True or False? The distance formula gives the same answer whichever point you call the first.

True. Both differences are squared , so swapping the points only changes their signs and squaring removes the difference. A distance is a length, so it could not depend on which end you chose to measure from.

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Coordinate Geometry (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontBasic Coordinate Geometry
Complete the formula for the distance $d$ between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$d^{2} = {(x_{1} -______)}^{2} + {(______- y_{2})}^{2}$

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Cards in this collection (15)

Complete the formula for the distance $d$ between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$d^{2} = {(x_{1} -______)}^{2} + {(______- y_{2})}^{2}$
The completed formula is:
$d^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$
It is Pythagoras' theorem: the two differences are the shorter sides of a right-angled triangle and $d$ is the hypotenuse.
Find the exact distance between $A (3, - 4)$ and $B (- 5, 8)$ .
The differences are $8$ and $- 12$ , so $d = \sqrt{8^{2} + {(- 12)}^{2}} = \sqrt{208}$ .
Taking out the square factor $16$ gives $4 \sqrt{13}$ , which is about $14.4$ .
True or False?
The distance formula gives the same answer whichever point you call the first.
True.
Both differences are squared, so swapping the points only changes their signs and squaring removes the difference.
A distance is a length, so it could not depend on which end you chose to measure from.
How do you find the gradient of the line joining two points?
Divide the change in $y$ by the change in $x$ : $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
Take the two differences in the same order, top and bottom, or the sign of the gradient comes out wrong.
What are the coordinates of the point dividing $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in the ratio $m : n$ ?
$(\frac{n x_{1} + m x_{2}}{m + n}, \frac{n y_{1} + m y_{2}}{m + n})$ , where $m$ multiplies the second point's coordinates and $n$ the first.
That crossing over is what makes a larger $m$ pull the dividing point towards $(x_{2}, y_{2})$ .
How is the midpoint formula related to dividing a line in a ratio?
The midpoint is simply the ratio case $1 : 1$ , which reduces the formula to $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .
So there is really only one formula to remember here, with the midpoint as its simplest instance.
What does each of the three forms of a straight-line equation show at a glance?
Three forms, each revealing something different:
- $y = m x + c$ : the gradient $m$ and the $y$ -intercept $(0, c)$
- $y - y_{1} = m (x - x_{1})$ : the gradient and a point the line passes through
- $a x + b y = c$ : a form with integer coefficients, from which both intercepts and the gradient are quickly found
True or False?
In $a x + b y = c$ , the number $c$ is the $y$ -intercept.
False.
The $y$ -intercept is $(0, \frac{c}{b})$ , found by setting $x = 0$ and then dividing by $b$ .
Only in the form $y = m x + c$ does the constant give the intercept directly.
What is the gradient of the line $a x + b y = c$ ?
The gradient is $- \frac{a}{b}$ , which comes from rearranging into $y = - \frac{a}{b} x + \frac{c}{b}$ .
Reading $a$ off as the gradient is the commonest slip: the coefficient of $y$ has to be divided out first.
Given two points, which form of the line equation is easiest to start from?
The point-gradient form $y - y_{1} = m (x - x_{1})$ , using the gradient and either one of the two points.
Rearrange afterwards into whatever form is asked for, multiplying through by any denominators to clear the fractions.
Find the equation of the line through $(- 2, 5)$ and $(6, - 7)$ in the form $a x + b y = c$ .
The gradient is $\frac{- 7 - 5}{6 - (- 2)} = - \frac{3}{2}$ , so $y - 5 = - \frac{3}{2} (x + 2)$ .
Multiplying through by $2$ gives $2 y - 10 = - 3 x - 6$ , which rearranges to $3 x + 2 y = 4$ .
Complete the conditions on the gradients $m_{1}$ and $m_{2}$ of two lines:
$parallel: m_{1}______m_{2}, perpendicular: m_{1} \times m_{2} =______$
The completed conditions are:
$parallel: m_{1} = m_{2}, perpendicular: m_{1} \times m_{2} = - 1$
Both work in either direction: equal gradients prove two lines parallel, and parallel lines must have equal gradients.
If a line has gradient $\frac{2}{5}$ , what is the gradient of a line perpendicular to it?
It is $- \frac{5}{2}$ , the negative reciprocal: turn the fraction upside down and change the sign.
Checking, $\frac{2}{5} \times (- \frac{5}{2}) = - 1$ , as it has to be.
How do you test whether two given equations describe perpendicular lines?
Rearrange both into $y = m x + c$ and multiply the two coefficients of $x$ together.
For $3 x - 5 y = 7$ and $y = \frac{1}{4} - \frac{5}{3} x$ the gradients are $\frac{3}{5}$ and $- \frac{5}{3}$ , whose product is $- 1$ .
Why does the negative reciprocal rule not work for $x = p$ and $y = q$ ?
Because a vertical line $x = p$ has no gradient at all, so there is nothing to multiply.
The two lines are still perpendicular, one being horizontal and the other vertical, but that has to be seen directly rather than got from the rule.

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