Sec 3 · G3 (Express)

Coordinate geometry

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1. Gradient

Gradient of AB

\displaystyle m = \dfrac{y_2 - y_1}{x_2 - x_1}

Line through A(1, 2) and B(3, 8): run 3 − 1 = 2 across, rise 8 − 2 = 6 up, with a right angle where they meet.

e.g.

\displaystyle m = \dfrac{8 - 2}{3 - 1}

\displaystyle m = \dfrac{6}{2}

\displaystyle m = 3

$\dfrac{\text{rise}}{\text{run}}$

From y = mx + c

\displaystyle y = mx + c

e.g.

\displaystyle 3x + 2y = 6

\displaystyle 2y = -3x + 6

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

gradient $m$
$y$ -intercept $c$

Four lines: rising to the right has a positive gradient, falling a negative gradient, horizontal a zero gradient, vertical an undefined gradient.

📖 Worked example

Find the gradient of the line joining $A(-3, 3)$ and $B(1, 2)$ .

Step 1 of 2

Put the coordinates into the gradient formula

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{2 - 3}{1 - (-3)}

📖 Worked example

Find the gradient of the line $-2x - 4y = 15$ .

Step 1 of 2

Make y the subject

-4y = 2x + 15

y = -\dfrac{1}{2}x - \dfrac{15}{4}

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2. Length of a line segment

Length of AB

\displaystyle AB = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}

A(1, 2) and B(4, 6): 4 − 1 = 3 across, 6 − 2 = 4 up, right angle at (4, 2), AB = 5.

e.g.

\displaystyle AB = \sqrt{(4 - 1)^{2} + (6 - 2)^{2}}

\displaystyle AB = \sqrt{25}

\displaystyle AB = 5

Pythagoras' theorem on the run and the rise

📖 Worked example

Find the length of the line segment joining $A(-1, -3)$ and $B(0, -8)$ , correct to 3 significant figures.

Step 1 of 2

Put the coordinates into the length formula

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AB = \sqrt{(0 - (-1))^2 + (-8 - (-3))^2}

= \sqrt{26}

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3. Equation of a straight line

Gradient and a point: substitute to find c

\displaystyle m = 3, \text{ through } (2, 1)

e.g.

\displaystyle 1 = 3(2) + c

\displaystyle c = -5

so $y = 3x - 5$

Two points: gradient first

\displaystyle m = \dfrac{y_2 - y_1}{x_2 - x_1}

through $(1, 3)$ and $(3, 7)$ : $m = 2$ , then $c = 1$
$y = 2x + 1$

📖 Worked example

Find the equation of the line with gradient $-2$ that passes through $(2, -2)$ .

Step 1 of 3

Read off the gradient

m = -2

📖 Worked example

Find the equation of the line passing through $A(-2, 4)$ and $B(2, 1)$ .

Step 1 of 3

Find the gradient: put the coordinates into the gradient formula

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{1 - 4}{2 - (-2)}

\phantom{m} = -\dfrac{3}{4}

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4. Parallel lines

Parallel lines have the same gradient

\displaystyle m_1 = m_2

Parallel lines y = 2x + 1 and y = 2x − 3, marked with arrows, crossing the y-axis at (0, 1) and (0, −3).

$y = 2x + 1$ and $y = 2x - 3$ : both $m = 2$

Parallel to y = 2x + 1, through (3, 4)

\displaystyle m = 2

e.g.

\displaystyle 4 = 2(3) + c

\displaystyle c = -2

$y = 2x - 2$

📖 Worked example

Find the equation of the line parallel to $y = -x - 5$ that passes through $(-2, -4)$ .

Step 1 of 3

Same gradient as the given line

m = -1

📖 Worked example

Find the equation of the line parallel to $2x + y = 5$ that passes through $(-2, -3)$ .

