AOS1 Topic 1: Coordinate Geometry (Part 2)
Coordinate geometry (also known as analytic geometry ) is the study of geometry using coordinate points. It allows for the calculation of distances between points, division of lines in a given ratio, finding midpoints, calculating the area of triangles in the Cartesian plane, and more.
Coordinate Geometry Terms
1. Cartesian Coordinate System:
Consists of two perpendicular axes:
The x-axis (horizontal)
The y-axis (vertical)
The point where the axes intersect is called the origin, denoted as \((0, 0)\). Any point in the plane can be described by an ordered pair \((x, y)\), where
\(x\) is the horizontal distance from the origin
\(y\) is the vertical distance from the origin.
2. Quadrants
The Cartesian plane is divided into four quadrants:
- Quadrant I : $(+x, +y)$
- Quadrant II : $(-x, +y)$
- Quadrant I : $(-x, -y)$
- Quadrant I : $(+x, -y)$
3. Distance between two points
The distance between two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is given by:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Example:
Find the distance between points $A(2,3)$ and $B(5,7)$.
\[AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Note: It has to be $+5$ since distance cannot be negative
4. Midpoint of a line segment
The midpoint of the line segment joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is the point with coordinates:
\[ \left( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \right) \]
Example:
Find the midpoint of the line segment joining $A(2,3)$ and $B(4,7)$.
\[ Midpoint = ( \frac{2+4}[2}, \frac{3+7}{2} ) = (3,5) \]
5. Gradient of a straight line
The gradient \(m\) of a straight line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Example:
Find the gradient of the line passing through points $A(1,2)$ and $B(4,8)$.
\[ m = \frac{8-2}{4-1} = \frac{6}{3} = 2 \]
6. Equation of a straight line
a. Gradient-intercept form:
A straight line with gradient \(m\) and y-axis intercept \(c\) has the equation:
\[ y = mx + c \]
Example:
Find the equation of a line with gradient $3$ and y-intercept $−2$.
\[y=3x−2\]
b. Point-gradient form:
The equation of a straight line passing through a given point \( (x_1, y_1) \) and having gradient \(m\) is:
\[ y - y_1 = m(x - x_1) \]
Example:
Find the equation of a line passing through point $(2,3)$ with gradient $4$.
\[ y-3 = 4(x-2) \]
\[y-3=3x-8\]
\[y=4x-5\]
c. Two-point form:
The equation of a straight line passing through two given points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
Example:
Find the equation of the line passing through points $A(1,2)$ and $B(3,8)$.
Calculate the gradient
\[ m = \frac{8-2}{3-1} = \frac{6}{2} = 3 \]
Use point-gradient form with point $A(1,2)$:
\[ y-2 = 3(x-1) \]
\[ y-2 = 3x-3\]
\[y = 3x-1\]
d. Intercept form:
The straight line passing through the two points \( (a, 0) \) and \( (0, b) \) has the equation:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
Example:
Find the equation of the line passing through $(2,0)$ and $(0,3)$.
\[ \frac{x}{2} + \frac{y}{3} = 1 \]
Multiplying both sides by $6$: \[3x +2y =6\]
7. Tangent of the angle of slope
For a straight line with gradient \(m\), the angle of slope is found using:
\[ m = \tan \theta \]
where \( \theta \) is the angle that the line makes with the positive direction of the x-axis.
Example:
Find the angle of slope for a line with gradient $1$.
\[ 1= \tan( \theta) \]
\[ \theta = 45^o \]
8. Perpendicular straight lines
If two straight lines are perpendicular to each other, the product of their gradients is \(-1\),
\[ m_1 \times m_2 = -1 \]
(unless one line is vertical and the other horizontal).
Example:
Find the gradient of a line perpendicular to a line with gradient $2$.
\[ m_1 \times m_2 = -1 \]
\[ 2 \times m_2 = -1 \]
\[ m_2 = \frac{-1}{2} \]
Coordinate geometry is a powerful tool for solving geometric problems and proving geometric theorems using algebraic methods. It bridges the gap between algebra and geometry, providing a deeper understanding of both subjects.
Created with GeoGebra ® , by pirsquared, Link
Coordinate Geometry Theorems
1. Midpoint Theorem: To Find Mid-point of a Line Connecting Two Points
Consider the same points A and B, which have coordinates \((x_1, y_1)\) and \((x_2, y_2)\), respectively. Let M(x,y) be the midpoint of lying on the line connecting these two points A and B. The coordinates of point M is given as:
\[ M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Proof:
The midpoint divides the line segment into two equal parts. Using the distance formula between $A$ and $M$, and $M$ and $B$, you can show both distances are equal.
