² = 25 at P(−3, 4)

\displaystyle m_{\text{tangent}} = \dfrac{3}{4}

e.g.

\displaystyle y - 4 = \dfrac{3}{4}(x + 3)

\displaystyle y = \dfrac{3}{4}x + \dfrac{25}{4}

$m_{OP} = -\dfrac{4}{3}$ ; the tangent is perpendicular to it

📖 Worked example

The point $P(-3, 4)$ lies on the circle $x^2 + y^2 = 25$ . Find the equation of the tangent to the circle at $P$ .

Step 1 of 2

The radius goes from the centre to P; the tangent is perpendicular to it. Put the coordinates into the gradient formula

m_{\text{radius}} = \dfrac{y_2 - y_1}{x_2 - x_1}

m_{\text{radius}} = \dfrac{4 - 0}{-3 - 0}

= -\dfrac{4}{3}

m_{\text{tangent}} = \dfrac{3}{4}

📖 Worked example

Find the number of points at which the line $y = 2x$ meets the circle $x^2 + y^2 = 10$ .

Step 1 of 2

Substitute the line into the circle, expand, and make one side zero

x^2 + (2x)^2 = 10

x^2 + 4x^2 = 10

5x^2 - 10 = 0

Practise this →

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