Step 1 of 3

Same gradient as the given line

m = -2

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5. Finding coordinates

Where	Put	Example: $y = 2x - 6$
meets the $x$ -axis	$y = 0$	$(3, 0)$
meets the $y$ -axis	$x = 0$	$(0, -6)$

Where two lines meet: the y-values are equal

\displaystyle 2x + 1 = -x + 7

Lines y = 2x + 1 and y = −x + 7 crossing at (2, 5).

e.g.

\displaystyle 2x + 1 = -x + 7

\displaystyle 3x = 6

\displaystyle x = 2

then $y = 2(2) + 1$ , so they meet at $(2, 5)$

A point on a line fits its equation

\displaystyle (k, 5) \text{ on } y = 2x - 1

e.g.

\displaystyle 5 = 2k - 1

\displaystyle k = 3

ax + by = c meets the x-axis

\displaystyle 2x + 3y = 12

e.g.

\displaystyle 2x + 3(0) = 12

\displaystyle 2x = 12

\displaystyle x = 6

the point is $(6, 0)$

📖 Worked example

Find the coordinates of the point where the lines $y = -x + 5$ and $y = -3x + 3$ meet.

Step 1 of 2

Where they meet, the y-values are equal

-x + 5 = -3x + 3

x = -1

A point from a given area: ½ × base × height

\displaystyle A(-2, 0),\ B(2, 0),\ \text{area } ABC = 8

e.g.

\displaystyle \dfrac{1}{2} \times 4 \times h = 8

\displaystyle h = 4

$C$ above the $x$ -axis: $y = 4$
$C$ on $y = 2x - 2$ : $4 = 2x - 2$ , so $C(3, 4)$

Triangle cut off by the axes

\displaystyle 3x + 4y = 24

e.g.

\displaystyle \dfrac{1}{2} \times 8 \times 6

\displaystyle = 24

$P(8, 0)$ and $Q(0, 6)$ : lengths, so no minus signs

📖 Worked example

The points $A(-2, 0)$ and $B(2, 0)$ lie on the $x$ -axis. The point $C$ lies on the line $y = 2x - 2$ , above the $x$ -axis. The area of triangle $ABC$ is $8$ units². Find the coordinates of $C$ .

Step 1 of 5

The base AB lies along the x-axis: its length

AB = 2 - (-2)

AB = 4

📖 Worked example

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

Step 1 of 3

AB is parallel to the y-axis: its length is the difference of the y-coordinates

AB = 0 - (-3)

AB = 3

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Sec 3 · G3 (Express)

Coordinate geometry

Practise coordinate geometry with step-by-step help. Free account.

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1. Gradient

Gradient of AB

\displaystyle m = \dfrac{y_2 - y_1}{x_2 - x_1}

e.g.

\displaystyle m = \dfrac{8 - 2}{3 - 1}

\displaystyle m = \dfrac{6}{2}

\displaystyle m = 3

$\dfrac{\text{rise}}{\text{run}}$

From y = mx + c

\displaystyle y = mx + c

e.g.

\displaystyle 3x + 2y = 6

\displaystyle 2y = -3x + 6

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

gradient $m$
$y$ -intercept $c$

📖 Worked example

Find the gradient of the line joining $A(-3, 3)$ and $B(1, 2)$ .

Step 1 of 2

Put the coordinates into the gradient formula

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{2 - 3}{1 - (-3)}

📖 Worked example

Find the gradient of the line $-2x - 4y = 15$ .

Step 1 of 2

Make y the subject

-4y = 2x + 15

y = -\dfrac{1}{2}x - \dfrac{15}{4}

Practise this →

2. Length of a line segment

Length of AB

\displaystyle AB = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}

e.g.

\displaystyle AB = \sqrt{(4 - 1)^{2} + (6 - 2)^{2}}

\displaystyle AB = \sqrt{25}

\displaystyle AB = 5

Pythagoras' theorem on the run and the rise

📖 Worked example

Find the length of the line segment joining $A(-1, -3)$ and $B(0, -8)$ , correct to 3 significant figures.

Step 1 of 2

Put the coordinates into the length formula

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AB = \sqrt{(0 - (-1))^2 + (-8 - (-3))^2}

= \sqrt{26}

Practise this →

3. Equation of a straight line

Gradient and a point: substitute to find c

\displaystyle m = 3, \text{ through } (2, 1)

e.g.