2. Angle Formula: To Find The Angle Between Two Lines
To find the angle $\theta$ between two lines with their gradients:
\[ \tan(\theta) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]
Special Cases:
Case 1: When the two lines are parallel to each other,
\(m_1 = m_2 = m\)
Substituting the value in the equation above,
\[ \tan(\theta) = \left| \frac{m - m}{1 + m \cdot m} \right| = 0 \]
Case 2: When the two lines are perpendicular to each other,
\(m_1 \cdot m_2 = -1\)
Substituting the value in the original equation,
\[ \tan(\theta) = \left| \frac{-1 - m_1 m_2}{1 + m_1 m_2} \right| = \text{undefined} \]
\[ \Rightarrow \theta = 90^\circ \]
3. Section Formula: To Find a Point Which Divides a Line into $m:n$ Ratio
Consider a line $A$ and $B$ having coordinates \((x_1, y_1)\) and \((x_2, y_2)\), respectively. Let P be a point that divides the line in the ratio $m:n$, then the coordinates of the point $P$ is given as:
When the ratio $m:n$ is internal (when $P$ lies between $AB$) :
\[ P \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]
When the ratio $m:n$ is external (when $P$ lies outside of $AB$):
\[ P \left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n} \right) \]
Example:
Find the point $P$ that divides the line segment joining $A(2,3)$ and $B(4,7)$ in the ratio $1:1$ (midpoint).
\[ P( \frac{1 \times 4 + 1 \times 2}{1+1}, \frac{1 \times 7 + 1 \times 3}{1+1}) = ( \frac{6}{2}, \frac{10}{2}) = (3,5) \]
4. Area of a Triangle in Cartesian Plane
The area of a triangle in coordinate geometry whose vertices are \((x_1, y_1)\), \((x_2, y_2)\) and \((x_3, y_3)\) is:
\[ \text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]
If the area of a triangle whose vertices are \((x_1, y_1)\), \((x_2, y_2)\) and \((x_3, y_3)\) is zero, then the three points are collinear.
Example:
Find the area of a triangle with vertices at $A(1,2)$, $B(4,6)$ and $C(5,3)$
\[ \text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]
\[ \text{Area} = \frac{1}{2} | 1(6 - 3) + 4(3 - 2) + 5(2 - 6)| \]
\[ \text{Area} = \frac{1}{2} | 1(3) + 4(1) + 5(-4)| \]
\[ \text{Area} = \frac{1}{2} | 3 + 4 -20 | \]
\[ \text{Area} = \frac{1}{2} | -13 | \]
\[ \text{Area} = \frac{13}{2} \text{units} \]
Example 1
- the distance AB
- the midpoint of line segment AB
- the gradient of line AB
- the equation of line AB
- the equation of the line parallel to AB which passes through the point (1, 5)
- the equation of the line perpendicular to AB which passes through the midpoint of AB
Example 2
Example 3
Example 4
Find the equations of the following straight lines:
Example 5
- Find the gradient of the given line:
Rewrite the equation \(2x + 5y - 10 = 0\) in slope-intercept form:\(5y = -2x + 10\)\(y = -\frac{2}{5}x + 2\)The gradient \(m\) is \(-\frac{2}{5}\).
- Use the point-gradient form of the equation of a line:
Point-gradient form: \(y - y_1 = m(x - x_1)\)Given point \((x_1, y_1) = (-1, 3)\)Substitute the values:\(y - 3 = -\frac{2}{5}(x - (-1))\)\(y - 3 = -\frac{2}{5}(x + 1)\)\(y - 3 = -\frac{2}{5}x - \frac{2}{5}\)\(y = -\frac{2}{5}x - \frac{2}{5} + 3\)\(y = -\frac{2}{5}x - \frac{2}{5} + \frac{15}{5}\)\(y = -\frac{2}{5}x + \frac{13}{5}\)So, the equation is \(y = -\frac{2}{5}x + \frac{13}{5}\).
Example 6
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Find the point of intersection:Solve the equations \(y = x\) and \(x + y = 6\) simultaneously:Substitute \(y\) with \(x\) in \(x + y = 6\):\(x + x = 6\)\(2x = 6\)\(x = 3\)Substitute \(x = 3\) back into \(y = x\):\(y = 3\)So, the point of intersection is \((3, 3)\).
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Find the gradient of the perpendicular line:The equation of the given line is \(3x + 6y = 12\). Rewrite this in slope-intercept form:\(6y = -3x + 12\)\(y = -\frac{1}{2}x + 2\)The gradient \(m\) of this line is \(-\frac{1}{2}\).The gradient of the line perpendicular to this will be the negative reciprocal of \(-\frac{1}{2}\), which is \(2\).