Analytical Geometry (Grade 10) — Mathematics Grades 8–11

Use the distance, gradient and midpoint formulae to solve problems on the Cartesian plane.

Gr 10 Term 3 - Analytical GeometryTerm 3 - 2 weeks (CAPS)

Practise Analytical Geometry in the app →

What gets asked

Calculate the distance between two given points.
Calculate the gradient of the line segment joining two points.
Decide whether two line segments are parallel or perpendicular.
Calculate the midpoint of a line segment.

You must be able to

Substitute two points into the distance formula and simplify the surd.
Subtract y-values over x-values, in the same order, to find a gradient.
Check for equal gradients to confirm lines are parallel, or a product of minus one for perpendicular.
Average the x-values and average the y-values to find a midpoint.

Traps that cost marks

Mixing up the coordinates, such as subtracting an x-value from a y-value. Keep x's with x's and y's with y's: $(x_{2} - x_{1})$ and $(y_{2} - y_{1})$ .
Calling the gradient of a vertical line zero, instead of undefined. A vertical line has no run, so you would divide by zero. Its gradient is undefined, not zero.
Using the gradients-multiply-to-minus-one test on a horizontal and a vertical line. That test needs two numbers. A vertical line's gradient is undefined, so check by shape instead.

Worked example

Calculate the gradient of the line joining

A (- 2; 1)

and

B (4; 5)

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
$m = \frac{5 - 1}{4 - ( - 2 )}$
$m = \frac{4}{6} = \frac{2}{3}$

Points and distance

Two numbers fix a point, and Pythagoras turns any two points into a length.

Cartesian plane: The grid made by a horizontal $x$ -axis and a vertical $y$ -axis crossing each other.
coordinates: The pair $(x; y)$ saying how far across and how far up a point sits.
origin: The point $(0; 0)$ , where the two axes cross.
right-angled triangle: A triangle with one angle of exactly $9 0^{\circ}$ .
isosceles: Having two equal sides.
equilateral: Having all three sides equal.

Every point on the Cartesian plane has two coordinates. Count from the origin, across first and then up. So $(3; 5)$ is not the same point as $(5; 3)$ . Watch the scale on each axis too. If one square stands for $2$ units, count units and not squares.

The distance between two points comes straight from Pythagoras. The gap across and the gap up are the short sides of a right-angled triangle, and the join between the points is the long side. That gives $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

Keep $x$ with $x$ and $y$ with $y$ . Which one you subtract from which makes no difference, because squaring wipes out the sign. Take one square root at the very end, of the total, and never of each piece on its own.

To name a triangle, work out all three side lengths first. Two equal sides make it isosceles. Three equal sides make it equilateral. For a right angle, find the longest side: the other two squared must add up to it squared.

When a distance is given and a coordinate is missing, fill in the formula and square both sides. Square each whole side, not each term inside it. Squaring usually hands back two answers, and both may be real points.

Rules to remember

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
right-angled when $a^{2} + b^{2} = c^{2}$ , with $c$ the longest side

Examples

Find the distance between

A (1; 2)

and

B (5; 5)

Worked answer

Across: $5 - 1 = 4$ . Up: $5 - 2 = 3$ .
Square each and add: $4^{2} + 3^{2} = 25$ .
One square root at the end: $25 = 5$ .
So $A B$ is $5$ units long.

Answer: $5$ units

The square root came once, at the end, and only after the adding.

Show that

P (0; 0)

Q (4; 0)

and

R (2; 5)

form an isosceles triangle.

Worked answer

$P Q$ runs along the $x$ -axis, so $P Q = 4$ .
$P R = 2^{2} + 5^{2} = 29$ .
$QR = (- 2)^{2} + 5^{2} = 29$ .
Two sides came out equal, so the triangle is isosceles.

Answer: isosceles, because $P R = QR$

The two slanted sides were worked out, not guessed from the picture.

The distance from

A (1; 3)

B (x; 3)

6

units. Find

x

Worked answer

Both points have $y = 3$ , so the gap up is $0$ .
The distance is then the gap across: $(x - 1)^{2} = 6$ .
Square both sides: $(x - 1)^{2} = 36$ .
So $x - 1 = 6$ or $x - 1 = - 6$ .
That gives $x = 7$ or $x = - 5$ .

