All of Straight Line Graphs in 20 Minutes! | Chapter 5 | A-Level Pure Maths Revision

Main ideas / lessons covered (Chapter 5: Straight Line Graphs)

1) Equation of a line (using gradient)

Given two coordinate points, ((x_1, y_1)) and ((x_2, y_2)), first find the gradient (m):

[ m=\frac{y_2-y_1}{x_2-x_1} ]

This is interpreted as change in (y) over change in (x).

Then use the point-slope form:

[ y-y_1=m(x-x_1) ]

If needed, substitute and simplify to write in slope-intercept form:

[ y=mx+c ]

Worked example (key steps): Find the line through ((5,7)) and ((3,-1)).

• Gradient: [ m=4 ]

• Substitute into (y-y_1=m(x-x_1)) to obtain: [ y=4x-13 ]

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2) Perpendicular lines (finding the perpendicular gradient)

If a line has gradient (m), then a line perpendicular to it has gradient:

[ m_\perp=-\frac{1}{m} ]

Then use the perpendicular gradient to form the equation of the line through a given point.

Worked example (key steps): Perpendicular to (y=2-3x) through ((6,5)).

• Original line gradient: (-3)

• Perpendicular gradient: [ m_\perp=\frac{1}{3} ]

• Use (y=\frac{1}{3}x+c)

• Substitute ((6,5)) to solve:
  • (c=3)

Result:

[ y=\frac{1}{3}x+3 ]

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3) Intersection of two lines + triangle area using coordinate geometry

To find the intersection and the area of a triangle formed:

1. Rearrange each line into a usable form (commonly (y=mx+c)).
2. Find the intersection point (A) by solving the two equations simultaneously.
3. Find point (B) where one line meets the (x)-axis:
   • set (y=0) and solve for (x).

4. Use triangle area: [ \text{Area}=\frac{1}{2}\times(\text{base})\times(\text{height}) ]

   • Height is the (y)-coordinate of (A)
   • Base is the horizontal distance along the (x)-axis from the origin to (B)

Worked example (key steps): Lines:

• (l_1: 4x-y=0 \Rightarrow y=4x)
• (l_2: 2x+3y-21=0 \Rightarrow y=-\frac{2}{3}x+7)

Intersection (A):

• Substitute (y=4x) into (2x+3y-21=0): [ x=\frac{3}{2},\quad y=6 ]

Point (B) (where (l_2) meets the (x)-axis):

• Set (y=0): [ 2x-21=0 \Rightarrow x=\frac{21}{2}=10.5 ]

Triangle area: [ \frac{1}{2}\times 10.5 \times 6=\frac{63}{2} ]

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4) Modelling with straight lines (linear relationship in context)

A common model is:

[ d=at+b ]

where:

• (d) = depth
• (t) = time
• (a) = gradient (rate of change of depth per second)
• (b) = intercept (depth at (t=0))

Worked example (container leaking):

Given:

• Start: (t=0), (d=19.1)
• After 100 seconds: (d=6.1)

Find the gradient:

[ a=\frac{6.1-19.1}{100-0}=\frac{-13}{100}=-0.13 ]

Model (using (b=19.1)):

[ d=-0.13t+19.1 ]

Interpretation:

• (a=-0.13): depth decreases by (0.13) units per second
• (b=19.1): initial depth

Find when the container is empty ((d=0)):

[ 0=-0.13t+19.1 ] [ t=146.9\text{ seconds} ]

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