IB Mathematics AA HL Flashcards- Equation of a straight line

Straight lines may be drawn with a graph of a linear function. The form that you have probably seen the most in previous years is \( y = mx + c \):
This is the gradient-intercept form, and \(m\) & \(c\) tell us certain properties about the line. As \(x\) changes, \(y\) changes by ‘\(mx\)’. ∴ gradient = \( \frac{Δy}{Δx} = \frac{Δmx}{Δx} = m \). So \(m\) is the gradient/slope. If you then set \(x = 0\) in any function, you can find y-intercepts. In any \(y = mx + c\), you get \(y = m(0) + c\), so \(y = c\) is a y-intercept.

[q]

E.G.  Draw a graph of function: \( y = 2x – 1 \)
– Gradient = 2 \( \left( \frac{2}{1} \right) \)
– y-intercept at y = -1

\( ax + by + d = 0 \) This is another common form for a straight line, and we can derive it from:
\( y = mx + c \),
\(- mx + y – c = 0\), then if \(m\) is a fraction, we can multiply everything by the denominator to get integer coefficients.

[a]

PERPENDICULAR
If we want to go from a slope of one line to the slope of a perpendicular line, then it must change from:

\[
\text{RISE} \quad \text{RUN} \quad \Rightarrow \quad \text{RUN} \quad \text{RISE} \quad \text{or} \quad m = – \frac{1}{m}
\]

[q]

E.G. Find the equation of a line perpendicular to \( y = \frac{3}{2}x – 4 \), passing through (10, -3):
Slope: \( \frac{3}{2} \Rightarrow – \frac{2}{3} \)
→ \( y – y_1 = m(x – x_1) \)
→ \( y + 3 = – \frac{2}{3} (x – 10) \)
→ \( y = – \frac{2}{3}x + 1 \)

[a]

INTERSECTIONS
Two lines’ intersection point can be found by ‘setting them equal to each other’, which is essentially just substitution.

E.G. Find the intersection point of \( y = 2x – 1 \) & \( y = -x + 8 \):
Solve \( -x + 8 = 2x – 1 \), \( 9 = 3x \), \(x = 3\).
Plug in: \(y = 2(3) – 1 \Rightarrow y = 5\) ∴ (3,5)

[x] Exit text