Answer: $x = 7$ or $x = - 5$

Squaring opened two doors, and each one lands $6$ units away.

Traps

Mixing up the coordinates, such as subtracting an x-value from a y-value. Keep x's with x's and y's with y's: $(x_{2} - x_{1})$ and $(y_{2} - y_{1})$ .
Taking the square root of each squared difference on its own, then adding those. Add the two squares first. One square root, at the end, of the total.
Calling a triangle right-angled because the plotted points look that way. Test it instead. The two shorter sides squared must add to the longest one squared.
Plotting $(3; 5)$ where $(5; 3)$ was given. Across comes first, then up. The first number is always the $x$ -value.

Learn and practise “Points and distance” in the app →

Gradient of a line

One number says which way a line leans and how hard, and it decides which lines are parallel.

gradient: How steep a line is: how far it rises for each step across.
rise: The change in $y$ as you move from one point to the other.
run: The change in $x$ over that same move.
parallel: Running in the same direction, so the two lines never meet.
perpendicular: Crossing at exactly $9 0^{\circ}$ .
undefined: No value exists, because working it out would mean dividing by zero.

The gradient of a line is its rise divided by its run. From two points that is $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ . The $y$ -values go on top. Take the two points in the same order above and below, or the sign comes out wrong.

The sign tells you the direction. A positive gradient climbs from left to right, and a negative one falls. A flat line has a rise of $0$ , so its gradient is $0$ . A line straight up the page has a run of $0$ , so its gradient is undefined. That is not the same as $0$ .

The size tells you the steepness, and you ignore the sign while judging it. A gradient of $- 4$ is steeper than a gradient of $2$ , even though $- 4$ is the smaller number.

Parallel lines have equal gradients, and that is the whole test. Two segments that simply do not meet inside a diagram may still be heading for each other off the page. Perpendicular lines have gradients whose product is $- 1$ : flip the fraction over and change its sign.

To find a missing coordinate, write both gradients using the unknown. Set them equal for parallel, or make their product $- 1$ for perpendicular, then solve. Afterwards check that neither segment came out vertical, because a vertical segment has no gradient to use.

Rules to remember

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
parallel: $m_{1} = m_{2}$
perpendicular: $m_{1} \times m_{2} = - 1$ , and both rules need gradients that exist — a vertical line has none

Examples

Find the gradient of the line through

P (- 3; 2)

and

Q (5; 6)

Worked answer

Take $P$ first, both above and below.
$m = \frac{6 - 2}{5 - ( - 3 )} = \frac{4}{8}$ .
$\frac{4}{8} = \frac{1}{2}$ , so the line climbs one up for every two across.

Answer: $m = \frac{1}{2}$

Subtracting in the same order above and below kept the sign right.

A (1; 2)

and

B (4; 8)

make one segment.

C (0; 5)

and

D (2; 4)

make another. Are they perpendicular?

Worked answer

$m_{A B} = \frac{8 - 2}{4 - 1} = 2$ .
$m_{C D} = \frac{4 - 5}{2 - 0} = - \frac{1}{2}$ .
Now multiply: $2 \times (- \frac{1}{2}) = - 1$ .
The product is $- 1$ , so the two segments are perpendicular.

Answer: yes, they are perpendicular

The second gradient was the first one flipped over with its sign changed.

P (2; 3)

and

Q (6; k)

lie on a line parallel to a line of gradient

\frac{1}{2}

. Find

k

Worked answer

Parallel means equal gradients, so $\frac{k - 3}{6 - 2} = \frac{1}{2}$ .
The bottom works out to $4$ , giving $\frac{k - 3}{4} = \frac{1}{2}$ .
Multiply both sides by $4$ : $k - 3 = 2$ .
So $k = 5$ .

Answer: $k = 5$

One condition gave one equation, and that was enough for one unknown.

