Overview: 

 

• Negative Reciprocal 

• Distance between two points 

• Midpoint between two points 

• Straight Lines 

• Circle Equations 

• Tangents of Circles 

 

 

Negative Reciprocal 

 

 Relationship between the gradient of parallel lines: 

 

• Parallel lines have the same gradient 

 

Relationship between the gradient of perpendicular lines: 

 

• Perpendicular lines have gradient that is the negative reciprocal of the original line 

 

How to calculate the negative reciprocal: 

 

• Let's consider the number 3 

• In order to find the negative reciprocal of the gradient we must first multiply by –1

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• Then we need to take the reciprocal of –3: 

 

 

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• And we have our negative reciprocal  

 

Practice Question: 

 

 

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Solution: 

 

 

 

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Distance between Two Points 

 

The equation for finding the distance between two points: 

 

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Practice Question: 

 

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Solution: 

 

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Midpoint between Two Points 

 

 The equation for finding the midpoint between two points: 

 

 

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 Practice Question: 

 

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Solution: 

 

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Straight Lines 

 

General Formula for a straight Line: 

 

 

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• m = Gradient  

• c = y-intercept 

 

Sketching Straight Line Graphs: 

 

• We can used the general formula in order to sketch out graphs  

• If we know the equation we know the y-intercept and the gradient 

• For example: 

 

 

 

• Here the gradient = 2 and the y-intercept = 4 ​

• Hence we can draw the graph: 

 

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Practice Question: 

 

 

 

Solution: 

 

 

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Circle Equations 

 

General Formula for a Circle: 

 

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• Where (a , b) is the centre and r is the radius 

 

 How to find the equation of a circle: 

 

• Consider the circle 

 

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• Here we can see the centre of the circle is (-4 , 4) and the radius is 4 

• We can therefore apply these to the general formula: 

 

 

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• This is the equation of the circle 

 

 

Tangents of Circles 

 

 

Finding a tangent to a circle: 

 

• Consider the question: 

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•  Here we know that the centre of the circle is (3 , 7) 

• This means we can find the gradient of the line between (3 , 7) and (11 , 1): 

 

 

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• We know that the tangent is perpendicular to the line between (3 ,7) and (11 , 1) 

• Therefore, the gradient of the tangent is the negative reciprocal of the gradient of the line between (3 ,7) and (11 , 1) 

 

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• Now we can use the point (11,1) and the gradient of the tangent we just found in order to find the y-intercept of the tangent: 

 

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• Therefore the equation of the tangent at (11,1) is: 

 

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Practice Question: 

 

 

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Solution: 

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