IB MYP 4-5 Maths- Coordinate geometry- Study Notes - New Syllabus
IB MYP 4-5 Maths- Coordinate geometry – Study Notes
Standard
- Coordinate geometry
IB MYP 4-5 Maths- Coordinate geometry – Study Notes – All topics
Coordinate Geometry: Distance, Midpoint, and Gradient
Coordinate Geometry: Distance, Midpoint, and Gradient
Coordinate Geometry deals with points on the Cartesian plane and their relationships using algebra and geometry.
Distance Between Two Points
The distance between two points in a Cartesian plane is the straight-line length joining them, calculated using Pythagoras’ theorem.
Formula:
\( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
Example:
Find the distance between \( A(3,4) \) and \( B(7,9) \).
▶️ Answer/Explanation
\( d = \sqrt{(7 – 3)^2 + (9 – 4)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.4 \)
Example :
Find the diagonal of a rectangle with vertices at \( (0,0) \) and \( (8,6) \).
▶️ Answer/Explanation
\( d = \sqrt{(8 – 0)^2 + (6 – 0)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \)
Midpoint of a Line Segment
The midpoint is the point exactly halfway between two given points on a line segment.
Formula:
\( M = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) \)
Example :
Find the midpoint of the line joining \( (2,5) \) and \( (6,9) \).
▶️ Answer/Explanation
\( M = \left( \dfrac{2+6}{2}, \dfrac{5+9}{2} \right) = (4, 7) \)
Example :
A point divides the line joining \( (4,2) \) and \( (8,10) \) into two equal parts. Find its coordinates.
▶️ Answer/Explanation
\( M = \left( \dfrac{4+8}{2}, \dfrac{2+10}{2} \right) = (6, 6) \)
Gradient (Slope) of a Line
The gradient (or slope) of a line measures its steepness, given by the ratio of vertical change to horizontal change between two points.
Formula:
\( m = \dfrac{y_2 – y_1}{x_2 – x_1} \)
Example:
Find the gradient of the line through \( (1,2) \) and \( (4,8) \).
▶️ Answer/Explanation
\( m = \dfrac{8 – 2}{4 – 1} = \dfrac{6}{3} = 2 \)
Example:
A line passes through \( (5,7) \) and \( (9,19) \). Find its slope.
▶️ Answer/Explanation
\( m = \dfrac{19 – 7}{9 – 5} = \dfrac{12}{4} = 3 \)
Equation of a Line
The equation of a straight line in slope-intercept form is expressed as:
\( y = mx + c \)
where:
- \( m \) = gradient (slope) of the line
- \( c \) = y-intercept (where the line crosses the y-axis)
Other Forms:
- Point-slope form: \( y – y_1 = m(x – x_1) \)
- Two-point form: \( y – y_1 = \dfrac{y_2 – y_1}{x_2 – x_1}(x – x_1) \)
Example:
Find the equation of the line passing through \( (2, 3) \) with gradient \( 4 \).
▶️ Answer/Explanation
Use point-slope form: \( y – y_1 = m(x – x_1) \)
\( y – 3 = 4(x – 2) \)
\( y – 3 = 4x – 8 \)
\( y = 4x – 5 \)
Example:
Find the equation of the line through \( (1, 2) \) and \( (5, 10) \).
▶️ Answer/Explanation
Gradient: \( m = \dfrac{10 – 2}{5 – 1} = \dfrac{8}{4} = 2 \)
Equation: \( y – 2 = 2(x – 1) \)
\( y – 2 = 2x – 2 \)
\( y = 2x \)
Perpendicular Bisector of a Line Segment
A perpendicular bisector of a line segment passes through its midpoint and is perpendicular to the segment.
Steps to Find:
- Find the midpoint of the segment: \( M = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) \)
- Find the gradient of the segment: \( m = \dfrac{y_2 – y_1}{x_2 – x_1} \)
- Gradient of perpendicular bisector = negative reciprocal: \( m_{\text{perp}} = -\dfrac{1}{m} \)
- Use point-slope form with midpoint and \( m_{\text{perp}} \) to find the equation.
Example:
Find the equation of the perpendicular bisector of the line joining \( (2, 4) \) and \( (6, 8) \).
▶️ Answer/Explanation
Step 1: Midpoint \( M = \left( \dfrac{2+6}{2}, \dfrac{4+8}{2} \right) = (4, 6) \)
Step 2: Gradient of segment \( m = \dfrac{8-4}{6-2} = \dfrac{4}{4} = 1 \)
Step 3: Gradient of perpendicular bisector \( m_{\text{perp}} = -1 \)
Step 4: Equation: \( y – 6 = -1(x – 4) \)
\( y – 6 = -x + 4 \)
\( y = -x + 10 \)
Example:
Find the perpendicular bisector of the segment joining \( (0, 0) \) and \( (4, 2) \).
▶️ Answer/Explanation
Midpoint: \( (2, 1) \)
Original slope: \( \dfrac{2-0}{4-0} = \dfrac{1}{2} \)
Perpendicular slope: \( -2 \)
Equation: \( y – 1 = -2(x – 2) \)
\( y – 1 = -2x + 4 \)
\( y = -2x + 5 \)