Traps

Writing the change in $x$ over the change in $y$ . The rise goes on top. Put the $y$ -values above and the $x$ -values below.
Calling the gradient of a vertical line zero, instead of undefined. A vertical line has no run, so you would divide by zero. Its gradient is undefined, not zero.
Flipping one gradient over but forgetting to change its sign. Perpendicular needs both. From $2$ you get $- \frac{1}{2}$ , not $\frac{1}{2}$ .
Using the gradients-multiply-to-minus-one test on a horizontal and a vertical line. That test needs two numbers. A vertical line's gradient is undefined, so check by shape instead.

Learn and practise “Gradient of a line” in the app →

Midpoint formula

Halfway between two points is just the average of their numbers, and it unlocks a parallelogram.

midpoint: The point exactly halfway along the line joining two points.
endpoint: One of the two points at the ends of a line segment.
vertex: A corner of a shape. Several of them together are called vertices.
diagonal: A line joining two corners that are not next to each other.
parallelogram: A four-sided shape whose opposite sides are parallel.

A midpoint is an average. Add the two $x$ -values and halve them, then add the two $y$ -values and halve those. Both halvings have to happen. Halving only one of them is the most common slip here.

Sometimes you are given one endpoint and the midpoint, and the other endpoint is missing. Work backwards. The midpoint's $x$ is the average, so the two $x$ -values add to twice it. Double the midpoint and take off the endpoint you have. Then do the same with $y$ .

The diagonals of a parallelogram cut each other exactly in half, so both diagonals share one midpoint. Given three corners, that fact hands you the fourth vertex. Take care which pairs are the diagonals: joining the wrong pair gives a crossed shape instead.

Rules to remember

$x_{M} = \frac{x _{1} + x _{2}}{2}$
$y_{M} = \frac{y _{1} + y _{2}}{2}$
in a parallelogram the diagonals share a midpoint

Examples

Find the midpoint of

A (- 3; 2)

and

B (7; 8)

Worked answer

Across: $\frac{- 3 + 7}{2} = 2$ .
Up: $\frac{2 + 8}{2} = 5$ .
So the midpoint is $M (2; 5)$ .

Answer: $M (2; 5)$

Both pairs were added, and both sums were halved.

P (1; 4)

is one endpoint of a segment and

M (5; 2)

is its midpoint. Find the other endpoint

Q

Worked answer

Double the midpoint's $x$ , then take off the endpoint you know: $2 \times 5 - 1 = 9$ .
Do the same with the $y$ -values: $2 \times 2 - 4 = 0$ .
So $Q$ is the point $(9; 0)$ .
Check: $\frac{1 + 9}{2} = 5$ and $\frac{4 + 0}{2} = 2$ .

Answer: $Q (9; 0)$

Doubling alone was not enough; the known endpoint still had to come off.

A (1; 1)

B (5; 2)

and

C (6; 7)

are three vertices of parallelogram

A B C D

. Find

D

Worked answer

In $A B C D$ the diagonals are $A C$ and $B D$ .
Midpoint of $A C$ : $\frac{1 + 6}{2} = 3, 5$ and $\frac{1 + 7}{2} = 4$ .
$B D$ must have that same midpoint, so $\frac{5 + x}{2} = 3, 5$ .
That gives $x = 2$ , and $\frac{2 + y}{2} = 4$ gives $y = 6$ .
So $D$ is the point $(2; 6)$ .

Answer: $D (2; 6)$

Working from the diagonals, not the sides, is what kept the shape from crossing.

Traps

Halving one of the two sums and leaving the other one alone. Both sums get halved. Add and halve for $x$ , then add and halve for $y$ .
Subtracting the coordinates instead of adding them and halving. A midpoint is an average, so it adds. Subtracting gives you a gap, not a point.
Doubling the midpoint and stopping there. Then take off the endpoint you already have: $2 \times 5 - 1 = 9$ .
Treating a diagonal as a side, so the fourth corner makes a crossed shape. In $A B C D$ the diagonals are $A C$ and $B D$ . Match those two midpoints.

Learn and practise “Midpoint formula” in the app →

Naming and proving shapes

A picture suggests the name of a shape, but only distances and gradients can prove it.

quadrilateral: A closed shape with four straight sides.
rectangle: A parallelogram with a right angle at one corner.
rhombus: A parallelogram with all four sides equal.
square: A rectangle with all four sides equal.

