Coordinate Geometry & Straight Lines Flashcards

All 20 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 formula for the gradient of the line through $(x_1,y_1)$ and $(x_2,y_2)$?
e.g. through $(5,-4)$ and $(7,2)$

$m = \dfrac{y_2-y_1}{x_2-x_1}$
e.g. $\dfrac{2-(-4)}{7-5} = 3$

In the equation $y = mx + c$, what do $m$ and $c$ represent?

$m$ is the gradient and $c$ is the $y$-intercept (where the line crosses the $y$-axis).

What is true of the gradients of two parallel lines?

They are equal: $m_1 = m_2$ (same gradient, different $y$-intercept).

What is the relationship between the gradients of two perpendicular lines?

$m_1 \times m_2 = -1$ — each gradient is the negative reciprocal of the other.
e.g. $\tfrac45$ and $-\tfrac54$

True or false? The gradient of the line $y = 3x + 4$ is $3x$.

False. The gradient is $3$ — a number, with no $x$ in it.

What are the steps to find an unknown coordinate when the gradient is given?
e.g. $P(1,k)$, $Q(9,6)$ and the gradient of $PQ$ is $2$

1. Set up the gradient equation: $\dfrac{6-k}{9-1} = 2$
2. Multiply out: $6 - k = 16$
3. Solve: $k = -10$

What is the equation of a straight line with gradient $m$ passing through the point $(x_1,y_1)$?
e.g. gradient $3$ through $(2,5)$

$y - y_1 = m(x - x_1)$
e.g. $y - 5 = 3(x - 2)$

What are the steps to find the equation of the line through two given points?
e.g. $(1,2)$ and $(3,8)$

1. Gradient: $m = \dfrac{8-2}{3-1} = 3$
2. Use $y - y_1 = m(x - x_1)$: $y - 2 = 3(x - 1)$
3. Rearrange: $y = 3x - 1$

What are the steps to find the gradient and $y$-intercept of a line given in the form $ax + by + c = 0$?
e.g. $2x + 5y - 10 = 0$

1. Make the $y$ term the subject: $5y = -2x + 10$
2. Divide by the coefficient of $y$: $y = -\tfrac25 x + 2$
3. Read off: gradient $-\tfrac25$, $y$-intercept $2$

What are the steps to find the equation of the line perpendicular to a given line through a given point?
e.g. perpendicular to $y = 2x + 1$ through $(4,3)$

1. Gradient of the given line: $2$
2. Negative reciprocal: $-\tfrac12$
3. Use $y - y_1 = m(x - x_1)$: $y - 3 = -\tfrac12(x - 4)$, i.e. $y = -\tfrac12 x + 5$

What are the steps to find the point of intersection of two straight lines algebraically?
e.g. $y = 2x + 1$ and $y = x + 4$

1. Set the expressions equal: $2x + 1 = x + 4$
2. Solve: $x = 3$
3. Substitute back: $y = 7$, so the point is $(3,7)$

What are the steps to find where a straight line crosses the axes?
e.g. $y = -2x + 6$

1. Put $x = 0$ for the $y$-axis: $(0,6)$
2. Put $y = 0$ for the $x$-axis: $0 = -2x + 6$, so $(3,0)$

True or false? Any two distinct points on a straight line give the same value for the gradient.

True. The gradient of a straight line is constant, so any two points on it can be used in $\frac{y_2-y_1}{x_2-x_1}$.

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

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