Coordinate Geometry Flashcards

All 7 cards in this deck

True or false? To find the distance between two points you add the difference in $x$ to the difference in $y$.

False. You square both differences, add them, then take the square root (Pythagoras).

What are the steps to find the distance between two points, giving the answer in the form $a\sqrt{b}$?
e.g. the distance between $S(5,-4)$ and $T(7,2)$

1. Square the coordinate differences: $(7-5)^2 + (2-(-4))^2 = 2^2 + 6^2$
2. Add: $4 + 36 = 40$
3. Square root and simplify: $\sqrt{40} = 2\sqrt{10}$

What is the formula for the midpoint of the line segment joining $(x_1,y_1)$ and $(x_2,y_2)$?
e.g. the midpoint of $(1,3)$ and $(9,7)$

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$
e.g. $(5,5)$

What are the steps to find the point $P$ that divides $AB$ in a given ratio?
e.g. $A(1,2)$, $B(9,6)$ with $AP:PB = 1:3$

1. Fraction of the way along: $\dfrac{1}{1+3} = \dfrac14$
2. Add that fraction of each difference to $A$: $1 + \tfrac14(9-1) = 3$ and $2 + \tfrac14(6-2) = 3$
3. $P = (3,3)$

True or false? If $P$ divides $AB$ in the ratio $1:3$, then $P$ is one third of the way from $A$ to $B$.

False. There are $1+3=4$ parts, so $P$ is one quarter of the way from $A$ to $B$.

What are the steps to show that three points form a right-angled triangle using gradients?
e.g. $A(0,2)$, $B(4,6)$, $C(10,0)$

1. Gradients of two sides: $AB = 1$, $BC = -1$
2. Multiply them: $1 \times (-1) = -1$
3. Product is $-1$, so $AB$ and $BC$ are perpendicular — right angle at $B$

What is the perpendicular bisector of the line segment $AB$?

The line that passes through the midpoint of $AB$ and is perpendicular to $AB$.