Geometry Toolkit (DP IB Applications & Interpretation (AI))

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  • State the expression for the midpoint of two points open parentheses x subscript 1 comma space y subscript 1 close parentheses and open parentheses x subscript 2 comma space y subscript 2 close parentheses.

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  • State the expression for the midpoint of two points open parentheses x subscript 1 comma space y subscript 1 close parentheses and open parentheses x subscript 2 comma space y subscript 2 close parentheses.

    The expression for the midpoint of two points open parentheses x subscript 1 comma space y subscript 1 close parentheses and open parentheses x subscript 2 comma space y subscript 2 close parentheses is open parentheses fraction numerator x subscript 1 plus x subscript 2 over denominator 2 end fraction comma fraction numerator y subscript 1 plus y subscript 2 over denominator 2 end fraction close parentheses.

    This is given in your exam formula booklet.

  • What is the formula for calculating the distance between two points open parentheses x subscript 1 comma space y subscript 1 close parentheses and open parentheses x subscript 2 comma space y subscript 2 close parentheses?

    The distance between two points is calculated using the formula, d equals square root of open parentheses open parentheses x subscript 1 minus x subscript 2 close parentheses squared plus open parentheses y subscript 1 minus y subscript 2 close parentheses squared close parentheses end root

    Where:

    • d is the distance between two points

    • open parentheses x subscript 1 comma space y subscript 1 close parentheses is the set of coordinates for a known point

    • open parentheses x subscript 2 comma space y subscript 2 close parentheses is the set of coordinates for another known point

    This is given in your exam formula booklet.

  • What is the formula for finding the gradient of a line between two points open parentheses x subscript 1 comma space y subscript 1 close parentheses and open parentheses x subscript 2 comma space y subscript 2 close parentheses?

    The gradient can be found using the formula m equals fraction numerator open parentheses y subscript 2 minus y subscript 1 close parentheses over denominator open parentheses x subscript 2 minus x subscript 1 close parentheses end fraction

    Where:

    • m is the gradient of the line

    • open parentheses x subscript 1 comma space y subscript 1 close parentheses is the set of coordinates for a known point

    • open parentheses x subscript 2 comma space y subscript 2 close parentheses is the set of coordinates for another known point

    This is given in your exam formula booklet.

  • Define a perpendicular bisector.

    A perpendicular bisector of a line segment cuts the line segment in half at a right angle.

  • True or False?

    Two lines are perpendicular if the product of their gradients is -1.

    True.

    Two lines are perpendicular if the product of their gradients is -1.

    E.g. if a line has a gradient of 2, then a line that is perpendicular to it will have a gradient of negative 1 half as 2 cross times negative 1 half equals negative 1.

  • What are the two key pieces of information needed to find the equation of a straight line?

    To find the equation of a straight line, you need the gradient of the line and the coordinates of a point on the line.

  • How do you find the gradient of a perpendicular bisector?

    The gradient of a perpendicular bisector can be found by dividing -1 by the gradient of the original line segment.

    It is the negative reciprocal of the gradient of the original line segment.

  • After finding the gradient of a perpendicular, how do you use a point on the line to find its full equation?

    After finding the gradient of a perpendicular, you can use a point on the line to find its full equation by:

    • Substituting the coordinates of the point and the gradient into the gradient-intercept form y equals m x plus c (then solving to find the value of c)

    • Or substituting the coordinates of the point and the gradient into the point-gradient form y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses (then rearranging that equation into whatever form is required)

    It is often easiest to substitute into the point-gradient form.

  • What is the point-gradient form of a straight line equation?

    The point-gradient form is y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses.

    Where:

    • m is the gradient of the line

    • open parentheses x comma space y close parentheses is the set of coordinates for any point on the line

    • open parentheses x subscript 1 comma space y subscript 1 close parentheses is the set of coordinates for a known point on the line

    This is given in your exam formula booklet.

  • True or False?

    a x plus b y plus d equals 0 is the equation of a straight line.

    True.

    The equation of a straight line is usually given in one of three main forms:

    • y equals m x plus c,

    • y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses,

    • or a x plus b y plus d equals 0.

    These are all given in your formula booklet.

  • What does the notation [AB] represent in coordinate geometry?

    The notation [AB] represents the line segment between points A and B.

  • Define the term arc.

    An arc is a part of the circumference of a circle.

  • Define the term radius.

    The radius is the distance from the centre of a circle to the circumference.

    Radius also refers to a line segment from the centre of a circle to the circumference.

  • Define the term sector.

    A sector is a part of a circle enclosed by two radii and an arc.

  • True or False?

    A minor arc has an angle at the centre less than 180°.

    True.

    A minor arc has an angle at the centre less than 180°.

  • State the equation for the length of an arc.

    The equation for the length of an arc is l equals theta over 360 cross times 2 pi r

    Where:

    • l is the length of the arc

    • theta is the angle of the sector

    • r is the radius of the sector

    This formula is in the exam formula booklet.

  • True or False?

    The perimeter of a sector is just the arc length.

    False.

    The perimeter of a sector is the arc length plus two radii.

  • State the equation for the area of a sector.

    The equation for the area of a sector is A equals theta over 360 cross times pi r squared

    Where:

    • A is the area of the sector

    • theta is the angle of the sector

    • r is the radius of the sector

    This formula is in the exam formula booklet.

  • What is the fraction used to calculate sector area or arc length?

    The fraction used to calculate sector area or arc length is the angle at the centre divided by 360°, theta over 360.

  • True or False?

    The area of a major sector is always larger than the area of a minor sector in the same circle.

    True.

    The area of a major sector is always larger than the area of a minor sector in the same circle.