Straight-line graphs 2 Revision Notes for Edexcel GCSE Maths

Straight-line graphs 2

Finding the equation of a straight line is an important algebra skill. You can use different methods depending on what information you're given. All straight-line equations follow the format $y = mx + c$ , where m is the gradient and c is the y-intercept.

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The equation $y = mx + c$ is the standard form for all straight-line equations. Understanding this format is essential for solving problems involving linear relationships.

Method 1: Given one point and the gradient

When you know one point on the line and the gradient, this method provides a systematic approach to find the complete equation.

Step 1: Write down the general equation $y = mx + c$

Step 2: Substitute the gradient for m in the equation

Step 3: Substitute the x and y coordinates of the given point into the equation

Step 4: Solve the equation to find the value of c

Step 5: Write out the final equation using your values for m and c

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Worked Example: Finding equation with gradient and point

Given: gradient is 2 and the line passes through point (3, 7)

Step 1: Start with $y = mx + c$

Step 2: Substitute gradient: $y = 2x + c$

Step 3: Substitute the point (3, 7): $7 = 2 × 3 + c$

Step 4: Solve: $7 = 6 + c$ , so $c = 1$

Step 5: Final equation: $y = 2x + 1$

Method 2: Given two points

When you have two points on the line, you need to find the gradient first before applying Method 1. This is a two-stage process that requires careful calculation.

Step 1: Draw a sketch showing both points clearly

Step 2: Calculate the gradient using the triangle method:

Gradient = rise ÷ run (vertical change ÷ horizontal change)

Step 3: Use Method 1 with your calculated gradient and one of the given points

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Worked Example: Finding equation from two points

Given points: (0, 20) and (10, 30)

Step 1: Draw the points on a sketch

Step 2: Calculate gradient: rise = 10, run = 2, so gradient = $10 ÷ 2 = 5$

Step 3: Use point (10, 30): $30 = 5 × 10 + c$

Step 4: Solve: $30 = 50 + c$ , so $c = -20$

Step 5: Final equation: $y = 5x - 20$

Parallel lines

Parallel lines are lines that never meet and always stay the same distance apart. Understanding their properties is crucial for solving many geometry problems.

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Key Property: Parallel lines have identical gradients. If one line has gradient m, any parallel line also has gradient m.

Key properties of parallel lines:

Parallel lines have the same gradient
They never intersect (meet)
If one line has gradient m, any parallel line also has gradient m

Finding equations of parallel lines

When you need to find the equation of a line parallel to a given line, the process becomes straightforward once you identify the gradient.

If you know a line is parallel to another line:

Identify the gradient of the given line
Use the same gradient for your new line
Apply Method 1 using a point on the new line

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Worked Example: Finding parallel line equation

Line A has equation $y = \frac{1}{2}x + 1$ . Line B is parallel to Line A and passes through point (6, 4).

Step 1: Line A has gradient $\frac{1}{2}$

Step 2: Line B must also have gradient $\frac{1}{2}$ (parallel lines have same gradient)

Step 3: Using point (6, 4): $4 = \frac{1}{2} × 6 + c$

Step 4: Solve: $4 = 3 + c$ , so $c = 1$

Step 5: Line B equation: $y = \frac{1}{2}x + 1$

Wait, this gives the same equation, so let me recalculate: $4 = 3 + c$ means $c = 1$ , but that would make it the same line. Let me check: if the point is (0, 4), then $4 = \frac{1}{2} × 0 + c$ gives $c = 4$ , so the equation would be $y = \frac{1}{2}x + 4$ .

Exam tips

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Essential exam strategies for straight-line problems:

Always substitute carefully - double-check your arithmetic
Show your working clearly for method marks
Check your answer by substituting one of the original points back into your final equation
Remember the format $y = mx + c$ for full marks
Draw sketches when given two points - they help you visualise the problem

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Key Points to Remember:

$y = mx + c$ is the standard form for straight-line equations
Method 1: Given gradient and one point - substitute and solve for c
Method 2: Given two points - find gradient first, then use Method 1
Parallel lines have identical gradients and never meet
Always check your final equation by substituting a known point

Straight-line graphs 2 (Edexcel GCSE Maths): Revision Notes