What does the sign of a gradient tell you about a line?

A positive gradient goes uphill from left to right, and a negative gradient goes downhill. The size of the gradient, ignoring its sign, is how steep the line is.

Why must the subtractions in the gradient formula run the same way?

Because swapping the order in only one of them flips the sign of the gradient, giving the wrong answer. Either point may come first, as long as both the top and the bottom follow the same choice.

Besides using two points, how else might you get a line's gradient?

From a line it is parallel or perpendicular to, since either of those fixes the gradient. Collinear points also help, because points lying on the same straight line share a gradient.

True or False? Every pair of straight lines has a point of intersection.

False. Parallel lines never meet, so their equations have no simultaneous solution. It is worth comparing the gradients before trying to solve, since equal gradients mean there is nothing to find.

GCSEFurther MathsAQAFlashcards3. Coordinate GeometryCoordinate Geometry & Straight Lines

Coordinate Geometry & Straight Lines (AQA GCSE Further Maths): Flashcards

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FrontCoordinate Geometry
Complete the distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$d = \sqrt{{(x_{1} - x_{2})}^{2} +______}$

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Cards in this collection (26)

Complete the distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$d = \sqrt{{(x_{1} - x_{2})}^{2} +______}$
The completed formula is:
$d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$
It is Pythagoras' theorem applied to the difference in the $x$ -coordinates and the difference in the $y$ -coordinates.
Find the distance between $(3, - 4)$ and $(- 5, 2)$ .
The differences are $3 - (- 5) = 8$ and $- 4 - 2 = - 6$ .
So $d = \sqrt{8^{2} + {(- 6)}^{2}} = \sqrt{100} = 10$ units.
True or False?
It does not matter which point you call $(x_{1}, y_{1})$ when finding a distance.
True.
Both differences are squared, and $3^{2}$ is the same as ${(- 3)}^{2}$ , so the sign is lost either way.
This is not true of a midpoint, where the signs genuinely do matter.
How do you find the midpoint of two points?
Take the average of the two $x$ -coordinates and the average of the two $y$ -coordinates.
As a formula that is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .
Find the midpoint of $(- 4, 3)$ and $(8, - 12)$ .
Average each coordinate separately: $\frac{- 4 + 8}{2} = 2$ and $\frac{3 + (- 12)}{2} = - 4.5$ .
So the midpoint is $(2, - 4.5)$ .
Which point divides $A B$ in the ratio $m : n$ ?
The point dividing $A B$ in the ratio $m : n$ lies $\frac{m}{m + n}$ of the way from $A$ to $B$ .
So a ratio of $2 : 3$ means $\frac{2}{5}$ of the way along, and a midpoint is the ratio $1 : 1$ .
$A$ is $(- 4, 3)$ and $B$ is $(8, - 12)$ . Find the point dividing $A B$ in the ratio $1 : 2$ .
The point is $\frac{1}{3}$ of the way from $A$ to $B$ , and the differences are 12 in $x$ and $- 15$ in $y$ .
Adding a third of each to $A$ gives $(- 4 + 4, 3 - 5) = (0, - 2)$ .
True or False?
Dividing $A B$ in the ratio $1 : 2$ gives the same point as dividing $B A$ in the ratio $1 : 2$ .
False.
The ratio is measured from the first letter, so one point is a third of the way from $A$ and the other a third of the way from $B$ .
Those are different points unless the ratio happens to be $1 : 1$ .
What does the sign of a gradient tell you about a line?
A positive gradient goes uphill from left to right, and a negative gradient goes downhill.
The size of the gradient, ignoring its sign, is how steep the line is.
True or False?
A line with gradient $- 5$ is steeper than one with gradient $- 4$ .
True.
Steepness depends on the size of the gradient, not on its sign.
Both lines go downhill, but the $- 5$ line falls more sharply.
Complete the gradient of the line joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$m = \frac{y_{2} - y_{1}}{______}$
