maths 1 outcome 1

1.1.1 the gradient formula

$(y_2 - y_1)/(x_2 - x_1)$, $x_1$ ≠ $x_2$

this is simply the distance in the vertical direction divided by the distance in the horizontal direction. If $x_1$ did equal $x_2$ what would the line look like?

1.1.2 the distance formula

 √$[(x_2 - x_1)^2 + (y_2 - y_1)^2]$

this is the Pythagorean theorem applied to the sides of a right-angled triangle where the hypotenuse is the straight line between two points and the sides are the vertical and horizontal distances between respective co-ordinates of the points. Note the expressions $y_2$ - $y_1$ and $x_2$ - $x_1$ are again used to express these distances.

1.1.3 that the gradient of a straight line is equal to the tan of the angle between the line and the positive direction of the x axis
again think of a right-angled triangle and the definition of $tan$ from SOHCAHTOA. $tan$ is opposite over adjacent, here the vertical distance divided by the horizontal distance. But this is just how we have defined the gradient

1.1.4 recognise the term locus

1.1.5 know the equation of line of the form ax + by + c = 0

1.1.6 know the equation of line of the form y – b = m(x – a)
given a point on a line and the gradient of the line you can write down the equation of the line.

1.1.7 determine the equation of a straight line from 2 points or 1 point and the gradient
for the case where you are given two points first work out the gradient via the gradient formula and then 1.1.6

1.1.8 know that the gradients of parallel lines are equal

1.19 know that lines with gradients $m_1$ and $m_2$ are perpendicular ⇔ $m_1$ x $m_2$ = -1

1.1.10 solve problems using the above properties of straight lines

1.1.11 know concurrency properties of medians, altitudes, angle bisectors and perpendicular bisectors