5.1 Linear Graphs
The Coordinate System
The Cartesian coordinate system uses two perpendicular axes to locate points in a plane.
The Cartesian coordinate system uses two perpendicular axes to locate points in a plane.
š” Key Terms:
ā¢ x-axis: Horizontal axis (left-right)
ā¢ y-axis: Vertical axis (up-down)
ā¢ Origin: The point $(0, 0)$ where axes cross
ā¢ Coordinates: Written as $(x, y)$ - always x first!
ā¢ y-axis: Vertical axis (up-down)
ā¢ Origin: The point $(0, 0)$ where axes cross
ā¢ Coordinates: Written as $(x, y)$ - always x first!
Click on the graph to see coordinates
ā” The Four Quadrants:
ā¢ Quadrant I: $(+, +)$ - top right
ā¢ Quadrant II: $(-, +)$ - top left
ā¢ Quadrant III: $(-, -)$ - bottom left
ā¢ Quadrant IV: $(+, -)$ - bottom right
ā¢ Quadrant II: $(-, +)$ - top left
ā¢ Quadrant III: $(-, -)$ - bottom left
ā¢ Quadrant IV: $(+, -)$ - bottom right
šÆ Identify the Quadrant
Finding the Midpoint
The midpoint is the point exactly halfway between two points. Find it by averaging the coordinates.
The midpoint is the point exactly halfway between two points. Find it by averaging the coordinates.
ā” Midpoint Formula:
$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Example 1: Find the midpoint of $(2, 4)$ and $(6, 10)$
$M = \left( \frac{2 + 6}{2}, \frac{4 + 10}{2} \right)$
$M = \left( \frac{8}{2}, \frac{14}{2} \right)$
$M = (4, 7)$
$M = \left( \frac{8}{2}, \frac{14}{2} \right)$
$M = (4, 7)$
Example 2: Find the midpoint of $(-3, 5)$ and $(7, -1)$
$M = \left( \frac{-3 + 7}{2}, \frac{5 + (-1)}{2} \right)$
$M = \left( \frac{4}{2}, \frac{4}{2} \right)$
$M = (2, 2)$
$M = \left( \frac{4}{2}, \frac{4}{2} \right)$
$M = (2, 2)$
Midpoint: (3, 4)
šÆ Find the Midpoint
Midpoint = (
,
)
Finding the Distance Between Two Points
Use the distance formula, which is based on Pythagoras' theorem.
Use the distance formula, which is based on Pythagoras' theorem.
ā” Distance Formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
š” Why does this work?
The line between two points is the hypotenuse of a right triangle!
ā¢ Horizontal distance = $|x_2 - x_1|$
ā¢ Vertical distance = $|y_2 - y_1|$
ā¢ By Pythagoras: $d^2 = (\Delta x)^2 + (\Delta y)^2$
ā¢ Horizontal distance = $|x_2 - x_1|$
ā¢ Vertical distance = $|y_2 - y_1|$
ā¢ By Pythagoras: $d^2 = (\Delta x)^2 + (\Delta y)^2$
Example 1: Find the distance between $(1, 2)$ and $(4, 6)$
$d = \sqrt{(4-1)^2 + (6-2)^2}$
$d = \sqrt{3^2 + 4^2}$
$d = \sqrt{9 + 16} = \sqrt{25} = 5$
$d = \sqrt{3^2 + 4^2}$
$d = \sqrt{9 + 16} = \sqrt{25} = 5$
Example 2: Find the distance between $(-2, 3)$ and $(4, -1)$
$d = \sqrt{(4-(-2))^2 + (-1-3)^2}$
$d = \sqrt{6^2 + (-4)^2}$
$d = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.21$
$d = \sqrt{6^2 + (-4)^2}$
$d = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.21$
Distance: 5
šÆ Find the Distance
What is Gradient?
The gradient (or slope) measures how steep a line is. It tells us how much $y$ changes for each unit change in $x$.
The gradient (or slope) measures how steep a line is. It tells us how much $y$ changes for each unit change in $x$.
ā” Gradient Formula:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$
š” Understanding Gradient:
ā¢ Positive gradient: Line goes up from left to right ā
ā¢ Negative gradient: Line goes down from left to right ā
ā¢ Zero gradient: Horizontal line ā
ā¢ Undefined gradient: Vertical line |
The larger the absolute value, the steeper the line!
ā¢ Negative gradient: Line goes down from left to right ā
ā¢ Zero gradient: Horizontal line ā
ā¢ Undefined gradient: Vertical line |
The larger the absolute value, the steeper the line!
Example 1: Find the gradient between $(2, 3)$ and $(6, 11)$
$m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$
This means: for every 1 unit right, go 2 units up.
This means: for every 1 unit right, go 2 units up.
Example 2: Find the gradient between $(1, 5)$ and $(4, -1)$
$m = \frac{-1 - 5}{4 - 1} = \frac{-6}{3} = -2$
Negative gradient: line slopes downward.
Negative gradient: line slopes downward.
Equation: y = 2x + 1
šÆ Find the Gradient
The Equation of a Straight Line
Every straight line can be written in the form $y = mx + c$.
Every straight line can be written in the form $y = mx + c$.
ā” Understanding y = mx + c:
ā¢ $m$ = gradient (slope)
ā¢ $c$ = y-intercept (where line crosses y-axis)
$y = mx + c$
ā¢ $m$ = gradient (slope)
ā¢ $c$ = y-intercept (where line crosses y-axis)
Example 1: What is the gradient and y-intercept of $y = 3x - 2$?