\displaystyle 1 = 3(2) + c

\displaystyle c = -5

so $y = 3x - 5$

Two points: gradient first

\displaystyle m = \dfrac{y_2 - y_1}{x_2 - x_1}

through $(1, 3)$ and $(3, 7)$ : $m = 2$ , then $c = 1$
$y = 2x + 1$

📖 Worked example

Find the equation of the line with gradient $-2$ that passes through $(2, -2)$ .

Step 1 of 3

Read off the gradient

m = -2

📖 Worked example

Find the equation of the line passing through $A(-2, 4)$ and $B(2, 1)$ .

Step 1 of 3

Find the gradient: put the coordinates into the gradient formula

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{1 - 4}{2 - (-2)}

\phantom{m} = -\dfrac{3}{4}

Practise this →

4. Parallel lines

Parallel lines have the same gradient

\displaystyle m_1 = m_2

$y = 2x + 1$ and $y = 2x - 3$ : both $m = 2$

Parallel to y = 2x + 1, through (3, 4)

\displaystyle m = 2

e.g.

\displaystyle 4 = 2(3) + c

\displaystyle c = -2

$y = 2x - 2$

📖 Worked example

Find the equation of the line parallel to $y = -x - 5$ that passes through $(-2, -4)$ .

Step 1 of 3

Same gradient as the given line

m = -1

📖 Worked example

Find the equation of the line parallel to $2x + y = 5$ that passes through $(-2, -3)$ .

Step 1 of 3

Same gradient as the given line

m = -2

Practise this →

5. Finding coordinates

Where	Put	Example: $y = 2x - 6$
meets the $x$ -axis	$y = 0$	$(3, 0)$
meets the $y$ -axis	$x = 0$	$(0, -6)$

Where two lines meet: the y-values are equal

\displaystyle 2x + 1 = -x + 7

e.g.

\displaystyle 2x + 1 = -x + 7

\displaystyle 3x = 6

\displaystyle x = 2

then $y = 2(2) + 1$ , so they meet at $(2, 5)$

A point on a line fits its equation

\displaystyle (k, 5) \text{ on } y = 2x - 1

e.g.

\displaystyle 5 = 2k - 1

\displaystyle k = 3

ax + by = c meets the x-axis

\displaystyle 2x + 3y = 12

e.g.

\displaystyle 2x + 3(0) = 12

\displaystyle 2x = 12

\displaystyle x = 6

the point is $(6, 0)$

📖 Worked example

Find the coordinates of the point where the lines $y = -x + 5$ and $y = -3x + 3$ meet.

Step 1 of 2

Where they meet, the y-values are equal

-x + 5 = -3x + 3

x = -1

A point from a given area: ½ × base × height

\displaystyle A(-2, 0),\ B(2, 0),\ \text{area } ABC = 8

e.g.

\displaystyle \dfrac{1}{2} \times 4 \times h = 8

\displaystyle h = 4

$C$ above the $x$ -axis: $y = 4$
$C$ on $y = 2x - 2$ : $4 = 2x - 2$ , so $C(3, 4)$

Triangle cut off by the axes

\displaystyle 3x + 4y = 24

e.g.

\displaystyle \dfrac{1}{2} \times 8 \times 6

\displaystyle = 24

$P(8, 0)$ and $Q(0, 6)$ : lengths, so no minus signs

📖 Worked example

The points $A(-2, 0)$ and $B(2, 0)$ lie on the $x$ -axis. The point $C$ lies on the line $y = 2x - 2$ , above the $x$ -axis. The area of triangle $ABC$ is $8$ units². Find the coordinates of $C$ .

Step 1 of 5

The base AB lies along the x-axis: its length

AB = 2 - (-2)

AB = 4

📖 Worked example

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

Step 1 of 3

AB is parallel to the y-axis: its length is the difference of the y-coordinates

AB = 0 - (-3)

AB = 3

Practise this →

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