The completed formula is:
$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
This is the change in $y$ divided by the change in $x$ , sometimes called rise over run.
Why must the subtractions in the gradient formula run the same way?
Because swapping the order in only one of them flips the sign of the gradient, giving the wrong answer.
Either point may come first, as long as both the top and the bottom follow the same choice.
Find the gradient of the line joining $(- 1, 4)$ and $(7, 28)$ .
The change in $y$ is $28 - 4 = 24$ , and the change in $x$ is $7 - (- 1) = 8$ .
So the gradient is $\frac{24}{8} = 3$ .
Complete the two gradient relationships:
Parallel lines have $______$ gradients.
Perpendicular lines have gradients whose product is $______$ .
The completed relationships are:
Parallel lines have equal gradients.
Perpendicular lines have gradients whose product is $- 1$ , so $m_{1} \times m_{2} = - 1$ .
The rule works both ways, so a product of $- 1$ also proves two lines are perpendicular.
What is the gradient of a line perpendicular to one with gradient $- \frac{2}{3}$ ?
The gradient is $\frac{3}{2}$ , the negative reciprocal: flip the fraction over and change the sign.
Checking, $- \frac{2}{3} \times \frac{3}{2} = - 1$ .
How do you check whether two given lines are parallel or perpendicular?
Rearrange both equations into the form $y = m x + c$ , since only then can you read off each gradient.
An equation like $2 x + 3 y = 6$ tells you nothing about its gradient until it has been rearranged.
Complete what each letter means in $y = m x + c$ :
$m$ is the $______$ , and $c$ is the $______$ .
The completed statement is:
$m$ is the gradient, and $c$ is the $y$ -axis intercept.
An alternative form is $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
What two things do you need to find the equation of a straight line?
You need a gradient $m$ and a point $(x_{1}, y_{1})$ that the line passes through.
Those go into $y - y_{1} = m (x - x_{1})$ , which can then be rearranged into either standard form.
Besides using two points, how else might you get a line's gradient?
From a line it is parallel or perpendicular to, since either of those fixes the gradient.
Collinear points also help, because points lying on the same straight line share a gradient.
What does the point of intersection of two lines represent?
The solution of their two equations solved simultaneously, since that point lies on both lines at once.
Its coordinates are the $x$ and the $y$ that satisfy both equations together.
True or False?
Every pair of straight lines has a point of intersection.
False.
Parallel lines never meet, so their equations have no simultaneous solution.
It is worth comparing the gradients before trying to solve, since equal gradients mean there is nothing to find.
How do you find where a line crosses the $x$ -axis?
Substitute $y = 0$ , because the $x$ -axis is itself the line $y = 0$ .
For the $y$ -axis substitute $x = 0$ instead, so $2 x + 3 y = 24$ crosses at $(12, 0)$ and $(0, 8)$ .
True or False?
$y = 4 x + 9$ and $y = 4 x + 9$ are parallel lines.
False.
They are the same line, and parallel lines have to be distinct.
Parallel means equal gradients with different intercepts, as in $y = 3 x + 7$ and $y = 3 x - 4$ .
Find the line parallel to $y = 3 x + 7$ passing through $(2, 1)$ .
The gradient is unchanged, so the line has the form $y = 3 x + d$ .
Substituting $(2, 1)$ gives $1 = 6 + d$ , so $d = - 5$ and the line is $y = 3 x - 5$ .
Find the line perpendicular to $y = 2 x - 2$ passing through $(2, - 3)$ .
The perpendicular gradient is $- \frac{1}{2}$ , so the line has the form $y = - \frac{1}{2} x + c$ .
Substituting $(2, - 3)$ gives $- 3 = - 1 + c$ , so $c = - 2$ and the line is $y = - \frac{1}{2} x - 2$ .
Write $y = - \frac{1}{2} x - 2$ in the form $a x + b y + c = 0$ with integer coefficients.
Move every term to one side, then clear the fraction by multiplying through by 2.
That gives $x + 2 y + 4 = 0$ .

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