Comparing with $y = mx + c$:
ā¢ Gradient $m = 3$
ā¢ Y-intercept $c = -2$ (crosses y-axis at $(0, -2)$)
ā¢ Gradient $m = 3$
ā¢ Y-intercept $c = -2$ (crosses y-axis at $(0, -2)$)
Example 2: Find the equation of the line with gradient 2 passing through $(0, 5)$
ā¢ $m = 2$
ā¢ Passes through $(0, 5)$, so $c = 5$
ā¢ Equation: $y = 2x + 5$
ā¢ Passes through $(0, 5)$, so $c = 5$
ā¢ Equation: $y = 2x + 5$
Example 3: Find the equation of the line passing through $(1, 4)$ and $(3, 10)$
Step 1: Find gradient
$m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$
Step 2: Use $y = mx + c$ with one point
$4 = 3(1) + c$
$4 = 3 + c$
$c = 1$
Answer: $y = 3x + 1$
$m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$
Step 2: Use $y = mx + c$ with one point
$4 = 3(1) + c$
$4 = 3 + c$
$c = 1$
Answer: $y = 3x + 1$
Line Explorer
y = x
šÆ Identify m and c
m =
c =
šÆ Find the Equation
What Makes Lines Parallel?
Parallel lines never meet - they have the same gradient.
Parallel lines never meet - they have the same gradient.
ā” Rule for Parallel Lines:
If two lines are parallel: $m_1 = m_2$
The y-intercepts can be different, but gradients must be equal!
The y-intercepts can be different, but gradients must be equal!
Example 1: Are $y = 3x + 2$ and $y = 3x - 5$ parallel?
Line 1: $m_1 = 3$
Line 2: $m_2 = 3$
$m_1 = m_2$ ā Yes, they are parallel!
Line 2: $m_2 = 3$
$m_1 = m_2$ ā Yes, they are parallel!
Example 2: Find the equation of the line parallel to $y = 2x + 1$ passing through $(3, 4)$
Step 1: Parallel means same gradient
$m = 2$
Step 2: Use point $(3, 4)$ to find $c$
$4 = 2(3) + c$
$4 = 6 + c$
$c = -2$
Answer: $y = 2x - 2$
$m = 2$
Step 2: Use point $(3, 4)$ to find $c$
$4 = 2(3) + c$
$4 = 6 + c$
$c = -2$
Answer: $y = 2x - 2$
y = x + 2 and y = x - 1
šÆ Find Parallel Line Equation
What Makes Lines Perpendicular?
Perpendicular lines meet at right angles (90Ā°). Their gradients have a special relationship.
Perpendicular lines meet at right angles (90Ā°). Their gradients have a special relationship.
ā” Rule for Perpendicular Lines:
If two lines are perpendicular: $m_1 \times m_2 = -1$
Or equivalently: $m_2 = -\frac{1}{m_1}$ (negative reciprocal)
Or equivalently: $m_2 = -\frac{1}{m_1}$ (negative reciprocal)
š” Finding Perpendicular Gradients:
ā¢ If $m_1 = 2$, then $m_2 = -\frac{1}{2}$
ā¢ If $m_1 = -3$, then $m_2 = \frac{1}{3}$
ā¢ If $m_1 = \frac{2}{5}$, then $m_2 = -\frac{5}{2}$
Flip the fraction and change the sign!
ā¢ If $m_1 = -3$, then $m_2 = \frac{1}{3}$
ā¢ If $m_1 = \frac{2}{5}$, then $m_2 = -\frac{5}{2}$
Flip the fraction and change the sign!
Example 1: Are $y = 2x + 1$ and $y = -\frac{1}{2}x + 3$ perpendicular?
$m_1 = 2$ and $m_2 = -\frac{1}{2}$
Check: $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$ ā
Yes, they are perpendicular!
Check: $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$ ā
Yes, they are perpendicular!
Example 2: Find the equation of the line perpendicular to $y = 3x + 2$ passing through $(6, 1)$
Step 1: Find perpendicular gradient
Original: $m_1 = 3$
Perpendicular: $m_2 = -\frac{1}{3}$
Step 2: Use point $(6, 1)$ to find $c$
$1 = -\frac{1}{3}(6) + c$
$1 = -2 + c$
$c = 3$
Answer: $y = -\frac{1}{3}x + 3$
Original: $m_1 = 3$
Perpendicular: $m_2 = -\frac{1}{3}$
Step 2: Use point $(6, 1)$ to find $c$
$1 = -\frac{1}{3}(6) + c$
$1 = -2 + c$
$c = 3$
Answer: $y = -\frac{1}{3}x + 3$
y = 2x and y = -0.5x (perpendicular)
šÆ Find the Perpendicular Gradient
šÆ Parallel, Perpendicular, or Neither?
Real Life Uses:
ā¢ Architecture: Ensuring walls are perpendicular to floors
ā¢ Road design: Parallel roads, perpendicular intersections
ā¢ Computer graphics: Rendering 3D objects
ā¢ Navigation: GPS and mapping systems
ā¢ Architecture: Ensuring walls are perpendicular to floors
ā¢ Road design: Parallel roads, perpendicular intersections
ā¢ Computer graphics: Rendering 3D objects
ā¢ Navigation: GPS and mapping systems