Plotting the points gives you a guess. It never gives you a reason. Every name has to be earned by a calculation: distances for how long the sides are, and gradients for which way they point.

Match the test to the name of the quadrilateral. Both pairs of opposite sides parallel makes a parallelogram. Add four equal sides and it is a rhombus. Add a right angle instead and it is a rectangle. Four equal sides alone is not a square, because a rhombus has those too.

So a rectangle takes two checks. Show that both pairs of opposite sides have equal gradients. Then show that two sides meeting at a corner have gradients multiplying to $- 1$ . Use sides for this and never a diagonal, because a diagonal points the wrong way.

In a longer problem the distance, gradient and midpoint formulae work together on one figure. Start from the coordinates every time. Never quote the thing you are proving as a reason inside the working that proves it.

Rules to remember

opposite sides parallel: $m_{A B} = m_{D C}$
right angle: $m_{A B} \times m_{B C} = - 1$
equal sides: $A B = B C$

Examples

Show that

A (0; 0)

B (4; 2)

C (3; 4)

and

D (- 1; 2)

form a rectangle.

Worked answer

$m_{A B} = \frac{2 - 0}{4 - 0} = \frac{1}{2}$ and $m_{D C} = \frac{4 - 2}{3 + 1} = \frac{1}{2}$ .
$m_{B C} = \frac{4 - 2}{3 - 4} = - 2$ and $m_{A D} = \frac{2 - 0}{- 1 - 0} = - 2$ .
Both pairs of opposite sides are parallel, so $A B C D$ is a parallelogram.
$\frac{1}{2} \times (- 2) = - 1$ , so $A B$ and $B C$ meet at a right angle.
A parallelogram with a right angle is a rectangle.

Answer: $A B C D$ is a rectangle

Gradients did both jobs: equal ones for parallel, and a product of $- 1$ for the angle.

A learner says

K (0; 0)

L (3; 4)

M (8; 4)

and

N (5; 0)

make a rectangle, because all four sides are

5

. Is that right?

Worked answer

$K L = 3^{2} + 4^{2} = 5$ , and the other three sides also come to $5$ .
Four equal sides does make it a rhombus, so that much is true.
But $m_{K L} = \frac{4}{3}$ and $m_{L M} = 0$ , and $\frac{4}{3} \times 0 = 0$ .
That product is not $- 1$ , so there is no right angle here.
It is a rhombus, not a rectangle.

Answer: no, it is a rhombus

Equal sides were never a test for a right angle, so a second check was needed.

△ P QR

P (2; 4)

Q (0; 0)

and

R (6; 2)

M

is the midpoint of

QR

. Show that

P M

is perpendicular to

QR

Worked answer

$M$ is $\frac{0 + 6}{2} = 3$ across and $\frac{0 + 2}{2} = 1$ up, so $M (3; 1)$ .
$m_{QR} = \frac{2 - 0}{6 - 0} = \frac{1}{3}$ .
$m_{P M} = \frac{1 - 4}{3 - 2} = - 3$ .
$\frac{1}{3} \times (- 3) = - 1$ , so $P M$ is perpendicular to $QR$ .

Answer: $P M$ is perpendicular to $QR$

The midpoint had to come first, because without $M$ there was no second gradient.

Traps

Naming the shape from how the plotted points look, with no test at all. Say which test you used. A name with no calculation behind it earns nothing.
Checking only that all four sides are equal and calling the shape a square. A rhombus passes that test too. A square also needs a right angle.
Showing that opposite sides are equal and stopping, as though that proved a rectangle. That gives a parallelogram. Test one angle before you call it a rectangle.
Using the property you are proving as a reason inside the proof. Start from the coordinates. Only what you have already worked out may be quoted.

Learn and practise “Naming and proving shapes” in the app →

Grade 10 Analytical Geometry

What gets asked

You must be able to

Traps that cost marks

Worked example

Points and distance

Rules to remember

Examples

Traps

Gradient of a line

Rules to remember

Examples

Traps

Midpoint formula

Rules to remember

Examples

Traps

Naming and proving shapes

Rules to remember

Examples

Traps

About this material