Coordinate Geometry

Question 1

Midpoint of a line segments,

description

formula

Answer 1

A line segment is the part of a line that lies between 2 points. The midpoint of a line segment lies exactly halfway between the points.

Coordinates of midpoint X3, Y3 is:
([X1 + X2]/2 , [Y1 + Y2]/2)

Question 2

Distance between two points,

description

formula

Answer 2

Given 2 points of a graph makes it possible to find the distance between them using pythagoras theorem.

d = √(X2 - X1)^2 + (Y2 - Y1)^2

Question 3

Gradient,

description

solving given points

solving given an equation

note for linear gradient

Answer 3

The gradient of a line is a value describing how steep the line is, calculated from the gradient fraction ∆y / ∆x and assigned the symbol m.

m = (Y2 - Y1) / (X2 - X1)

ax + by = c
by = -ax + c
y = (-b / a)x + c
m = -b / a

Gradient of a straight line does not change.

Question 4

Collinear,

meaning

proof

Answer 4

If 3 or more points lie on the same straight line they are collinear.

The gradient between any of the given points must be the same for the line to be collinear (collinearity test)

Question 5

Parallel lines,

meaning/description

formula

Answer 5

Lines are parallel if they have the same gradient, (m1 = m2)

(Y2 - Y1) / (X2 - X1) = (Y4 - Y3) / (X4 - X3)

Question 6

Perpendicular lines,

meaning/description

formula

Answer 6

Lines are perpendicular if the angle between them is 90º, meaning the lines will intersect at right angles, (m1 x m2 = -1)

{(Y2 - Y1) / (X2 - X1)} x {(Y4 - Y3) / (X4 - X3)} = -1

Question 7

Equations of a straight line,

gradient intercept form

general equation form

finding the equation of a straight line;
given the gradient and a point on the line
given 2 points
derived from a graph with 2 points of intersection

Answer 7

y = mx + c

ax + by + c = 0

Method 1:
1. Substitute the gradient and coordinates of the point into the equation of line,
y - y1 = m(x + x1)
2. Rearrange the equation into either form

Method 2:
1. Find the gradient
2. Substitute the gradient and either point into the equation of line,
y - y1 = m(x + x1)
3. Rearrange the equation into either form

Method 3:

1. y intercept will be the value for c
2. Gradient found from evaluating distance between 2 points
3. Place found values into equation form y = mx + c

Question 8

Equation of line making angle of θ with the x-axis,

description

formula

Answer 8

If a line passes through a given point and makes an angle of θ with the x-axis, a right angle triangle can be constructed and using the pythagorean theorem the equation of line can be found.

Gradient (m):
tan(θ) = opposite / adjacent = y / x = {(Y2 - Y1) / (X2 - X1)}

Question 9

Equation of a circle,

description/formula

formula for circle centre at (0,0)

general equation of a circle

circle equation relation

Answer 9

The equation of a circle with centre (p,q) and radius r is: (x - p)^2 + (y - q)^2 = r^2

x^2 + y^2 = r^2

x^2 + y^2 + 2gx + 2fy + c = 0
With circle centre being (-g, -f)

A circle has many x values that map to many y values, meaning it is not a function

Question 10

Points of intersection and graphs,

tangent

Answer 10

When a line touches a curve at exactly one point/place we call that line a tangent to the curve.