IB MYP 4-5 Maths- y = mx + c, gradients and intercepts- Study Notes - New Syllabus
IB MYP 4-5 Maths- y = mx + c, gradients and intercepts – Study Notes
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- y = mx + c, gradients and intercepts
IB MYP 4-5 Maths- y = mx + c, gradients and intercepts – Study Notes – All topics
Equation of a Straight Line: \( y = mx + c \)
Equation of a Straight Line: \( y = mx + c \)
The equation of a straight line in a 2D Cartesian plane is:
\( y = mx + c \)
where:
- \( m \) = gradient (slope) of the line
- \( c \) = y-intercept (point where the line crosses the y-axis)
Gradient (m)
The gradient of a line measures its steepness. For two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the gradient is:
\( m = \dfrac{y_2 – y_1}{x_2 – x_1} \)
Intercepts
- y-intercept: The value of \( y \) when \( x = 0 \). In \( y = mx + c \), the y-intercept is \( c \).
- x-intercept: The value of \( x \) when \( y = 0 \): \( x = -\dfrac{c}{m} \)
Other Forms of the Line
- Slope-Intercept Form: \( y = mx + c \)
- Point-Slope Form: \( y – y_1 = m(x – x_1) \)
- Two-Point Form: \( y – y_1 = \dfrac{y_2 – y_1}{x_2 – x_1}(x – x_1) \)
- Intercept Form: \( \dfrac{x}{a} + \dfrac{y}{b} = 1 \), where \( a \) and \( b \) are the x- and y-intercepts.
Important Notes
- If \( m > 0 \), the line slopes upward.
- If \( m < 0 \), the line slopes downward.
- If \( m = 0 \), the line is horizontal.
- If undefined, the line is vertical (equation \( x = a \)).
Example:
Find the gradient and y-intercept of the line \( y = 3x – 5 \).
▶️ Answer/Explanation
Compare with \( y = mx + c \):
\( m = 3 \), \( c = -5 \).
Gradient: 3, y-intercept: -5.
x-intercept: Set \( y = 0 \): \( 0 = 3x – 5 \Rightarrow x = \dfrac{5}{3} \).
Example :
Find the equation of a line passing through \( (2, 5) \) with gradient \( -3 \).
▶️ Answer/Explanation
Use point-slope form: \( y – y_1 = m(x – x_1) \)
\( y – 5 = -3(x – 2) \)
\( y – 5 = -3x + 6 \)
\( y = -3x + 11 \)
Example:
Find the equation of a line through \( (1, 2) \) and \( (4, 8) \).
▶️ Answer/Explanation
Gradient: \( m = \dfrac{8 – 2}{4 – 1} = \dfrac{6}{3} = 2 \)
Equation: \( y – 2 = 2(x – 1) \)
\( y – 2 = 2x – 2 \)
\( y = 2x \)
Example:
Find the equation of a line passing through points (2, 3) and (6, 11) in all forms.
▶️ Answer/Explanation
Step 1: Find the gradient (m)
\( m = \dfrac{11 – 3}{6 – 2} = \dfrac{8}{4} = 2 \).
(a) Slope-Intercept Form:
\( y = mx + c \)
Substitute \( m = 2 \) and point (2, 3):
\( 3 = 2(2) + c \Rightarrow 3 = 4 + c \Rightarrow c = -1 \).
Equation: \( y = 2x – 1 \).
(b) Point-Slope Form:
\( y – y_1 = m(x – x_1) \)
\( y – 3 = 2(x – 2) \).
(c) Two-Point Form:
\( y – 3 = \dfrac{11 – 3}{6 – 2}(x – 2) \Rightarrow y – 3 = 2(x – 2) \).
(d) Intercept Form:
From slope-intercept form \( y = 2x – 1 \):
X-intercept: set y = 0 → \( 0 = 2x – 1 \Rightarrow x = 0.5 \).
Y-intercept: set x = 0 → y = -1.
\( \dfrac{x}{0.5} + \dfrac{y}{-1} = 1 \).