What is true of the gradients of two parallel lines?

They are equal , and the statement works both ways. Equal gradients mean the lines are parallel, and parallel lines must have equal gradients.

How do you find the equation of a line through two given points?

Find the gradient first, from the two points. Then put that gradient and either point into the point-gradient form, and rearrange into the form required.

Why does the distance formula work?

It is Pythagoras' theorem applied to a right-angled triangle drawn under the line. The horizontal gap between the points is one shorter side, the vertical gap is the other, and the line itself is the hypotenuse.

IGCSEFurther MathsCambridge (CIE)Additional MathsFlashcardsCoordinate GeometryStraight-Line Graphs

Straight-Line Graphs (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontLinear Graphs
Complete the gradient formula for a line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$m = \frac{______}{______}$

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Cards in this collection (18)

Complete the gradient formula for a line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$m = \frac{______}{______}$
The completed formula is:
$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
The change in $y$ goes on top, so the gradient measures how far the line rises for each unit it moves to the right.
What is true of the gradients of two parallel lines?
They are equal, and the statement works both ways.
Equal gradients mean the lines are parallel, and parallel lines must have equal gradients.
True or False?
The gradient formula gives the same answer whichever of the two points you call the first one.
True.
Swapping the points changes the sign of both the top and the bottom, and those two sign changes cancel.
So $\frac{6 - 1}{- 2 - 8}$ and $\frac{1 - 6}{8 - (- 2)}$ both give $- \frac{1}{2}$ .
How do you test whether two lines are parallel?
Rearrange both equations into the form $y = m x + c$ .
Then compare the coefficients of $x$ : if they are the same, the lines are parallel.
What can you read straight off the form $y = m x + c$ ?
The gradient is $m$ and the $y$ -intercept is $(0, c)$ .
This is called the gradient-intercept form, and it is why rearranging an equation into it makes two features immediately visible.
Find the equation of the line parallel to $y = 3 x + 7$ that passes through $(2, 1)$ .
A parallel line has the same gradient, so it takes the form $y = 3 x + d$ .
Substituting $(2, 1)$ gives $1 = 6 + d$ , so $d = - 5$ and the line is $y = 3 x - 5$ .
What is the point-gradient form of a straight line?
It is $y - y_{1} = m (x - x_{1})$ , which uses the gradient $m$ and any one point $(x_{1}, y_{1})$ on the line.
It is usually the easiest starting point, because you can then rearrange into whatever form the question asks for.
What is true of the gradients of two perpendicular lines?
Their product is $- 1$ , so each is the negative reciprocal of the other.
So a line of gradient 2 is perpendicular to one of gradient $- \frac{1}{2}$ , since $2 \times - \frac{1}{2} = - 1$ .
How do you find the equation of a line through two given points?
Find the gradient first, from the two points.
Then put that gradient and either point into the point-gradient form, and rearrange into the form required.
True or False?
Every pair of perpendicular lines has gradients that multiply to give $- 1$ .
False.
The lines $x = p$ and $y = q$ are perpendicular, but a vertical line has no gradient at all, so there is nothing to multiply.
The rule covers every case except a horizontal line paired with a vertical one.
Complete the midpoint of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$(\frac{______}{2}, \frac{______}{2})$
The completed midpoint is:
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
Note that the coordinates are added, unlike the gradient and distance formulas, where they are subtracted.
Fill in the gradient of a perpendicular line:
If a line has gradient $\frac{3}{4}$ , any line perpendicular to it has gradient $______$ .
The completed sentence is:
If a line has gradient $\frac{3}{4}$ , any line perpendicular to it has gradient $- \frac{4}{3}$ .
Turn the fraction upside down and change the sign, which is what taking the negative reciprocal means.
Why can the midpoint be thought of as an average?
Because it is the mean of the two $x$ -coordinates alongside the mean of the two $y$ -coordinates.
That puts it the same distance from each end, which is what a midpoint has to be.
You have found the perpendicular line $y = - \frac{1}{2} x - 2$ , but the question wants the form $a x + b y + c = 0$ with integer coefficients. What do you do?
Multiply every term by 2 to clear the fraction, then move everything to one side:
$x + 2 y + 4 = 0$
Always check which form is asked for, since a correct equation written the wrong way round may not be accepted.
Complete the distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$d = \sqrt{(______)^{2} + (______)^{2}}$
The completed formula is:
$d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$
Because each bracket is squared, it does not matter which way round you subtract.
How do you test whether two lines are perpendicular?
Rearrange both equations into the form $y = m x + c$ .
Then multiply the two coefficients of $x$ : if the product is $- 1$ , the lines are perpendicular.
Why does the distance formula work?
It is Pythagoras' theorem applied to a right-angled triangle drawn under the line.
The horizontal gap between the points is one shorter side, the vertical gap is the other, and the line itself is the hypotenuse.
Why keep working with $d^{2}$ rather than square rooting straight away?
Because square roots are rarely exact, so rounding one early carries an error into everything that follows.
Take the square root only when you need the final answer, and leave it as a surd if an exact value is wanted.

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